Please help me find the mistake in my related rates kite problem

Question

I have been having utter fits with this assignment. It was due months ago, but because I submitted it on time, I can resubmit until the end of the semester for full points.

A kite 100 ft above the ground moves horizontally. The kite-holder stands still while letting out string at a rate of 4 ft/sec. The angle made by the string and the ground gradually decreases. At what rate is this angle changing when 200 ft of string has been let out?"

I calculated this myself, several times using several different approaches, and I always come back to the same answer: $$\frac {d\theta}{dt} = -\frac 1 {100} \frac {rad} {sec}$$

This really seems correct! In fact, in class, we did a problem (seemingly) just like this one, but let the rate = 8 ft/sec. And in class, we determined that the rate was $$\frac {d\theta}{dt} = -\frac 1 {50} \frac {rad} {sec}$$

So, concerning the one at hand, my study buddy got a grade back on it, and the teacher apparently said that he did it incorrectly, BUT got the right answer somehow. The correct answer is, supposedly,

$$\frac {d\theta}{dt} = \frac {\sqrt 3}{50} \frac {rad} {sec}$$

?!?!?!?!?!

I even asked this question on Chegg, for Pete's sake, and I got the exact same answer as my own! Where does this newfangled answer even come from??

Is there something blantantly obvious I'm missing here? I mean... maybe?? But what?

Please confirm or discredit my insanity. I don't know what to believe because I don't know how to believe anymore.

Cheers,

-Jon

Answer 1

$$\sin\theta=\frac{100}{l}$$ $$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$ $$\frac{\sqrt{3}\times 100 }{200}.\frac{d\theta}{dt}=-\frac{1}{100}$$ $$\frac{d\theta}{dt}=-\frac{1}{50\sqrt{3}}\text{rad per sec}$$

Answer 2

Funny, I get yet a third answer:

Let $l$ be the distance from the person to the kite. Initially $l=100$ and increases at $4$ feet per second.

The relationship between the angle $\theta$ and $l$ is $$\sin \theta = \frac{100}{l}.$$

Therefore $$\frac{dl}{dt}\sin \theta + l\cos\theta \frac{d\theta}{dt} = 0.$$

Now after $200$ feet of string has been released, $l=300$ and $\sin\theta=\frac{1}{3}$. We can compute $\cos\theta$ by solving the triangle with leg $100$ and hypotenuse $300$: $\cos\theta = \frac{\sqrt{300^2-100^2}}{300} = \frac{2\sqrt{2}}{3}.$ Therefore $$\frac{d\theta}{dt} = -4\cdot \frac{1}{3} \cdot \frac{1}{300\cdot \frac{2\sqrt{2}}{3}} = -\frac{\sqrt{2}}{300}.$$

Answer 3

Given The Dead Legend's calculations,

$$\sin\theta=\frac{100}{l}$$ $$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$ $$\frac{\sqrt{3}\times 100 }{200}.\frac{d\theta}{dt}=-\frac{1}{100}$$ $$\frac{d\theta}{dt}=-\frac{1}{50\sqrt{3}}\text{rad per sec}$$

I finally saw what I did wrong: I was taking

$$cos\theta = \frac {\sqrt 3}{2} $$

and not

$$cos\theta = \frac {100\sqrt 3}{200} $$

Which makes sense now.

However, I still don't think the answer is right.

dl/dt = 4 ft/sec, l = 200. I followed The Dead Legend's calculations and got back

$$\frac{d\theta}{dt} = - \frac{\sqrt 3}{150}$$

I plugged it into a calculator and got back the same results.

---

So how did my study buddy get the right answer? The only possible answer is that he didn't. My teacher must have thought he did.

When I reviewed my buddy's notes, he followed the same approach as The Dead Legend.

1. He derived $$\sin\theta=\frac{100}{l}$$
2. He got back $$\cos\theta \frac{d\theta}{dt}=-\frac{100}{l^2}.\frac{dl}{dt}$$

And everything went swimmingly (except for a dropped minus sign), where he got the equation down to

$$\frac {400}{40000} \cdot \frac {200}{100 \sqrt 3}$$

Which could be simplified to

$$\frac {1}{100} \cdot \frac {200}{100 \sqrt 3}$$

Then he simplified 200/100√3.

Here is where he went wrong.

$$\frac {200}{100 \sqrt 3}$$ $$= \frac {2}{\sqrt 3}$$

$$= 2 \sqrt 3$$

Oops.

When I rationalized the denominator, I got back $$= \frac {2 \sqrt 3}{3}$$

And from THERE, I finished his calculation (despite the curiously absent minus sign.)

$$\frac {1}{100} \cdot \frac {2 \sqrt 3}{3}$$ $$= \frac {2 \sqrt 3}{300}$$ $$= \frac {\sqrt 3}{150}$$

Tack on the dropped minus...

$$= - \frac {\sqrt 3}{150}$$

There it is, folks. The elusive answer, if indeed everything was done correctly.

One way or another, my brain hurts. So glad I'm getting a different teacher for Calc II.