Derivative Archives | Jake's Math Lessons

1. Draw a sketch

As always, we’ll start by drawing a quick sketch of all of the information that is being described in the problem. To do this we should first think about what information we have. First of all, we need to think about the shape that’s being formed with the ladder.

Since the ladder is standing on the ground and leaning up against a vertical wall, we can say that a triangle would be formed by the 3 objects in the problem. More specifically we know that the vertical wall forms a 90 degree angle with the ground. Therefore, the triangle formed by the ground, the wall, and the ladder would be a right triangle.

On top of this, the problem also gives us a few pieces of information about the dimensions of the triangle and how they are changing. It actually tells us about how fast the ladder is moving, but since the ladder is what forms the triangle, we can deduce how the dimensions of the triangle are changing.

We are given 3 pieces of information about the position of the ladder as well as how the ladder is moving at the specific instant we are looking at.

Bottom of the ladder is 3 m away from the wall.
Top of the ladder is moving down the wall at a rate of 0.15 m/s.
Bottom of the ladder is moving away from the wall at a rate of 0.2 m/s.

Adding these labels to our drawing from above would give us something like this:

related rates ladder sliding away from wall — Ladder sliding down and away from a wall

This sketch gives us a pretty good idea of what is going on in this problem. Not only that, but we will be able to use this to get an idea of what kind of equation we will need to come up with.

2. Come up with your equation

Before we come up with our equation we want to sort through the information we are given and asked to find. This is important because we need to decide what measurements and variables we want in the equation.

What are we looking for?

The question asks us to find the length of the ladder. Therefore, we will need to find the length of the hypotenuse of the triangle in our drawing. Because of this we want to be sure to include the hypotenuse of our triangle in our equation.

What do we know about?

Looking back up at our labeled drawing, you can see that we really only have information about the bottom and side of our triangle. We know the length of the bottom side of the triangle and the rate of change of this side.

And we were also given information about the rate of change of the left side of the triangle. But remember that our equation in this step cannot include rates of change. Instead, the fact that we know this rate of change tells us that we can use the left side length of the triangle in our equation, not its rate of change.

We aren’t given any information about the angles in the triangle other than the fact that it’s a right triangle. As a result, we probably don’t want our equation to involve the angles of this triangle.

Since we know that our equation either needs to include, or can include the lengths of the sides of the triangle we should label them. Let’s go back to our drawing and label the side of the triangle. You can label them whatever you’d like, but I’ll go with x, y, and z.

triangle from ladder with sides labelled — Right triangle labeled with sides x, y, and z

Putting it into an equation

At this point we’ve figured out that we need an equation that related the sides of a right triangle.

What do you know about the relationship between the sides of a right triangle, but neither of the other two angles?

You’re probably thinking Pythagorean Theorem. If you are, you’re right! Pythagorean Theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. Remember this can only be applied to the sides of a right triangle, so noticing that is actually very important. In other words we know $$z^2=x^2+y^2.$$

3. Implicit differentiation

Now that we have come up with our equation, we need to apply implicit differentiation to take the derivative of both sides of our equation with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$

Before we do this though I want to point something out. Let’s look at each of the letters in this equation and consider how we need to treat them when we differentiate with respect to time.

Consider z first. z represents the length of the hypotenuse of the triangle. This is the side that is formed by the ladder. As time changes what happens to the length of the ladder? Nothing. It doesn’t change at all. It’s constant. Therefore we can treat z like a constant. If z is a constant and never changes, then $z^2$ would be constant too. It doesn’t change as time changes.

So when we take the derivative of z with respect to time, the derivative will be 0. The derivative of any constant is 0.

Unfortunately x and y won’t be as convenient. Looking back at our drawings you can see that the sides labelled x and y are changing over time. As the ladder slides down and away from the wall, these two sides of the triangle change in length. Therefore, when we take the derivative of $x^2$ and $y^2$ we will need to treat x and y as functions of time. Doing this means that we will need to use the chain rule, where x and y are the inside function and they are being plugged into another function that squares them.

Back to the derivative

Knowing how to treat each letter in our equation, let’s go ahead and take the derivative of both sides with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$ $$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$

4. Solve for the desired rate of change

Now all we need to do is plug in all of the information we have and solve for the right variable. However, this one is a little weird. The reason I say this is that we are actually not looking for a rate of change.

Remember the question told us to find the length of the ladder. Which means we need to find the value of z. The differential equation we just ended up with doesn’t even have a z in it, so how can we use it to find z?

Well, we’re actually going to need to go back to our original equation. $$z^2=x^2+y^2$$ We know the value of x based on information we were given, but we don’t know the value of y yet. If we could figure out what y was, then we could use this equation, plug in the value for x and y, then solve for z.

How do we find y?

This is what we will use our differential equation from the previous step for. That equation has a y in it, and we know the value of all the other variables.

We are told that the moment we are considering is when the bottom of the ladder is 3 m from the wall. Since that corresponds to the side of our triangle labelled x, we know $\mathbf{x=3}$ .
We are also told that the ladder is moving away from the wall at a rate of 0.2 m/s. Therefore, x must be getting longer, or increasing, at that rate. So $\mathbf{\frac{dx}{dt}=0.2}$ .
And finally, the ladder is sliding down the wall at a rate of 0.15 m/s. So y must be getting shorter, or decreasing, at that rate. This means $\mathbf{\frac{dy}{dt}=-0.15}$ .

Plugging it all into our equation

Knowing all of the values in our equation aside from y, we can plug these in and solve for y.

$$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ $$0 = 2(3)(0.2) + 2y(-0.15)$$ $$0=1.2-0.3y$$ $$0.3y=1.2$$ $$y=4$$

Now that we know x and y, we can plug them back into our original equation and solve for z.

$$z^2=x^2+y^2$$ $$z^2=(3)^2+(4)^2$$ $$z^2=25$$ $$z=5$$

So the ladder must be 5 m long!

L’HOSPITAL’S RULE – HOW TO – With Examples

L’Hospital’s Rule really just tells us one thing that makes evaluating certain limits a lot easier. Limits that meet 3 specific requirements can be made much simpler using L’Hospital’s Rule. First let me introduce L’Hospital’s Rule, then we can go over the 3 conditions that you need to check before you can apply it to any given limit.

What does L’Hospital’s Rule tell us?

To find a limit of a function that is a fraction, we can take the derivative of the top of the fraction and the derivative of the bottom of the fraction and make a new fraction out of the derivatives.

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Notice that we don’t use the quotient rule here. The reason for this is that we are taking the limit of this fraction and not taking the derivative of this fraction. The limit is the key piece that allows us to avoid the quotient rule and take the derivative of each piece of the fraction separately to create another limit that is equal to the original.

The hope is that the fraction resulting from the derivatives will be easier to evaluate than the original limit was.

How do we know when to use L’Hospital’s Rule?

Before we can apply L’Hospital’s rule to any given limit, we need to confirm that these three conditions are met:

f(x) and g(x) are differentiable on some open interval that includes $\mathbf{x=a}$ . This will basically just mean that both the numerator and denominator are differentiable at $x=a$ .
$\mathbf{g'(x) \neq 0}$ near $\mathbf{x=a}$ . Note that it doesn’t matter if $g'(x)=0$ AT $x=a$ as long as you can pick some interval (as small as is necessary) around $x=a$ where $g'(x) \neq 0$ for all x‘s in that interval besides $x=a$ . You likely won’t need to worry about running into a function that you can’t pick a small enough interval around $x=a$ to make this work.
As x $\rightarrow$ a, f(x) AND g(x) $\mathbf{\rightarrow 0}$ — OR — f(x) AND g(x) $\mathbf{\rightarrow \pm \infty}$

That’s really all there is to it. Let’s jump into some practice problems and I will show you how to apply L’Hospitals Rule.

Example 1

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x}$$

If we tried to use limit properties to evaluate this limit, we would see that both the top and the bottom of this fraction go to either positive or negative infinity as x goes to infinity.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} \rightarrow \frac{\infty}{- \infty}$$

So L’Hospital’s Rule might help…

Now we just need to confirm that the other two conditions are met.

$f(x)=e^x$ is an exponential function and is differentiable everywhere, for any value of x. And $g(x)=-x^2 + 1000x$ is also differentiable everywhere since it’s a polynomial. Since it’s differentiable everywhere, it is also differentiable for any infinitely large x value.

$g'(x) = -2x+1000$ will go to $-\infty$ as x approaches $\infty$ . $g'(x) \neq 0$ for any infinitely large x value since it just continues to go to $- \infty$ .

So we know that this limit meets all 3 requirements needed to apply L’Hospital’s Rule.

Now we know we can apply L’Hospital’s Rule

Taking the derivative of the top and bottom of the fraction individually tells us that:

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \lim_{x \to \infty} \frac{e^x}{-2x+1000}$$

Now we can evaluate this new limit instead. But if we do this, we will notice that we will still end up in the same situation that we had before.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} \rightarrow \frac{\infty}{- \infty}$$

But what if it didn’t really help make the limit easier?

Which puts us in a perfect situation to consider using L’Hospital’s Rule to evaluate this new limit as well. By the same logic as before, we can confirm that the first two conditions are met as well as x gets infinitely large. Since all 3 required conditions are met we can go ahead and apply L’Hospital’s Rule a second time.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} = \lim_{x \to \infty} \frac{e^x}{-2}$$

Now we can simply use the basic limit properties to evaluate this last limit.

Now we have a much easier limit

$$\lim_{x \to \infty} \frac{e^x}{-2} = \ – \frac{1}{2} \lim_{x \to \infty} e^x = \ – \infty$$

So therefore,

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \ – \infty$$

A quick note on applying L’Hospital’s Rule twice

This is an interesting problem because it shows that you can apply L’Hospital’s Rule multiple times on the same problem. You just need to make sure that each time you apply it, the resulting limit still meets all 3 required conditions before applying it to the new limit. There is no limit to the number of times you can continue applying L’Hospital’s Rule over and over in the same problem as long as you are making sure that the limit you are applying it to meets all 3 conditions every time you apply it.

Example 2

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 3} \frac{x-3}{27-x^3}$$

Again, if we think about what value the top and bottom of this fraction will go towards as x approaches 3, we would see that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} \rightarrow \frac{0}{0}$$

Since we get another indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

We can check that $g'(x) = -3x^2$ doesn’t equal zero anywhere near $x=3$ . This is because $g'(x) = -3x^2$ is continuous everywhere and the only place where $g'(x)=-3x^2=0$ is when $x=0$ . As a result of these two things, we can pick some interval around $x=3$ that doesn’t include $x=0$ to satisfy condition #2.

Also, both f(x) and g(x) are polynomials and are therefore differentiable everywhere. So we know they will both be differentiable on any interval around $x=3$ .

Now we know we can apply L’Hospital’s Rule

Doing so by taking the derivative of the top and bottom of our fraction separately tells us that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \lim_{x \to 3} \frac{1}{-3x^2}$$

And doing this gives us an easier limit to deal with. Now we can simply apply the limit properties to evaluate. Applying the limit properties tells us that:

$$\lim_{x \to 3} \frac{1}{-3x^2} = \frac{1}{-3 \Big( \lim_{x \to 3}x \Big) ^2}= \frac{1}{-3(3)^2} = \ – \frac{1}{27}$$

So therefore we know that:

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \ – \frac{1}{27}$$

Example 3

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 0} \frac{|x|}{x^5+2x}$$

If we think about what value the numerator and denominator of this fraction will approach as x approaches 0 from both sides, we would see:

$$\lim_{x \to 0} \frac{|x|}{x^5+2x} \rightarrow \frac{0}{0}$$

Since we get an indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

Upon checking condition #1 however, we run into a problem. Condition #1 requires that f(x) and g(x) are both differentiable on some interval containing $x=0$ , including $x=0$ .

But $f(x) = |x|$ is not differentiable at $x=0$ . Therefore, we actually can’t apply L’Hospital’s Rule to evaluate this limit. I won’t go into the details here since we won’t be using L’Hospital’s Rule. But if you want to try evaluating this limit, I’d recommend considering both one-sided limits on their own and compare them to start. You can see a similar application here.

Example 4

$\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}$ | Solution

How to find the equation of a tangent line

One common application of the derivative is to find the equation of a tangent line to a function. Usually when you’re doing a problem like this, you will be given a function whose tangent line you need to find. And you will also be given a point or an x value where the line needs to be tangent to the given function.

Using these two pieces of information, you need to create an equation for a line that satisfies the required conditions. This process is very closely related to linear approximation (or linearization) and differentials.

When coming up with the equation of the line, there are a couple different approached you could take. You should decide which one to use based on your own personal preference. The only difference between the different approaches is which template for an equation of a line you prefer to use. Remember, there are two main forms that a line will take: $$y=mx+b$$ $$y=m(x-x_0)+y_0$$ Another thing to keep in mind is that the first form is generally easier when we are given the y-intercept of the line. The second form above is usually easier when we are given any other point that isn’t the y-intercept.

In both of these forms, x and y are variables and m is the slope of the line. In the first equation, b is the y-intercept. And in the second equation, $x_0$ and $y_0$ are the x and y coordinates of some point that lies on the line. This could be any point that lies on the line.

It is also important to notice that a line would be tangent to a function at a specific point if and only if the following two conditions are met.

The function and its tangent line need to go through the same point
and they both need to share the same slope at that shared point.

But how does the derivative apply?

You should always keep in mind that a derivative tells you about the slope of a function. So if we take a function’s derivative, then look at it at a certain point, we have some information about the slope of the function at that point. Since a tangent line has to have the same slope as the function it’s tangent to at the specific point, we will use the derivative to find m.

So let’s jump into a couple examples and I’ll show you how to do something like this.

Example 1

Find the equation of the tangent line to the function $\mathbf{y=x^3+4x-6}$ at the point (2, 10).

In order to find this tangent line, let’s consider the two conditions that need to be met for our line to be a tangent line at the specified point.

The tangent line and the given function need to go through the same point. Since the problem told us to find the tangent line at the point $(2, \ 10)$ , we know this will be the point that our line has to go through.
The tangent line and the function need to have the same slope at the point $(2, \ 10)$ . In order to find this slope we will need to use the derivative. Let’s start with this.

Finding the slope of the tangent line

Remember that the derivative of a function tells you about its slope. So to find the slope of the given function $y=x^3+4x-6$ we will need to take its derivative. This will just require the use of the power rule. $$y’=3x^2+4$$

But how can we use this to find the slope of the tangent line when it has variables in it?

This is where the specific point we need to consider comes into play. We know that the tangent line and the function need to have the same slope at the point $(2, \ 10)$ . Therefore, they need to have the same slope when $x=2$ . In order to find the slope of the given function y at $x=2$ , all we need to do is plug 2 into the derivative of y.

Therefore, the slope of our line would simply be $$y'(2)=3(2)^2+4=16.$$ And because of this we also know the slope of our tangent line will be $$m=16.$$ So we know this will guarantee that our tangent line has the right slope, now we just need to make sure it goes through the right point.

Making sure the tangent line contains the given point

Since we do know a point that has to lie on our line, but don’t know the y-intercept of the line, it would be easier to use the following form for our tangent line equation. $$y=m(x-x_0)+y_0$$

And since we already know $m=16$ , let’s go ahead and plug that into our equation. $$y=16(x-x_0)+y_0$$

Now to finish our tangent line equation, we just need the x and y coordinates of a point that lies on this line. Then we can simply plug them in for $x_0$ and $y_0$ . Well, we were given this information! We were told that the line we come up with needs to be tangent at the point $(2, \ 10)$ . Therefore, our tangent line needs to go through that point. This tells us our tangent line equation must be $$y=16(x-2)+10$$ $$y=16x-32+10$$ $$y=16x-22$$

And that’s it! We know that the line $y=16x-22$ will go through the point $(2, 10)$ on our original function. And we know that it will also have the same slope as the function at that point.

We can even use Desmos to check this and see what our function and tangent line look like together.

tangent line to a function — $y=x^3+4x-6$ and $y=16x-22$

Example 2

Find the equation of the line that is tangent to the function $f(x) = xe^x$ when $x=0$ .

To start a problem like this I suggest thinking about the two conditions we need to meet.

The tangent line and the given function need to intersect at $\mathbf{x=0}$ . This time we weren’t given the y coordinate of this point so we will need to figure that out.
Then we need to make sure that our tangent line has the same slope as f(x) when $\mathbf{x=0}$ .

Finding the slope of the tangent line

I personally think that it’s a little easier to find the slope of the tangent line first, but you can start with making sure the other condition is met if you prefer.

When you’re asked to find something to do with slope, your first thought should be to use the derivative. The derivative of a function tells you about it’s slope. Since we need the slope of f(x), we’ll need its derivative.

Looking at our function $f(x)=xe^x$ you can see that it is the product of two simpler functions. To find it’s derivative we will need to use the product rule. I’m not going to show every step of this, but if you aren’t 100% sure how to find this derivative you should click the link in the last sentence. $$f'(x) = e^x + xe^x$$ $$f'(x) = e^x \big(1+x \big)$$

Now consider the fact that we need our tangent line to have the same slope as f(x) when $x=0$ . To find the slope of f(x) at $x=0$ we just need to plug in 0 for x into the equation we found for f'(x). $$f'(0) = e^{(0)} \big( 1 + (0) \big)$$ $$f'(0) = 1(1)=1$$

So we know that the slope of our tangent line needs to be 1.

Making sure the tangent line contains the required point

Now we just need to make sure that our tangent line shares the same point as the function when $x=0$ . In order to do this, we need to find the y value of the function when $x=0$ . This would be the same as finding f(0). $$f(0) = (0)e^{(0)} = 0$$

Since this is the y value when $x=0$ , we can also say that this is the y-intercept. We know the y intercept of our tangent line is 0. Since we figured out the y-intercept, it would be easiest to use the $y=mx+b$ form of the line for the tangent line equation.

We already found that the slope will be 1 and that the y-intercept will need to be 0, so we can plug these values in for m and b. Doing this tells us that the equation of our tangent line is $$y=(1)x+(0)$$ $$y=x.$$

Again, we can see what this looks like and check our work by graphing these two functions with Desmos.

tangent lines example 2 — $y=xe^x$ and $y=x$

Finding the Tangent Line Equation with Implicit Differentiation

Depending on the curve whose tangent line equation you are looking for, you may need to apply implicit differentiation to find the slope.

Example 3

Find the equation of the line that is tangent to the curve $\mathbf{y^3+xy-x^2=9}$ at the point (1, 2).

First we need to apply implicit differentiation to find the slope of our tangent line.

$$\frac{d}{dx} \big[ y^3 + xy – x^2 \big] = \frac{d}{dx} [9]$$ $$3y^2 \frac{dy}{dx} + 1\cdot y + x \cdot \frac{dy}{dx} – 2x = 0$$ $$3y^2 \frac{dy}{dx} + x \frac{dy}{dx} = -y + 2x$$ $$\frac{dy}{dx} \big[ 3y^2 + x \big] = -y + 2x$$ $$\frac{dy}{dx} = \frac{-y+2x}{3y^2+x}$$

Now we can plug in the given point (1, 2) into our equation for $\mathbf{\frac{dy}{dx}}$ to find the slope of the tangent line.

$$m=\frac{-(2)+2(1)}{3(2)^2+(1)}=\frac{0}{13}=0$$

With this slope, we can go back to the point slope form of a line. Since we know the slope and a point that lies on this line, we can plug that information into the general point slope form for a line. This will leave us with the equation for a tangent line at the given point.

$$y=m(x-x_0)+y_0$$ $$y=0(x-1)+2$$ $$y=2$$

So the constant function $\mathbf{y=2}$ is tangent to the curve $\mathbf{y^3+xy-x^2=9}$ at the point (1, 2).

Example 4

Find the equation of the line that is tangent to the curve $\mathbf{16x^2 + y^2 = xy + 4}$ at the point (0, 2).

Again, we will start by applying implicit differentiation to find the slope of the tangent line.

$$\frac{d}{dx} \big[ 16x^2 + y^2 \big] = \frac{d}{dx} [xy + 4]$$ $$32x + 2y \frac{dy}{dx} = 1\cdot y + x \cdot \frac{dy}{dx}$$ $$2y \frac{dy}{dx} – x \frac{dy}{dx}= -32x + y$$ $$\frac{dy}{dx} \big[ 2y-x \big] = -32x+y$$ $$\frac{dy}{dx} = \frac{-32x+y}{2y-x}$$

Now we can plug in the given point (0, 2) into our equation for $\mathbf{\frac{dy}{dx}}$ to find the slope of the tangent line.

$$m = \frac{-32(0)+(2)}{2(2)-(0)}$$ $$m=\frac{2}{4}$$ $$m=\frac{1}{2}$$

$$y=m(x-x_0)+y_0$$ $$y=\frac{1}{2}(x-0)+2$$ $$y=\frac{1}{2}x+2$$

Tangent Line Equation Without Derivatives

There are some cases where you can find the slope of a tangent line without having to take a derivative. This is not super common because it does require being able to take advantage of additional information. Usually you will be able to do this if you know some geometrical fact about the curve whose tangent line equation you are looking for.

The most common example of this is finding the a line that is tangent to a circle.

Example 5

Find the equation of the line that is tangent to the circle $\mathbf{(x-2)^2+(y+1)^2=25}$ at the point (5, 3).

We already are given a point that we know needs to lie on our tangent line. This tells us that if we can find the slope of the tangent line, we would just be able to plug it all into the point slope form for a linear function and we would have a tangent line. So we just need to find the slope of the tangent line.

In order to find this slope we can take advantage of a geometrical fact about circles: a line connecting the center of a circle to its edge will be perpendicular to a line that is tangent to the circle at that edge point. Below you can see what this looks like on a graph of this circle, or at least a portion of it.

Graph of a circle and its tangent line — $\mathbf{(x-2)^2+(y+1)^2=25}$ and the tangent line at *(5, 3)*

$\mathbf{(x-2)^2+(y+1)^2=25}$ and the tangent line at *(5, 3)*

Therefore, if we know the slope of a line connecting the center of our circle to the point (5, 3) we can use this to find the slope of our tangent line. Based on the general form of a circle, we know that $\mathbf{(x-2)^2+(y+1)^2=25}$ is the equation for a circle that is centered at (2, -1) and has a radius of 5. So we need to find the slope of a line connecting the points (5, 3) and (2, -1). We can do this using the formula for the slope of a line between two points.

$$slope = \frac{y_2 – y_1}{x_2 – x_1}$$ $$slope = \frac{3 – (-1)}{5 – 2}$$ $$slope = \frac{4}{3}$$

Now we can simply take the negative reciprocal of $\mathbf{\frac{4}{3}}$ to find the slope of our tangent line. So we know the slope of our tangent line will be $\mathbf{- \frac{3}{4}}$ .

Knowing that the slope of our tangent line will be $\mathbf{- \frac{3}{4}}$ and that it will go through the point (5, 3), we can put this into the point slope form of a line to find the equation of our tangent line.

$$y=m(x-x_0)+y_0$$ $$y= \ – \frac{3}{4}(x-5)+3$$ $$y= \ – \frac{3}{4}x + \frac{15}{4}+3$$ $$y= \ – \frac{3}{4}x + \frac{15}{4}+ \frac{12}{4}$$ $$y= \ – \frac{3}{4}x + \frac{27}{4}$$

Hopefully all of this helps you gain a bit of a better understanding of finding tangent lines, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I’ll send you my FREE bonus study guide to help you survive calculus! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.

Examples of product, quotient, and chain rules

I have already discuss the product rule, quotient rule, and chain rule in previous lessons. But I wanted to show you some more complex examples that involve these rules. The reason for this is that there are times when you’ll need to use more than one of these rules in one problem. So let’s dive right into it!

Example 1

Find the derivative of $y \ = \ sin(x^2 \cdot ln \ x)$ .

At first glance of this problem, the first thing we should notice is that we can think of this function as one function plugged into another. There is a clear inner function and a clear outer function.

It can be broken down as $x^2 \cdot ln \ x$ being plugged into sin(x) for x. Since our function can be thought of as one function plugged into another, we will want to start out with the chain rule.

Chain rule

The first thing we need to do to apply the chain rule is to figure out our inside function and outside function. It’s usually easier to think about the insider function first.

Finding f and g

To find the inside function we just need to answer the question: what function is being plugged into another?

Looking at our function you can see that we are taking $x^2 \cdot ln \ x$ and plugging that into another function. So we will say $$g_1(x) = x^2 \cdot ln \ x$$

Now that we have decided on the inside function, we need to find the outside function. All we need to do here is look at the original function, and replace our inside function with a single x. So we will replace $x^2 \cdot ln \ x$ with just x. This gives us $$f_1(x) = sin(x).$$

Finding f’ and g’

Now that we have found f and g, we just need to take each of their derivatives to find f’ and g’.

Finding f’ should be simple here. $$f’_1(x) = cos(x)$$

Finding g’ will be a little more tricky.

Looking at our function $g_1(x)$ you can see that it is actually the product of two simpler functions, $x^2$ and $ln \ x$ . Therefore, we are going to have to use the product rule to find this derivative. You can kind of think of this as a smaller sub-problem within our problem, so we will come back to the chain rule after applying the product rule.

Product rule

We need to use the product rule to find the derivative of $$g_1(x) = x^2 \cdot ln \ x.$$ The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $f_2(x)$ and $g_2(x)$ .

Finding f and g

With the product rule it doesn’t really matter which function is f and which is g. As long as we correctly identify that our function is a product of two simpler functions, it’ll work out correctly. So we will say $$f_2(x) = x^2$$ $$g_2(x) = ln \ x.$$

Finding f’ and g’

Now we just need to find the derivatives of f and g. Since they are fairly simple functions, this shouldn’t be too difficult.

To find the derivative of f we just need to use the power rule. $$f’_2(x) = 2x.$$

And finding the derivative of g should be a derivative that you have memorized. Using Wolfram Alpha we can see that $$g’_2(x) = \frac{1}{x}.$$

Plugging into the formula

Now that we have found all the pieces we need, we can simply plug them all into the product rule formula. $$g’_1(x) = f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x^2 \cdot \frac{1}{x} \ + \ 2x \cdot ln \ x$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x \ + \ 2x \cdot ln \ x$$

Back to chain rule

Now that we know the derivative of $x^2 \cdot ln \ x$ we can go back up to the chain rule. We knew that $$g_1(x) = \ x^2 \cdot ln \ x$$ and by using the product we just found that $$g’_1(x) = \ x \ + \ 2x \cdot ln \ x.$$ Just as a quick reminder, we already found that $$f_1(x) = sin(x)$$ $$f’_1(x) = cos(x).$$ Now we just need to plug these four pieces into the formula for chain rule. $$h’_1(x) = \ f’_1 \big( g_1(x) \big) \cdot g’_1(x)$$ $$\frac{d}{dx} \Big[ sin \big(x^2 \cdot ln \ x \big) \Big] \ = \ cos \big( x^2 \cdot ln \ x \big) \cdot \big( x \ + \ 2x \cdot ln \ x \big)$$

And that’s it! We could factor out a like term out of one of these factors, but it wouldn’t really make the function any simpler so I won’t do that. We can also use Wolfram Alpha to check our answer. A quick note on that, Wolfram Alpha uses “log” instead of “ln” to describe a natural log.

Example 2

Find the derivative of $y = \frac{x \ sin(x)}{ln \ x}$ .

Looking at this function we can clearly see that we have a fraction. Therefore, we can break this function down into two simpler functions that are part of a quotient. So we can see that we will need to use quotient rule to find this derivative.

Quotient rule

As discussed in my quotient rule lesson, when we apply the quotient rule to find a function’s derivative we need to first determine which parts of our function will be called f and g.

Finding f and g

With the quotient rule, it’s fairly straight forward to determine which part of our function will be f and which part will be g. We will always say f is the numerator (top of our fraction) and g is the denominator (bottom of our fraction). So we can say $$f_1(x) \ = \ x \ sin(x)$$ $$g_1(x) \ = ln \ x.$$

Finding f’ and g’

Once we have determined which part of our function we are going to call f and which part will be g, we need to take each of their derivatives so we can use the quotient rule formula.

First we’ll start with finding f’. To find this we need to find the derivative of $f_1(x)= x \ sin(x)$ . Notice this function is actually a product of two simpler functions. So in order to find $\mathbf{f'_1(x)}$ we will actually need to use the product rule. This will create a smaller sub-problem for us so we will need to come back to the quotient rule in a moment.

Product rule

As we did in the previous example, or in my product rule lesson, we need to start by determining which piece of the function $f_1(x) = x \ sin(x)$ will be $f_2$ and which will be $g_2$ .

Finding f and g

When using the product rule it doesn’t really matter which piece of the product is called f and g. So we will say $$f_2(x) = x$$ $$g_2(x) = sin(x).$$

Finding f’ and g’

In order to use the product rule formula, we need to find the derivative of each of these pieces now. Fortunately, both of these pieces are simple functions to differentiate. $$f’_2(x) = 1$$ $$g’_2(x) = cos(x)$$

Plugging into the formula

Now we just need to plug the four pieces we’ve found into the product rule formula. $$f’_1(x) = \ f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \cdot cos(x) \ + \ 1 \cdot sin(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \ cos(x) \ + \ sin(x)$$

Back to the quotient rule

Now that we have used the product rule to find $$f’_1(x) = \ x \ cos(x) \ + \ sin(x)$$ we need to find $g'_1(x)$ so we can use the quotient rule formula. Remember $g_1(x) = ln \ x$ , which is a function whose derivative you should memorize. $$g’_1(x) = \frac{1}{x}.$$ So now we know all of the pieces we need to apply the quotient rule formula.

$$h'(x) \ = \ \frac{f’_1(x) \cdot g_1(x) \ – \ f_1(x) \cdot g’_1(x)}{g^2_1(x)}$$ $$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ \Big[ \big( x \ cos(x) + sin(x) \big) \cdot ln \ x \Big] \ – \ \Big[ x \ sin(x) \cdot \frac{1}{x} \Big]}{\big( ln \ x \big)^2}$$

And now we can just simplify by distributing through all of our parenthesis.

$$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ x \ ln(x) \ cos(x) \ + \ ln(x) \ sin(x) \ – \ sin(x)}{ln^2(x)}$$

And that’s the answer! Again, we can check this using Wolfram Alpha.

If you have any questions on any of this just let me know! You can email me at jakesmathlessons@gmail.com. You can also use the contact form below and I’ll add you to my email list and send you my calculus 1 study guide to help you boost your calculus scores! I’d also love to hear any suggestions for future posts so please don’t hesitate to reach out to me. If you want some more practice with derivatives go check out my other lessons and problems related to derivatives.

RELATED RATES – Square Problem

Each side of a square is increasing at a rate of 6 $\frac{cm}{s}$ . At what rate is the area of the square increasing when the area of the square is 16 $cm^2$ ?

If you haven’t already, you should check out my related rates lesson. I go through the steps that can be used to solve any related rates problem. I will use these steps here, but it might be useful for you to see a more detailed explanation of why we use these steps.

1. Draw a sketch

The first thing we will always want to do is draw a sketch of the situation described by the problem. This problem is relatively simple to draw out. Here we have a square whose sides are growing at a constant rate. We know the sides are growing at a rate of 6 $\frac{cm}{s}$ and we want to consider the moment when the area of the square is 16 $cm^2$ .

2. Come up with your equation

Now that we have our drawing laid out, we need to create an equation whose derivative we will need to find.

What are we looking for?

This question is asking us to find the rate at which the area of the square is increasing. So it tells us that we need to have some variable that represents the rate of change of the area at some point.

Of course, we will need to take the derivative of our equation soon. Doing this will introduce the ‘rate of change’ part. Therefore, we need to make sure that our equation we make includes the area of the square. As long as we do this, we will end up with the rate of change of the area once we take its derivative.

What do we know about?

There was only two pieces of information that the question directly told us.

Each side of a square is increasing at a rate of 6 $\frac{cm}{s}$ .
The area of the square in this instant is 16 $cm^2$ .

Putting it into an equation

So at this point, we have figured out that we need our equation to include the square’s area, and we know something about the square’s area and the rate of change of its sides.

Although, we don’t know anything about the actual side lengths at the given instant, we can figure that out if we have to. Therefore, it’s fine if our equation includes the length of the square’s sides.

To summarize, the only measurements of this square we have discussed so far are its side lengths and its area. So our equation should relate its area and its side lengths. Keep in mind, we don’t want to put anything that represents a rate of change into our equation. These will come when we take the derivative of our equation.

The simplest equation that relates the area of a square with its side lengths would likely be the formula for the area of a square that you are already familiar with. $$A=l^2$$ Where A is the area of the square, and l is the length of its sides.

3. Implicit differentiation

Once we have created our equation like we did above, we need to go ahead and take its derivative with respect to time. This will be done using implicit differentiation.

Since we will be taking the derivative with respect to time, we will need to treat the A and the l in our equation as functions of time. This will require using the chain rule to find their derivatives.

$$\frac{d}{dt} \big[ A \big] = \frac{d}{dt} \Big[ l^2 \Big]$$ $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$

4. Solve for the desired rate of change

Now all we have to do is solve for the rate of change the question is asking about. Just like I said earlier, we need to find the rate of change of the square’s area. Since this is exactly what $\frac{dA}{dt}$ represents and we have already isolated this, we don’t have much else to do. All we need to do is plug in the values we have for everything else in our equation.

The only other variables we need to plug in are l and $\frac{dl}{dt}$ . The tricky thing here is that we don’t directly know what l is at the moment in this problem. Instead we know that the area of the square is
16 $cm^2$ . Since we know that the relationship between the area of a square and its side lengths is $$A=l^2$$ we can find l in this instant. $$16=l^2$$ $$4=l$$ So we know that in this moment the length of the square’s sides is 4 cm.

Remember, the problem also told us that the side lengths are increasing at a rate of 6 $\frac{cm}{s}$ . This directly tells us the rate of change of the sides lengths. So we also know that $$\frac{dl}{dt}=6$$

Putting all of this into our equation will give us: $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$ $$\frac{dA}{dt} = 2(4)(6)$$ $$\frac{dA}{dt} = 48$$

So the area of this square is increasing at a rate of 48 $\mathbf{\frac{cm^2}{s}}$ when the area is 16 $\mathbf{cm^2}$ .

If you’re still having some trouble with related rates problems or just want some more practice you should check out my related rates lesson. At the bottom of this lesson there is a list of related rates problems that I have posted along with their solutions. I also have several other lessons and problems on the derivatives page you can check out. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

You can also enter your name and email using the form below and I will send you my calculus 1 study guide as a welcome gift!

RELATED RATES – Cone Problem (Water Filling and Leaking)

Water is leaking out of an inverted conical tank at a rate of 10,000 $\frac{cm^3}{min}$ at the same time water is being pumped into the tank at a constant rate. The tank has a height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 $\frac{cm}{min}$ when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Jake’s Math Lessons Complete Calculus 1 Package

Here we have another related rates problem. This is a pretty typical problem you would see in a calculus class. There is a lot going on in this one since we have a related rates with a cone filling and leaking water. Just like I said when I discussed related rates, these problems tend to follow a specific pattern. If you need a refresher on what the four steps are just click that link to my related rates lesson. Otherwise, we’ll jump right into it.

1. Draw a sketch

You should always start a related rates problem with a drawing of the real world situation that’s being described in the problem. The problem describes an “inverted conical tank.” This just means that the tank is in the shape of an up-side-down cone. Other than that, the other facts are quite simple.

Water is leaking out at a rate of 10,000 $\frac{cm^3}{min}$ .
Liquid is being pumped into the tank at an unknown constant rate.
Tank’s height is 6 m.
Diameter of the circular opening is 4 m.
Water level is rising at a rate of 20 $\frac{cm}{min}$ when the height of the water is 2 m.

So we need to put all of these facts into a drawing, which might look something like this.

The important thing to point out about our sketch is that we have two cones here.

One cone is the tank, which is the larger cone. This one will always stay the same size and is not changing. This means that its height, diameter of the base, and its volume are all constants.

The second cone is the water sitting in the bottom of the tank, which is the smaller cone. This one is changing as our liquid flows into and out of the tank. As time passes its dimensions change. Therefore, its height, diameter of the base, and volume are all functions of time.

2. Come up with your equation

Now that we have made our sketch, we need to come up with an equation that relates all of the different quantities we’re dealing with. All of the information we know about and the information we are looking for relates to a volume or measurements of some cone. The measurements are either of our water in the tank or the tank itself, but in both cases the measurements describe a cone.

What are we looking for?

The question asks us to find the rate at which water is being pumped into the tank. As it is pumped into the tank, this will impact the volume of the smaller cone which is the water sitting in the tank. Therefore, the information we are looking for will somehow relate to how quickly the volume of the liquid in the tank is changing. Or the rate of change of the volume of the small cone.

What do we know about?

We were given quite a bit of information so let’s break it into a few different pieces.

We know the height and base diameter of the tank. Using these, we can find its base radius and volume. Since the tank isn’t changing we know, or can easily find, any important measurement of the large cone.
We know how fast water is flowing out of the tank. This tells us a piece of information relating to the rate of change of the small cone. We will need to use this once we know how fast water is flowing into the tank.
The last information we already know is the height and the rate of change of the height of the small cone at a particular moment.

Putting it into an equation

As I mentioned above, we need to find the rate of change of the volume of the liquid in the tank. Since we know we will need to use implicit differentiation to get the rate of change, our equation needs to involve the volume of the small cone. Remember, the equation we come up with should include quantities and measurements, not rates of change yet.

So we are dealing with the volume of something that’s in the shape of a cone. Because of this we can start by looking at the formula for a volume of a cone then we’ll see if we can use this directly or if we will need to make some adjustments. The volume of a cone can be described by

$$V=\pi r^2 \frac{h}{3}$$

where r is the radius of the cone’s base and h is its height.

Clearly we would be able to use this to get the rate of change of the volume since the equation already includes the volume. But this means we need to know everything else in the equation.

Other than the volume, the equation requires that we know the radius and height of the cone. We also need to know the rate of change of these other parts.

By referring back to our original drawing, you can see that we clearly already know the height and the rate of change of the height of the liquid in the tank at this specific moment. Remember, these two values are changing as liquid goes into the tank, but we know their values at this specific instant.

But what about the radius?

We don’t know the radius yet, but we can find it by comparing the small and large cones and using similar triangles. However, the problem is that we don’t know the rate of change of the radius. We could figure it out, but it would be quite complicated and there is actually something else we can do instead.

So we want to avoid using the radius, or diameter, because we don’t want to have to find its rate of change. So what else can we do?

In order to get around this, we will need to write an equation that doesn’t include the radius. But remember, it’s fine if we use the height because we know its rate of change. The important fact we need to use here is that the two cones will always form similar triangles. This will be true no matter where the water level is, as it fills up.

Consider the following drawing which represents the conical tank shown from the side.

As shown in the sketch above, the top sides of these triangles are parallel. This means that each angle in the small triangle is the same as the corresponding angle in the large triangle. Therefore, these are similar triangles. Since they are similar triangles, we know one important fact:

$$\frac{x}{2}=\frac{4}{6}$$

So we can use this to solve for x.

$${x}=\frac{4}{6} \cdot 2$$

$$x=\frac{4}{3}$$

So we can see that in both of these triangles, the top side is two-thirds as long as the height of the triangle. In other words, if we multiply the height of the triangle by $\frac{2}{3}$ , this would give us the length of the top side of the triangle. Remember the top of the triangle is the diameter of the corresponding cone from our first drawing. This tells us the following relationship between the height and diameter of the cones:

$$\frac{2}{3}h=d$$

Since our goal here is to find out the relationship between the radius and the height, not the diameter and the height, we need to do one more thing. The diameter of a circle is always twice as much as the radius, so we know $d=2r$ . We can substitute this in for d and solve for r.

$$\frac{2}{3}h=2r$$

$$\frac{1}{3}h=r$$

Going back to our equation

Now we need to go all the way back to our volume equation that involved h and r.

$$V=\pi r^2 \frac{h}{3}$$

Since we now have this new way to write r in terms of h, we can substitute this into the volume equation and simplify from there.

$$V=\pi \bigg( \frac{1}{3}h\bigg)^2 \frac{h}{3}$$

$$V=\pi \cdot \frac{1}{9}h^2 \frac{h}{3}$$

$$V= \frac{1}{27} \pi h^3$$

Now we have an equation involving V and h. Since we are looking for the rate of change of V, and we already know about V, h, and the rate of change of h, this equation will work perfectly.

3. Implicit differentiation

We created our equation containing all of the necessary pieces. Remember earlier I said that we are going to need to find the value of $\frac{dV}{dt}$ in order to find how quickly liquid is going into the tank? Now is the time to do this.

As with any related rates problem, we need to take the derivative of our equation. Since we already have an equation for V, all we need to do to find $\frac{dV}{dt}$ is take the derivative of V with respect to time. Let’s do that now.

$$\frac{d}{dt} \big[ V \big] = \frac{d}{dt} \bigg[ \frac{1}{27} \pi h^3 \bigg]$$

The left side of this equation will be quite simple, but to find the right side we need to keep a couple things in mind. We are going to take the derivative with respect to time. Therefore, we need to treat h as a function of time rather than a variable. This means that we will need to use the chain rule to take this derivative. You can see a more detailed explanation of how to do this and why we are doing it in my implicit differentiation lesson.

Also remember that $\frac{1}{27} \pi$ is just a constant coefficient in front of our $h^3$ term.

$$\frac{dV}{dt}= \frac{3}{27} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

4. Solve for desired rate of change

The final step of a related rates problem is to solve of the rate of change the question asked for. Since $\frac{dV}{dt}$ is already isolated we don’t need to worry about solving for the variable we are looking for.

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

All we need to do now is plug in the other information missing from the equation to tell us all we need to know about $\frac{dV}{dt}$ . As you can see, this mean we need to figure out the values for h and $\frac{dh}{dt}$ . Luckily, both the height and the rate the height is increasing of the small cone were both given to us.

Looking back at our drawing, which I have put below, we can see that both of these values we need are already labeled.

Looking at our sketch we can see that h, which is shown as the height of our small cone, is 2 m. We can also see that $\frac{dh}{dt}$ , which is shown as the speed at which the water level is rising, is 20 $\frac{cm}{min}$ . But there is one problem here. We need to be careful of our units!

Notice that the height is measured in meters and its rate of change is measured in centimeters.

We need to convert one of these two measurements so that we are either using meters or centimeters throughout the whole problem. It doesn’t matter which one we choose as long as there are the same. I will go ahead and use meters. Don’t worry, if you need the answer in centimeters I’ll discuss how to convert to that in a minute. Since $100 \ cm = 1 \ m$ we know that

$$h=2m$$

$$\frac{dh}{dt} = 0.2 \ \frac{m}{min}$$

Putting it all together

So now we can plug these values back into our equation for $\frac{dV}{dt}$ .

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi (2)^2 \cdot 0.2$$

$$\frac{dV}{dt}= \frac{4}{9} \pi \cdot \frac{1}{5}$$

$$\frac{dV}{dt}= \frac{4 \pi}{45}$$

$$\frac{dV}{dt} \approx 0.279 \ \frac{m^3}{min}$$

But remember this isn’t quite our answer. We still have one more step. The question asked us to find the rate at which water is being pumped into the tank. We just found the rate at which the volume of water in the tank is changing.

There are three important values to think about here.

The rate water is being pumped into the tank.
Rate of change of the water actually in the tank.
The rate liquid is flowing out of the tank.

We are trying to find the rate water is being pumped into the tank and we already know the other two rates. It was given that water is flowing out of the tank at a rate of 10,000 $\frac{cm^3}{min}$ .

Be careful of the units

We know the rate at which the liquid is flowing out of the tank, but it uses centimeters. Since we decided earlier that we want to use meters instead, we need to convert this as well. This will be a little different though.

Think about a cube with each side length being 1 m. This cube would be 1m x 1m x 1m and would have a volume of 1 $m^3$ . Since we know that $1 \ m = 100 \ cm$ , we can also say that this same cube is 100cm x 100cm x 100cm and would actually have a volume of 1,000,000 $cm^3$ . Therefore, we can say

$$1m^3 = 1,000,000cm^3.$$

If we divide both sides of this equation by 100, we can see that

$$\frac{1}{100}m^3 = 10,000cm^3.$$

So we then can say that water is flowing out of the tank at a rate of 0.01 $\frac{m^3}{min}$ .

How it fits together

We know that $\frac{dV}{dt} \approx 0.279 \ \frac{m^3}{min}$ is a positive number. This means that the amount of water in the tank is increasing at this instant. Since some water is flowing out of the tank at a rate of 0.01 $\frac{m^3}{min}$ , it must be flowing into the tank faster than this. If it were not, the volume in the tank would be decreasing.

In fact, you can think of the rate of change of the volume of water in the tank as the rate of water flowing in minus the rate of water flowing out. Let’s say X represents the rate at which liquid is flowing into the tank and we can use this equation to represent the previous sentence.

$$X-0.01=0.279.$$

Solving for X tells us

$$X=0.289.$$

So water must be flowing into the tank at a rate of 0.289 $\frac{m^3}{min}$ . And this is the value the question was asking us to find!

We can also give our answer in other units

If we go back to our equation before we started plugging in anything, we can instead plug everything in terms of centimeters instead of meters. In this case we will have $h=200cm$ and $\frac{dh}{dt}=20 \frac{cm}{min}$ .

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi (200)^2 (20)$$

$$\frac{dV}{dt} \approx 279,252.68$$

Now just like above, we need to also consider how fast the water is flowing out of the tank. This time we don’t need to convert the units though since we are using centimeters.

$$Y-10,000= 279,252.68$$

$$Y= 289,252.68$$

So we can also say that water is flowing into the tank at a rate of 289,252.68 $\frac{cm^3}{min}$ . This is more likely going to be a better answer simply because it is more precise that the previous answer we found!

Enter your name and email below and I’ll send you my bonus calculus 1 study guide for FREE!

This was a good cone related rates example, but if you want some more practice you should check out my related rates lesson. At the bottom of this lesson there is a list of related rates problems and solutions. I also have several other lessons and problems on the derivatives page you can check out. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

Solution – A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

Jake’s Math Lessons Complete Calculus 1 Package

1. Draw a sketch

Here we have a related rates problem. As I said when I discussed related rates problems initially, the first thing I like to do with these problems is draw a sketch of the scene that is being described. If you want to refer back to that, you can click here. Otherwise, let’s sketch the problem described here.

A kite 100ft above the ground moves horizontally at a speed of 8 ft/s.

2. Come up with your equation

The next thing we need to do is set up our equation which will relate our different quantities. To do this, we will want to consider what value the question is asking us to find.

What are we looking for?

It asks “at what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?” Therefore, the value we are looking for is the “rate of change of the angle between the string and the horizontal.” This just means we will need to consider the angle between the string and the ground (the ground is the horizontal in this case). If you look back at our drawing, you will see that this angle is represented by $\theta$ . Since our goal is to find how fast $\theta$ is changing, we need $\theta$ to be in our original equation.

What do we know about?

Now we need to consider what other quantities or variables we know something about. Clearly we know something about the two sides of the triangle that are labeled as being 100 ft and 200 ft. And we can use these two sides to figure out the length of the third side, which is not labeled in our drawing.

Although we could simply call one of those sides $a$ and the other one $b$ and proceed from there, there is another option that may simplify our problem.

Consider the fact that the kite is moving horizontally. This means that the kite is not getting any further from or closer to the ground as it moves. Therefore, the side that is labeled 100 ft will actually be 100 ft at any point in this kite’s flight. Because of this we actually don’t need to designate a variable to this side of the triangle. Instead this side is simply a constant 100 ft.

Now we just need to use one of the other two sides of the triangle. We could technically use either one, but one will be a lot easier than the other. It looks like the hypotenuse would be the easier of the two, because we know it’s 200 ft at this moment. However, we don’t know exactly how fast it’s changing. We can figure that out but it wouldn’t be easy.

We do know exactly how fast the unlabeled side is changing. The question states that the kite is moving horizontally at a speed of 8 $\frac{ft}{s}$ . Since this unlabeled side is exactly horizontal, we know its rate of change is also 8 $\frac{ft}{s}$ . We can figure out its length using Pythagorean Theorem later, but this would certainly be easier than finding the rate of change of the hypotenuse. Therefore, I will go ahead and use the unlabeled side.

Since this unlabeled side is going to be changing we will need to designate a variable to this side of the triangle. As the kite moves away from the person flying it, the person holding the string has to let more string out and allow it to become longer. This means that this unlabeled side in our drawing will need to be described with a variable. We will call it side $a$ .

Putting it into an equation.

Now we have three different quantities we need to relate somehow:

Angle $\theta$ (this will be changing as the kite moves).
Side $a$ (this will be changing as the kite moves and the string is let out).
Side labeled 100 ft (this will not change and can be treated as a constant).

So we have two sides and an angle that we need to make an equation with. To do this, think about where these sides are in relation to the angle $\theta$ . The side labeled 100 ft is the side opposite to the angle $\theta$ and the side we’re calling $a$ is adjacent to the angel $\theta$ .

Usually when dealing with two sides and one angle of a triangle, you will want to use either sine, cosine, or tangent to relate the three. So which one should be used when we know the opposite side and the adjacent side to the angle in question?

Remember soh, cah, toa?

Sine Opposite Hypotenuse
Cosine Adjacent Hypotenuse
Tangent Opposite Adjacent

Since we have the opposite side and the adjacent side, we want to use tangent. Therefore we can say:

$$tan(\theta) = \frac{100}{a}$$

Since it will make finding the derivative easier, I am going to rewrite this as

$$tan(\theta) =100a^{-1}$$

3. Implicit differentiation

As with any related rates problem, we now need to take the derivative of both sides of the equation with respect to time. Since $\theta$ and $a$ are both functions of time, we will need to use chain rule for both sides of this equation. We know they are functions of time because they are both going to be dependent on the position of the kite as time progresses. We don’t have an explicit formula for either of these functions, but we know their values are dependent on time.

$$\frac{d}{dt}tan(\theta) =\frac{d}{dt}100a^{-1}$$

$$\frac{d}{dt}\frac{sin(\theta)}{cos(\theta)} =\frac{d}{dt}100a^{-1}$$

To find the derivative of the left side of this equation you will need to use the quotient rule and the chain rule. I’m not going to show all the steps of how to do this but if you want a refresher, you can read about the quotient rule here and the chain rule here. Using Wolfram Alpha, you can see that

$$\frac{d}{dx}tan(x)=\frac{1}{cos^2x}$$

Therefore, we can say that

$$\frac{d}{dt}tan(\theta)=\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt}$$

Plugging this back into the left side of our equation, we get

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =\frac{d}{dt}100a^{-1}$$

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

4. Solve for desired rate of change

The last step of any related rates problem is to solve for the desired rate of change. Now remember the thing we need to find is the rate of change of our angle $\theta$ . This is exactly what $\frac{d\theta}{dt}$ represents. So now we just need to solve for $\frac{d\theta}{dt}$ .

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

Now we just need to plug in the values for $a$ , $\frac{da}{dt}$ , and $\theta$ and we will have our answer. We don’t know all of these values but we can find them.

Finding a

As I mentioned before, we can find $a$ by using Pythagorean Theorem. Looking back at our drawing, we have a right triangle with side lengths of 100 ft, 200 ft, and $a$ . We know that

$$100^2 + a^2 = 200^2$$

$$10,000 + a^2 = 40,000$$

$$a^2 = 30,000$$

$$a = \sqrt{30,000}$$

$$a = 100\sqrt{3}$$

Finding $\mathbf{\frac{da}{dt}}$

This was actually given. We know that $a$ is the horizontal distance the kite is away from the person flying the kite. We know that the kite is moving horizontally at a speed of 8 $\frac{ft}{s}$ . Because of this we know that this is also the rate at which $a$ is changing. Since $\frac{da}{dt}$ is the rate of change of $a$ , we know

$$\frac{da}{dt} = 8$$

Finding $\mathbf{\theta}$

To find $\theta$ we will need to go back to the original equation we came up with before the implicit differentiation step:

$$tan(\theta) = \frac{100}{a}$$

Since we know $a$ , we can plug it in here and solve for $\theta$ .

$$tan(\theta) = \frac{100}{100\sqrt{3}}$$

$$tan(\theta) = \frac{1}{\sqrt{3}}$$

This angle is actually on the unit circle and by using this we know:

$$\theta = \frac{\pi}{6}$$

Note that $\theta$ will be in radians.

Now we can plug all of these into our equation for $\frac{d\theta}{dt}$ .

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

$$\frac{d\theta}{dt} =-100 \big(100\sqrt{3} \big)^{-2} \cdot 8 \cdot cos^2 \Bigg( \frac{\pi}{6} \Bigg)$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \Bigg( \frac{\sqrt{3}}{2} \Bigg)^2$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \frac{3}{4}$$

$$\frac{d\theta}{dt} =-\frac{1}{50}$$

So we can say that the angle between the string and the horizontal is decreasing at a rate of $\frac{1}{50} \ \frac{radians}{s}$ when 200 ft of string has been let out.

And that’s the answer to the question! Hopefully that wasn’t too bad, but if you have any questions I’d love to hear them. I know related rates problems can be challenging so you can email me any questions or suggestions at jakesmathlessons@gmail.com. If you have any other problems you’d like to see worked out go ahead and send me an email.

Enter your name and email below to get your FREE copy of my calculus 1 study guide to help you boost your test scores!

If you feel you need some more practice with related rates, you can check out the lesson where I discussed related rates for more examples.

Also, if you want to check out some other problems and get some practice with derivatives, go check out my derivatives page. You can see what other topics I’ve already covered and problems I’ve worked through. If you can’t find your problem there just let me know and I may post the solution to your problem.

Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $a$ and $b$ that make the following function differentiable for all values of $x$ .

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

Jake’s Math Lessons Complete Calculus 1 Package

When trying to solve a problem like this, there are actually two things you will need to consider for our function $f(x)$ . Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous. The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere. Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $f(x)$ is continuous at $x=-1$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$ Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $f(x)$ will be continuous everywhere as long as $a=3$ . However, this doesn’t really tell us that $f(x)$ is differentiable everywhere as well.

Making sure f(x) is differentiable everywhere

We now know that we will need to let $a=3$ in order for this function to be continuous and to have a chance of being differentiable. As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $x=-1$ . The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain. The only place we may have a problem is when we have to switch between the two functions.

What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $x$ . In other words, we need to be able to find its derivative no matter what $x$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $f(x)$ when $x=-1$ since we know it would exist for all other values of $x$ . By using the definition of a derivative, we need to make sure the following limit exists at $x=-1$ .

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $x=-1$ , we can plug in $-1$ for $x$ . Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously. If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $f(x)$ was continuous, we will need to consider each one sided limit separately in order to find this limit. And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

Setting up the limits

When $h$ is slightly less than $0$ , and we are considering the left sided limit, $f(-1+h)$ would need to be found using the $y=bx^2-3$ because this would involve inputting $x$ values which are less than $-1$ .

By the same reasoning, when $h$ is slightly greater than $0$ , and we are considering the right sided limit, $f(-1+h)$ would need to be found using the $y=3x+b$ because this would involve inputting $x$ values which are greater than $-1$ . Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal. Therefore, in order to show that the derivative of $f(x)$ exists at $x=-1$ , these two one-sided limits need to be equal to each other. Before setting them equal to each other, first we’ll simplify them a bit. First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

What does this tell us?

So now if we put both pieces together, we know that $a=3$ will ensure that $f(x)$ is continuous and then making $b=-\frac{3}{2}$ will also make sure $f(x)$ is differentiable at $x=-1$ . This would in turn make $f(x)$ differentiable for all values of $x$ , or make it differentiable everywhere.

As always, I want to hear your questions! Go check out my other lessons about derivatives and if you can’t get your question answered, I’d love to hear from you. Leave a comment below or email me at jakesmathlessons@gmail.com. If you have questions on this problem and solution or if you have another question you would like to see me answer, just ask it. Or if you have an entire topic you would like to see me write a lesson about, just let me know.

Enter your name and email below and I will also send you my calculus 1 study guide as a FREE bonus as a welcome gift!

Solution – Find two numbers whose sum is 23 and whose product is a maximum

Find two numbers whose sum is 23 and whose product is a maximum.

When you see a problem like this, it’s obvious our goal is to maximize some function. The more challenging part is figuring out what the function would be that we want to maximize. When we look at this problem however, we can actually break it down into two different facts we know about these two numbers, which we will call a and b.

The two numbers add up to 23.
We need to maximize their product.

Each of these two bullet points can actually be represented mathematically and they can be used together.

Fact #1

First of all, we know that these two numbers add up to 23. Or in other words,

$$a+b=23.$$

Fact #2

Also, we need to maximize the product of these numbers. Another way to put this is that we want to maximize the function f, which represents the product of a and b. So we need to maximize

$$f(a, \ b)=ab.$$

Putting the two facts together

The problem with this, is that it is difficult to maximize multivariable functions, and in fact, this function wouldn’t have a maximum unless we go back to our other equation. Since we know $a+b=23$ , we can actually use this to rewrite the function we need to maximize in terms of only one variable.

$$a+b=23$$

$$a=23-b$$

Now since we know $a=23-b$ , we can go to our function of a and b and replace each a with (23-b). Doing this will give us a new function with only b‘s in it, which we can then maximize.

$$f(a, \ b)=ab$$

Replacing a with (23-b) gives us a new function which we’ll call g.

$$g(b)=(23-b)b$$

Maximizing the function to find a and b

Now we just need to find the value of b that maximizes g. To do this, we just need to take its derivative with respect to b, set it equal to 0, and solve for b. This will always be the general process when you need to maximize or minimize a function. But first I will expand the function so it’s easier to take its derivative.

$$g(b)=23b-b^2$$

$$g'(b)=23-2b$$

Now set this equal to 0 and solve for b.

$$0=23-2b$$

$$2b=23$$

$$b=\frac{23}{2}$$

We can even check this using Wolfram Alpha. Below you can see a graph of $y=g(b)$ showing the maximum value at $b=\frac{23}{2}$ .

Find two numbers whose sum is 23 and whose product is a maximum

Now remember, we needed to find the value of a and b that would maximize their product. So we need to use $b=\frac{23}{2}$ to find a. To do this, we can just use the relationship we found earlier:

$$a=23-b$$

$$a=23-\bigg(\frac{23}{2}\bigg)$$

$$a=\frac{46}{2}-\frac{23}{2}$$

$$a=\frac{23}{2}$$

So now we know that $a=\frac{23}{2}$ and $b=\frac{23}{2}$ are the two numbers whose sum is 23 and whose product is as large as possible.

There are plenty of other lessons and solutions to help you make sense of derivatives on my derivatives page. Go check them out and as always, I’d love to here your questions! Leave a comment below or email me at jakesmathlessons@gmail.com with any questions or suggestions you may have. Every email and comment helps me gear this site toward what you want to see, so please don’t hesitate to reach out.

You can also enter your name and email below and I’ll send you my calcu 1 study guide full of useful tips and tricks to help you boost your test scores!

RELATED RATES – 4 Simple Steps

Related rates problems are one of the most common types of problems that are built around implicit differentiation and derivatives. Typically when you’re dealing with a related rates problem, it will be a word problem describing some real world situation.

Typically related rates problems will follow a similar pattern. They can usually be broken down into the following four related rates steps:

The first thing you will usually want to do after reading the problem is to draw a sketch of the situation being described.
Then you will need to come up with some equation that relates the different quantities described to you, which may be volumes, areas, or distances.
Once you have this equation, you’ll perform implicit differentiation on both sides of the equation, usually with respect to time.
Then you just need to solve for the desired rate of change that the question is asking about.

I always think the best way to learn a new concept is practice, practice, practice. So let’s jump into an example. If you want to skip ahead, there is a list of other examples at the bottom of this page with a link to their solutions.

Example 1

At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 PM?

1. Draw a sketch

Like I said before, the best thing to do first is draw a picture of what is being described. The problem tells us where these ships are, and the direction and speed at which they’re moving at noon, so I think we should start by drawing our scene at noon, seen in Figure 2.1.

Related rates ships' position at Noon. — Figure 2.1

Now let’s think about what will be happening between noon and 4:00 PM. We know the speed and direction both of these ships are traveling. We also know that they will be moving for 4 hours before we consider their position again.

Ship A: This ship is moving 35 km/h for 4 hours. Therefore, between noon and 4:00 PM we know it will move east

$$35\frac{km}{h}*4h=140km.$$

Ship B: This ship is moving 25 km/h for 4 hours. Therefore, between noon and 4:00 PM we know it will move north

$$25\frac{km}{h}*4h=100km.$$

Considering both of these facts, at 4:00 PM, our scene would look something like this:

Related rates ships' position at 4:00 PM. — Figure 2.2

2. Come up with your equation

Now, the next thing we need to do is come up with an equation that relates the different distances given based on where the boats are at 4:00 PM. We also want to consider what the question is asking before we come up with this equation. The question is asking us to find how fast the distance between the boats is changing. We won’t be able to find an equation for this right away, but this tells us that we need an equation that involves the distance between the two boats. From there, we can figure out how fast that distance is changing.

To do this, we will want to think of the drawing in Figure 2.2 as a triangle. The three vertices of the triangle would be ship A, ship B, and the point in the water where boat B was at noon, which is shown in the drawing above. Doing this gives us something like this:

Ships' position drawn as a triangle to come up with the relates rates equation. — Figure 2.3

It is important to remember that both ships are still moving, they don’t stop at 4:00. As a result of this, the sides of our triangle are not constants. Instead, they would be variables, so the triangle we may want to consider is this one:

Related rates triangle labeled for transition to Pythagorean Theorem. — Figure 2.4

From this, we can create our equation. What we want to think about is what variable has to do with the value the question is asking us to find and what variables do we know something about.

What are we looking for?

The question asks us to find how fast the distance between the ships is changing. By comparing Figure 2.4 to the prior drawings, we can see that side $z$ is the one that represents the distance between the two ships. Therefore, we will need to include $z$ in our equation which we will eventually differentiate.

What do we know about?

This is where we want to consider the actual numerical values we have figured out and how they relate to the different variables in our drawing. At 4:00 PM we know the values of $x$ and $y$ . We also know the speed at which the ships are moving in relation to a fixed point, which is the vertex of the triangle that makes up the right angle. Therefore, we not only know the values of $x$ and $y$ , but we also know their rates of change. These will simply be the speeds of the ships.

Putting it into an equation.

We need an equation that relates the thing we are looking for with the things we already know. Since we’re looking for some information about $z$ (the rate of change of $z$ ), and we know everything about $x$ , $y$ , and both of their rates of change at 4:00, we need an equation relating $x$ , $y$ , and $z$ .

Since we know the relationship between these variables is that they are the side lengths of a right triangle, the simplest equation we can use is Pythagorean Theorem. Based on this we know that

$$z^2=x^2+y^2.$$

3. Implicit differentiation

Now that we have our equation, we need to take its derivative. This is where the implicit differentiation comes in. Before we do this, let’s think about what we want to differentiate with respect to.

I mentioned in the beginning of this article that we will usually differentiate with respect to time. The reason for this is that $x$ , $y$ , and $z$ are all changing as time passes. If fact, since we know the ships’ initial positions and their velocities, we could actually write $x$ , $y$ , and $z$ as functions of time. Therefore, when we differentiate both sides of our equation, we will treat $x$ , $y$ , and $z$ as functions, and time (represented by $t$ ) will be the variable. If you want a bit more explanation on the next few step, I explained this a bit more here.

$$z^2=x^2+y^2$$

$$\frac{d}{dt}\big[ z^2\big]=\frac{d}{dt}\big[ x^2+y^2\big] $$

$$2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$

Now we can divide both sides by $2$ to simplify.

$$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$

4. Solve for desired rate of change

Now remember, the thing we are trying to find in this problem is the rate of change of $z$ . This is exactly what $\frac{dz}{dt}$ represents, so we will solve for this by dividing both sides by $z$ .

$$\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$$

All we have left to do now is plug in all the pieces on the right side of the equation and that would give us our answer. We are looking for the value of $\frac{dz}{dt}$ at 4:00 PM, so we need to use the values of all the other variables based on what they are at 4:00 PM also. We can gather most of the information we need from Figure 2.2, shown above. Here it is again:

Ships' movement compared to related rates triangle problem. — Figure 2.2

Clearly, we can see that $x=10km$ and $y=100km$ at 4:00. We also know the value of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ . The reason for this is that these two things represent the rate of change of $x$ and $y$ . Since $x$ and $y$ represent the distance between each of the ships and a fixed point, $\frac{dx}{dt}$ and $\frac{dy}{dt}$ would be given by the speed of each ship. It is also important to point out that is works because the ships are moving directly toward the fixed point or directly away from the fixed point. The length of $x$ is shrinking by $35\frac{km}{h}$ because ship A is moving at that speed. Therefore, we know

$$\frac{dx}{dt}=-35\frac{km}{h}.$$

Notice this value is negative. This is simply because $x$ is getting smaller at 4:00. Similarly, we know that $y$ is getting larger by $25\frac{km}{h}$ at 4:00 and therefore we can say that

$$\frac{dy}{dt}=25\frac{km}{h}.$$

Now the only thing we still need to figure out is the value of $z$ at 4:00 PM. To do this, let’s go back to our Pythagorean Theorem equation we looked at earlier. We know

$$z^2=x^2+y^2.$$

Since we already know the values of $x$ and $y$ at 4:00, we can just plug them into the Pythagorean Theorem equation to get the value of $z$ at 4:00.

$$z^2=10^2+100^2$$

$$z^2=100+10,000$$

$$z^2=10,100$$

$$z=\sqrt{10,100}$$

Now that we have all the pieces figured out, we can go back and plug them into our equation for $\frac{dz}{dt}$ .

$$\frac{dz}{dt}=\frac{10*(-35)+100*25}{\sqrt{10,100}}$$

$$\frac{dz}{dt}=\frac{2,150}{\sqrt{10,100}}\approx 21.393\frac{km}{h}$$

So we can say that the distance between the two ships is changing at about $21.393\frac{km}{h}$ at 4:00 PM.

Other Examples

Like I said before, the best way to gain an understanding of related rates problems is practice. Here are some more complete solutions of other fun related rates problems. Just click on the problem to see the full solution.

As always, I’d love to see your questions! You can leave a comment below or email your questions to me at jakesmathlessons@gmail.com. Whether you have questions about this lesson or want me to post a solution for another problem you’re stuck on, just send me a message.