How do I set up this related rates problem?

I seem to have problems with these sort of questions, can someone shed some light on this for me?

You are holding a reel with a line, attached to a balloon, spooling from it. The balloon was released from a spot on level ground 20ft from you and is rising straight up. How fast is the balloon rising when the reel indicateds that 25 feet of line is out and that more is spooling from it at 3 feet per second? Also, how high is the balloon at that same point?

What do I need to do to set up the equation here?

• Seems similar to this question you asked: math.stackexchange.com/questions/251723/… – Eric Angle Dec 12 '12 at 20:21
• @EricAngle It is very similar, I just struggle with how to work these related rates problems. I missed the week that we talked about them unfortunately. I can draw the picture but it still confuses me. – StrugglingWithMath Dec 12 '12 at 20:25

It helps a lot in problems like this to draw a picture - that way you learn about how to connect what you are given, with what you need to find. E.g. In this problem if you draw a sketch:

You'll see you have a right triangle where you are at one vertex, the balloon at another, and the the point on the ground directly below the ascending balloon being the third vertex. The hypotenuse has length $y = 25$, and has you and the balloon as endpoints.

Let $x$ be the length of the side corresponding to the balloon's ascent (a leg of the right triangle).

The remaining side (leg of right triangle) has length $20$.

HINT: Since you have a right triangle, you can use the Pythagorean Theorem to relate these sides to obtain $x$ when $y = 25$, and to get an equation relating $x$ with $y$.

You are asked to find $\dfrac{dx}{dt}$ when $y=25$; and you were given the information that $\dfrac{dy}{dt}=3$ when $y = 25$.

HINT $2$

$\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt},\;$ and you are given $\dfrac{dy}{dt}=3\;$ when $y=25.\;$

Find $\dfrac{dy}{dx}$ and evaluate at $y = 25$ after you obtain the equation relating $x$ and $y$ (using the first hint); then you can solve for $\dfrac{dx}{dt}\;$ when $y = 25$.

• Thanks for commenting, I'll need to look over implicit differentiation before coming back and giving this a try. I'll be back though (likely later this week -- not in time for the test though). Luckily I believe there will only be one or two implicit problems on the test. Expect a comment later this week when I come back. (I have to shift focus to one last topic before I take the test) – StrugglingWithMath Dec 12 '12 at 21:08
• Let me know if this makes sense now, when you've had a chance to review it. – Namaste Mar 21 '13 at 15:15

Draw a picture: you should have a right triangle with yourself as a vertex, the balloon as another, and the third being the point on the ground from which the balloon ascended. The hypotenuse, call its length $y$, has yourself and the balloon as endpoints. Call the length of the side corresponding to the balloon's ascent $x$. The remaining side has length $20$. Now find an equation that relates the sides of the triangles.

Note that you are asked to find ${dx\over dt}\Bigl|_{y=25}$; and you were given the information that ${dy\over dt}\Bigl|_{y=25} =3$.

This should get you going ...

And remember: draw a picture!

Edit:

It's important to draw the picture that represents what's going on at an arbitrary point in time, not at the specific point in time when $y=25$. In the above, $y$ and $x$ are functions of $t$. Once you have the equation relating $x$ and $y$, you then differentiate implicitly with respect to time, which gives you an equation involving the rates of change of $y$ and $x$ (which is what you're after). Only then do you consider what is going on when $y=25$.

• Is the balloon rising at a constant rate, or is it accelerating upward? – Anderson Green Dec 12 '12 at 20:12
• @StrugglingWithMath Draw the picture for the general case, not at at the moment when $y=25$. With the set up I suggested above, the ballon's speed is $dx\over dt$. – David Mitra Dec 12 '12 at 20:33
• @StrugglingWithMath So you have two variables: $y$ is the distance from you to the balloon, and $x$ is the distance from the balloon to the ground. – David Mitra Dec 12 '12 at 20:34
• @StrugglingWithMath No, use the Pythagorean Theorem. – David Mitra Dec 12 '12 at 20:41
• My issue here is I don't understand implicit differentiation well enough, I need to go back and go over it before I can go further with this question apparently. As far as the equation, it should be more along the lines of $x^2 + y^2 = c^2$ I suppose – StrugglingWithMath Dec 12 '12 at 20:58

I realize this question (and answers) are WAY OLD, but implicit differentiation is a subject that we non-mathematicians have a bit of trouble with. I thought I might offer a layman's procedure to handle implicit differentiation.

First, you know how to differentiate $y = x^2$ with respect to $x$. $$\frac {dy}{dx} = 2x$$

Here, while differentiating $y = x^2$, you can imagine that you applied the chain rule. You encountered two functions, $y$ and $x$. On the left side of the equation, upon encountering $y$, you first differentiated it with respect to itself. The derivative of $y^1 = 1 * y^0 = 1*1 = 1$. Then, because we are differentiating with respect to x, you then multiplied by the derivative with respect to x of the inner function $y$ which is $dy \over dx$. That gave you

$$1 * \frac {dy}{dx} =$$

or, simply, $dy \over dx$ on the left side of the equation. On the right side of the equation, upon encountering $x^2$, you first differentiated it with respect to itself, getting $2x$, and then, because we are differentiating with respect to x, multiplied by the derivative with respect to x of the "inner function," which is $dx \over dx$. Thus, on the right side, you obtained

$$= 2x * \frac {dx}{dx}$$ $$= 2x * 1$$ $$= 2x$$

Putting it together you obtained $\frac {dy}{dx} = 2x$.

The process for implicit differentiation is exactly the same.

So, differentiate the equation for a circle of radius 5: $x^2 + y^2 = 25$.

Assuming we are differentiating the equation with respect to x, when you encounter $x^2$ you first take the derivative of $x^2$ with respect to $x$ (which is $2x$) and then multiply by the derivative with respect to $x$ of the inner function which is $dx \over dx$ (which equals 1). So, the derivative of the $x^2$ term is

$$2x * \frac {dx}{dx}$$ $$2x * 1$$ $$2x$$

When you encounter $y^2$ you first take the derivative of $y^2$ with respect to $y$ (which is $2y$) and then multiply by the derivative of the inner function which is $dy \over dx$. (Remember, we are differentiating the entire equation with respect to $x$.) So, the derivative of the $y^2$ term is

$$2y * \frac {dy}{dx}$$

Of course, the derivative of the constant $25$ is $0$. Putting it all together, the derivative with respect to $x$ of $x^2 + y^2 = 25$ is

$$2x + 2y \frac {dy}{dx} = 0$$

Try differentiating with respect to $y$ and see what you get.

Suppose, we are not differentiating with respect to $x$. Suppose instead, we are differentiating with respect to $g$, or $V$, or $anything$ - even something that doesn't show up in the equation? We follow exactly the same procedure. Let's differentiate $y = x^2$ with respect to $t$:

$$y = x^2$$ $$1 * \frac {dy}{dt} = 2x \frac {dx}{dt}$$ $$\frac {dy}{dt} = 2x \frac {dx}{dt}$$

In words, take the derivative of $y^1$ with respect to itself (which is 1) and multiply that by the derivative of $y$ with respect to $t$ (which is $dy \over dt$). Set that equal to the derivative of $x^2$ with respect to itself (which is $2x$) and multiply that the derivative of $x$ with respect to $t$.

Similarly, let's differentiate $x^2 + y^2 = 25$ with respect to $p$:

$$2x \frac {dx}{dp} + 2y \frac {dy}{dp} = 0$$

Finally, let's differentiate the area of a circle with respect to time:

$$A = \pi r^2$$

$$1 * \frac {dA}{dt} = \pi * 2r * \frac {dr}{dt}$$ $$\frac {dA}{dt} = 2 \pi r \frac {dr}{dt}$$

The "layman's rule" is first differentiate with respect to the function itself ($y$ or $V$, or whatever symbol has been assigned) and then differentiate with respect to the inner function, namely the "with respect to" function.

Do not forget the other rules of differentiation ...

As a final observation, remember the other rules of differentiation such as the product and quotient rules, etc. Suppose, for example, that you are differentiating the volume of a right circular cone with respect to time. The formula is $V = \frac {\pi r^2 h}{3}$. In this case volume varies depending upon both $r$ and $h$, both of which may be changing. Because $r$ and $h$ are multiplied by each other, you will need to use the product rule when differentiating, like so:

$$V = \frac {\pi}{3}r^2h$$ $$\frac {dV}{dt} = \frac {\pi}{3}(2r \frac {dr}{dt}h + r^2 \frac {dh}{dt})$$

The procedure is the same. The only difference here is that we have applied the product rule ($r^{'}h + rh^{'}$) when differentiating.