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:

  1. Angle \(\theta\) (this will be changing as the kite moves).
  2. Side \(a\) (this will be changing as the kite moves and the string is let out).
  3. 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.

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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 f continuous everywhere

Consider the following piecewise function: $$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

Jake’s Math Lessons Complete Calculus 1 Package

 

Problem credit: Chapter 2.4 #36 in Single Variable Calculus: Concepts and Contexts by James Stewart.

How to find a and b for a piecewise function to be continuous everywhere.

When we see piecewise functions like this and our goal is to make sure it is continuous everywhere, there are two cases we want to consider separately:

  1. making each individual piece continuous on its reduced domain,
  2. and making the different pieces line up when our function switches between them.

First let’s consider each piece individually on the reduced domain where \(f\) is actually defined by that function.

Piece 1: \(y=\frac{x^2-4}{x-2}\) when \(x<2\)

When considering a function to find x values where the function is not continuous, two very common things to look for would be taking the square root of a negative number or dividing by zero.  Obviously we don’t need to worry about taking the square root of a negative number here since we don’t have any square roots.  However, we do have a fraction, and therefore we should consider if there is an x value that would cause us to divide by zero.

Looking at the denominator of our fraction, we see that we would divide by zero when we plug in some number for x that causes

$$x-2=0.$$

Clearly, this would happen when \(x=2\).  And in fact, this is the only value for x where \(y=\frac{x^2-4}{x-2}\) is not continuous.  However, remember that I said we want to see if there are any x values that cause the function not to be continuous within our limited domain for the specific piece of the function.  Since \(x=2\) is NOT in the domain \(x<2\), we can say that this piece of our function is continuous on its entire domain and won’t contribute any discontinuities to f.

Piece 2: \(y=ax^2-bx+3\) when \(2\leq x<3\)

Looking at this piece of our piecewise function, clearly we need to consider our constants a and b.  Since our function f is a function of x (indicated by f(x)), we can consider the other letters in this piece of our function (a and b) to be constants.  I discussed this in a bit more detail here, but it basically means that a and b are some set number, they do not change.

The important thing to realize though, is that regardless of what values we choose for a and b, this piece of our function will just be a polynomial.  More specifically, it will be a quadratic function.  No matter what our a and b end up being we will have an \(x^2\) term, an x term, and a constant term.  Since we know this piece of our function will be a polynomial no matter what our a and b are we can say that this piece of our function will be continuous for any a and b we select.

This is the case because polynomials are always continuous everywhere.  If this section of our function would be continuous everywhere, it would certainly be continuous on our limited domain of \(2\leq x<3\).  Therefore, we don’t need to worry about our a and b causing any discontinuities here.

Piece 3: \(y=2x-a+b\) when \(x\geq 3\)

This is actually going to be similar to piece 2 above.  We can see by looking at this function that no matter what a and b values we select here, they will both be constants which will combine to a single constant when we combine like terms.  Once we do this, we will be left with an x term and a constant term.  Therefore, this is also a polynomial (in this case it would also be a linear function), and would be continuous everywhere as a result.  Because of this, it must be continuous on our limited domain of \(x\geq 3\).

Now we have shown that we will not get any discontinuities within any of our restricted domains that make up our piecewise function, but what about on their edges.  We also have to make sure we are selecting an a and a b value so that we can switch from one part of our function to the other without jumping up or down and leaving a hole in our function.  If we do not select the correct a and b we may get discontinuities at the x values where we switch from one function to the other.

In order to prevent this, we need to make sure that our first two pieces of f(x) transition smoothly when \(x=2\) (this is the place we switch between these two functions) and we need to make sure that the second and third piece of our function have a smooth transition between them at \(x=3\).

Continuity at \(x=2\)

Remember by definition, a function f(x) is continuous at \(x=a\) if $$\lim_{x \to a} f(x) = f(a).$$

We need to apply this rule here to make sure that our function f is continuous at \(x=2\). Therefore, we need to pick an a and a b so that $$\lim_{x \to 2} f(x) = f(2).$$

As I explained here, a limit will only exist if both of its one-sided limits exist and are equal to each other.  Therefore, rather than finding the above limit, we actually want to find these two one-sided limits:

$$\lim_{x \to 2^{-}} f(x) = f(2)$$

$$\lim_{x \to 2^{+}} f(x) = f(2).$$

Let’s start with the left-sided limit.  Since we are approaching \(x=2\) from the left, we are looking at x values that are slightly less than 2.  Therefore, we need to use the piece of our function that is defined for \(x<2\).  So the limit we really want to find, and the equation we want to solve is:

$$\lim_{x \to 2^{-}} \frac{x^2-4}{x-2} = f(2).$$

When attempting to solve this limit, we see that we can not just plug in 2 for x because this would cause us to divide by zero.  As a result of this, I would first suggest simplifying the function whose limit we are trying to find.  Using the difference of squares rule on the top of our fraction we can rewrite this limit as

$$\lim_{x \to 2^{-}} \frac{(x+2)(x-2)}{x-2} = f(2).$$

Once we do this, we can see that the top and bottom of our fraction contain a \((x-2)\) term so we can use them to cancel each other out.  This leaves us with

$$\lim_{x \to 2^{-}} (x+2) = f(2).$$

\(y=x+2\) is a linear function and is continuous everywhere, so we can simply plug in \(x=2\) to find this limit.

$$4=f(2)$$

Now we need to figure out the right side of the equation.  To find f(2), we just need to plug in \(x=2\) into our function f(x), which is our original piecewise function.  The first thing we need to do is decide which piece defines our function when \(x=2\).  We know that \(f(x)=ax^2-bx+3\) when \(2\leq x<3\).  We can use this part to find f(2) because this is the piece that defines f when \(x=2\).  Plugging in 2 for x we get

$$4=ax^2-bx+3$$

$$4=a(2)^2-b(2)+3$$

$$4=4a-2b+3$$

$$1=4a-2b$$

Notice we have only one equation at this point which relates two unknown constants, making it impossible to find one unique solution for a and b.  To do this we will need another equation relating these two constants.

I would also like to point out that this equation came from using the equation involving only the left sided limit, and we still need to look at the right sided limit.

$$\lim_{x \to 2^{+}} f(x) = f(2)$$

If we look at this limit, we see that we would be approaching \(x=2\) from the right side, which would mean we are looking at x values slightly larger than 2.  For x values slightly larger than 2, but infinitely close to 2, we would use the \(y=ax^2-bx+3\) piece to define f.  We already found f(2) above, so putting these two facts together we see

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

Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit.

$$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$

$$4a-2b+3=4a-2b+3$$

Notice both sides of this equation are the same.  Because of this, this equation will actually be true no matter what we put in for a and b.  This doesn’t really help us at all in this case, but it was important to test it out and see what it told us.  So we know that f will be continuous at \(x=2\) as long as

$$1=4a-2b.$$

At this point we have found a set of a and b values that make f continuous at \(x=2\).  But we need to find just one a and one b that will accomplish this for all x values.

But this does not tell us anything about whether it would also be continuous at \(x=3\).  Checking that may give us another relationship between a and b that we can use to find the single unique solution for the two constants that will make f continuous everywhere.

Continuity at \(x=3\)

Making sure that f is continuous at \(x=3\) will be an extremely similar process to what we just did around \(x=2\).  Similar to above, we will need to make sure the following equations are true:

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

$$\lim_{x \to 3^{+}} f(x) = f(3).$$

Let’s start with the right sided limit.  This means that we are getting closer and closer to \(x=3\) and we are coming from x values that are slightly larger than 3.  Therefore, we will need to use the piece of our function that defines f for \(x\geq 3\).

$$\lim_{x \to 3^{+}} 2x-a+b = f(3)$$

Since a and b are both constants, \(y=2x-a+b\) is a linear function, and is continuous everywhere as a result.  Because of this, we can just plug 3 in for x to find this limit.

$$2(3)-a+b = f(3)$$

$$6-a+b = f(3)$$

To find f(3) we just need to plug 3 in for x into the piece of our function that defines it when \(x=3\), which is the third piece of f.

$$6-a+b = 2(3)-a+b$$

$$6-a+b = 6-a+b$$

We see here another case where this equation gives us no useful information.  This is because this equality will be true no matter what we plug in for a and b.  Let’s move onto the left sided limit and see what we get there.

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

Now that our x value is approaching 3 from the left side, it is slightly smaller than 3 and slowly increasing.  Since we are considering what our function is doing around \(x=3\) for x values slightly smaller than 3, we will want to use the piece of f that is defined for \(2\leq x<3\).  We already found f(3) above, so we will also plug that in.

$$\lim_{x \to 3^{-}} ax^2-bx+3 = 6-a+b$$

Just like before, since we know \(y=ax^2-bx+3\) is a polynomial, it is continuous everywhere and we can find this limit by plugging in \(x=3\).

$$a(3)^2-b(3)+3 = 6-a+b$$

$$9a-3b+3 = 6-a+b$$

$$10a-4b=3$$

Putting it all together

Now we have another relationship that relates a and b.  Alone, it isn’t very useful, but we can take the previous equation we found from ensuring f is continuous at \(x=2\) and solve the following system of equations:

$$\mbox{(1):         }4a-2b=1$$

$$\mbox{(2):         }10a-4b=3.$$

Now that we have two equations that relate these two unknown constants, we have enough information to solve for them.  You always need at least as many equations as you have variables (or unknown constants in this case).  There are a couple different ways to solve a system of equations like this, but I will use substitution.  First we will solve for b in equation (1), then plug that into equation (2).

$$4a-2b=1$$

$$-2b=-4a+1$$

$$\mbox{(3):         }b=2a-\frac{1}{2}$$

Now we will plug this into equation (2) and solve for a.

$$10a-4\Big(2a-\frac{1}{2}\Big)=3$$

$$10a-8a+2=3$$

$$2a+2=3$$

$$2a=1$$

$$a=\frac{1}{2}$$

Now we can plug this back into our equation for b, equation (3).

$$b=2\Big(\frac{1}{2}\Big)-\frac{1}{2}$$

$$b=1-\frac{1}{2}$$

$$b=\frac{1}{2}$$

Now we have shown that f(x) will be continuous at \(x=2\) and at \(x=3\) if \(a=\frac{1}{2}\) and \(b=\frac{1}{2}\).  We already showed that f is continuous for \(x<2\), \(2<x<3\), and \(x>3\) for all values of a and b.  Now we also know that f is continuous at \(x=2\) and \(x=3\) if \(a=\frac{1}{2}\) and \(b=\frac{1}{2}\).  Therefore, we can say that f would be continuous everywhere if \(a=\frac{1}{2}\) and \(b=\frac{1}{2}\).

In fact, we can see that f is continuous by plugging in the values we found for a and b and graphing it using Desmos.

find a and b for piecewise function to be continuous

This type of problem is one that you will likely run into a few times.  It actually comes in handy when you are trying to solve a similar type of problem that requires you to make a function differentiable everywhere.  If you would like to see this I have worked through one of these and you can see that by clicking this link: Solution – Find the values of a and b that make the function differentiable everywhere.

SHORTCUT – More examples

https://youtu.be/f2h-f2R13uA

Check out the other topics I’ve covered and the problems I’ve worked through.  You can see a list of related lessons on my limits page.

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