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

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

• Please find a title that is informative about the content of the post and does not mention your state of mind or your teacher.
– Pedro
May 8, 2017 at 5:56
• Use a bit of estimation to figure out where your answer should be. The answer should be around $\sin^{-1}(100/198) - \sin^{-1}(100/202) \approx 0.012$, which immediately rules out the second and third answers. The first answer is still incorrect, but at least plausible. May 8, 2017 at 6:00
• Thanks for the comments, guys! Calculus is not turning out to be my strong suit. May 8, 2017 at 6:01

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

• Here $\frac{dl}{dt}=4 ft.sec^{-1}$ and you can calculate $\cos\theta$ by making a triangle May 8, 2017 at 5:54
• Okay, I will look into this one more time in the morning: I really appreciate the help, thank you immensely! May 8, 2017 at 5:59
• I got the same answer with you :D May 8, 2017 at 6:02
• OOOOOH. I see where I went wrong. I was using the ANGLE of cos(theta), √3/2, not the value, 100√3/200. BUT, after plugging in all of the values into step 3 into a calculator, I get back -√3/150 rad/sec. I've triple checked it, and I can't get anything else back. What went wrong?? May 8, 2017 at 14:16
• Remember that $1/\sqrt{3} = \sqrt{3}/3$ ... May 8, 2017 at 14:59

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 $$\frac{d\theta}{dt} = -4\cdot \frac{1}{3} \cdot \frac{1}{300\cdot \frac{2\sqrt{2}}{3}} = -\frac{\sqrt{2}}{300}.$$

• You made a typo in the end.( Please see) May 8, 2017 at 6:03
• please note that the kite is 100 feet above ground, and then it moves horizontally. so $l$ should stay at 100, not increase to 300. May 8, 2017 at 6:04
• @user215939 the question is not super clear, but I assume the kite starts directly overhead, so that there is initially 100 feet of string between the person and the kite. The kite then moves horizontally until the line connecting the kite to the person is 200 feet longer. May 8, 2017 at 6:06
• @user7530 the question did not state the kite started directly overhead. the last sentence of the problem states "when 200 ft of string has been let out." so at the end, $l$ should be 200 May 8, 2017 at 6:11
• The wording could definitely be improved. Does "when 200 ft of string has been let out" mean a hypotenuse of 200 or the initial 100 plus the let out 200? Your answer is a totally reasonable and the first one I got. May 8, 2017 at 6:13

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.

• AAAAAND I am an exhausted idiot. My answer turned out to be the exact same thing as what The Dead Legend got, but in a different form. My apologies, everyone. This question is just a pain, and this class is just a pain. May 8, 2017 at 15:08