Related Rates: Moving Kite

Question

A kite 75 feet above the ground moves horizontally at a speed 4 ft/s. At what rate is the angle $\theta$, which is between the string and the horizontal ray, decreasing when 250 feet of string has been let out?

I discovered the answer is $\cfrac{d\theta}{dt}=-\cfrac{3}{625}$ radians per second. Initially I tried solving it using $\cos\theta=\cfrac{4t}{250}$, differentiating both sides to solve for $\cfrac{d\theta}{dt}$, but I got the wrong answer. It turns out I needed to implicitly differentiate $\cot\theta=\cfrac{4t}{75}$. Can anyone explain why this particular trig function is necessary? I understand generally it's crucial to pick the right trig function because you're usually solving for a value of a particular side of a right triangle, but in this case I figured as long as I pull out $\cfrac{d\theta}{dt}$ and isolate it then I should get the right answer regardless of the function I use. But that wasn't the case.

Any answers?

Answer 1

The problem is not with which trig function you picked since it should always work properly, i.e., you get the same answer, regardless of which function you use. Instead, the problem is that the $250$ is not a constant. Instead, consider the diagram above, where $A$ is the person's position and $C$ is where the kite is located, so $\left|AC\right|$ is the length of the string currently let out. Using the Pythagorean theorem gives that in general

$$\begin{equation}\begin{aligned} \left|AC\right| & = \sqrt{(4t)^2 + 75^2} \\ & = \sqrt{16t^2 + 5625} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This means $\left|AC\right| = 250$ is the length of the string that is let out only at a certain time, in particular at $4t = \sqrt{250^2 - 75^2} \implies t \approx 59.62$ seconds.

Using that $\cos \theta$ is the adjacent side length divided by hypotenuse gives, from \eqref{eq1A}, that the correct expression using $\cos$ would instead be

$$\cos \theta = \frac{4t}{\sqrt{16t^2 + 5625}} \tag{2}\label{eq2A}$$

I'll leave it to you to differentiate and solve this yourself to confirm that you then get the same result as you got using $\cot \theta = \frac{4t}{75}$.

Answer 2

It isn't necessary to work with the cotangent function in particular. Its use is suggested in the given solution only as a matter of convenience. We have a rate at which the horizontal position, rendered as a distance from the point directly over the kite-flyer, is increasing, $ \ \frac{dH}{dt} \ = \ +4 \ $ ft./sec. , so we will want to work with a trigonometric function of $ \ \theta \ $ that involves $ \ H \ $ in some way. While $ \ \cos \theta \ = \ \frac{H}{L} \ \ , $ is a reasonable choice, one that uses the constant altitude will be less troublesome, such as $ \ \tan \theta \ = \ \frac{75}{H} \ \ . $

Upon rearranging this equation to eliminate the ratio, we can then differentiate implicitly to obtain $$ H · \tan \theta \ \ = \ \ 75 \ \ \Rightarrow \ \ \frac{d}{dt} \ [ \ H · \tan \theta \ ] \ \ = \ \ \frac{d}{dt} \ [ \ 75 \ ] \ \ \Rightarrow \ \ \frac{dH}{dt} · \tan \theta \ + \ H · \sec^2 \theta · \frac{d \theta}{dt} \ \ = \ \ 0 $$

$$ \Rightarrow \ \ \frac{dH}{dt} · \frac{75}{H} \ \ = \ \ -H · \left(\frac{L}{H} \right)^2 · \frac{d \theta}{dt} $$ $$ \Rightarrow \ \ \frac{d \theta}{dt} \ \ = \ \ \frac{dH}{dt} · \frac{75}{H} · \left(-\frac{1}{H} \right) · \left(\frac{H}{L} \right)^2 \ \ = \ \ - \frac{dH}{dt} · \frac{75}{L^2} \ \ . $$

It works out that we don't need to know a value for $ \ H \ $ at all. The rate at which the horizontal position is changing is $ \ \frac{dH}{dt} \ = \ +4 \ $ ft./sec. at the time when $ \ L \ = \ 250 \ $ feet, so we find that $$ \frac{d \theta}{dt} \ \ = \ \ - (+4 \ \text{ft./sec.}) · \frac{75 \ \text{ft.}}{250^2 \ \text{ft.}^2 } \ \ = \ \ -\frac{300}{250·250} \ \frac{\text{(rad.)}}{\text{sec.}} \ \ = \ \ -\frac{3}{625} \ \text{rad./sec.} \ \ . $$

So we don't need to know a value for time $ \ t \ $ either.

The "problem" with using the cosine function here is that neither of the quantities in the ratio is constant (as John Omielan notes), so everything needs to be differentiated:

$$ L · \cos \theta \ \ = \ \ H \ \ \Rightarrow \ \ \frac{d}{dt} \ [ \ L · \cos \theta \ ] \ \ = \ \ \frac{d}{dt} \ [ \ H \ ] $$ $$ \Rightarrow \ \ \frac{dL}{dt} · \cos \theta \ + \ L · (-\sin \theta) · \frac{d \theta}{dt} \ \ = \ \ \frac{dH}{dt} \quad \quad \mathbf{[ \ 1 \ ]} $$

We need to use the "Pythagorean" relation $ \ H^2 \ = \ L^2 - 75^2 \ \ $ to "clear away" a certain factor, so that we can again avoid dealing with the value of $ \ t \ $ . Differentiating this equation implicitly produces $ \ 2·H·\frac{dH}{dt} \ = \ 2·L·\frac{dL}{dt} - 0 \ \Rightarrow \ \frac{dL}{dt} \ = \ \frac{H}{L}·\frac{dH}{dt} \ \ . $ Equation $ \ \mathbf{1} \ $ can then be written as $$ \left[ \ \frac{H}{L}·\frac{dH}{dt} \ \right] · \frac{H}{L} \ + \ L · \left(-\frac{75}{L} \right) · \frac{d \theta}{dt} \ \ = \ \ \frac{dH}{dt} $$

$$ \Rightarrow \ \ \frac{dH}{dt} · \left( \ 1 \ - \ \frac{H^2}{L^2} \ \right) \ \ = \ \ \frac{dH}{dt} · \left( \frac{L^2 \ - \ H^2}{L^2} \ \right) \ \ = \ \ \frac{dH}{dt} · \left( \frac{75^2}{L^2} \ \right) \ \ = \ \ -75 · \frac{d \theta}{dt} $$ $$ \Rightarrow \ \ \frac{d \theta}{dt} \ \ = \ \ \frac{dH}{dt} · \left( -\frac{75}{L^2} \ \right) \ \ = \ \ - (+4) · \frac{75}{250^2} \ \ , $$

yielding the same result as before after inserting the given simultaneous values of $ \ \frac{dH}{dt} \ $ and $ \ L \ \ . $