# Thread: Related Rates

1. ## Related Rates

Problem:
A girl flies a kite at a height of 300ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/s. How fast must she let out the string when the kite is 500 ft away from her?

The correct answer is 20ft/s

My procedures

$tan\theta = \frac{s}{300}$
$(sec^2x)(\frac{d\theta}{dt}) = (\frac{1}{300})(\frac{ds}{dt})$
$\frac{d\theta}{dt} = (\frac{1}{12}) ( cos^2\theta)$
$\frac{d\theta}{dt}=\frac{4}{75}$
-----------------------------------
$csc\theta = \frac{r}{300}$
$(-csc\theta cot\theta)(\frac{d\theta}{dt}) =( \frac{1}{300})(\frac{dr}{dt})$
$\frac{dr}{dt} = -\frac{320}{9}$

Can anyone tell me what I did wrong?

2. ## Re: Related Rates

Originally Posted by rofredtan
Problem:
A girl flies a kite at a height of 300ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/s. How fast must she let out the string when the kite is 500 ft away from her?

The correct answer is 20ft/s

My procedures

$tan\theta = \frac{s}{300}$
$(sec^2x)(\frac{d\theta}{dt}) = (\frac{1}{300})(\frac{ds}{dt})$
$\frac{d\theta}{dt} = (\frac{1}{12}) ( cos^2\theta)$
$\frac{d\theta}{dt}=\frac{4}{75}$
-----------------------------------
$csc\theta = \frac{r}{300}$
$(-csc\theta cot\theta)(\frac{d\theta}{dt}) =( \frac{1}{300})(\frac{dr}{dt})$
$\frac{dr}{dt} = -\frac{320}{9}$

Can anyone tell me what I did wrong?
$\frac{dr}{dt} = \frac{dr}{ds} \cdot \frac{ds}{dt}$

where $r = \sqrt{300^2 + s^2}$ ft and $\frac{ds}{dt} = 25$ ft/s.

Find dr/dt when s = 500.

3. ## Re: Related Rates

Still not getting the answer. My teacher teaches us as if we were algebra dummies or something like that, so probably he explained to us in a very complicated way, thinking that we got it?

$\frac{dr}{dt} = ?$
$\frac{ds}{dt} = \frac{25ft}{s}$
$r = \sqrt{300^2+s^2}$
$= \sqrt{340000}$

$tan\theta=\frac{s}{300}$
$(sec^2\theta)(\frac{d\theta}{dt}) = \frac{1}{300} (\frac{ds}{dt})$
$\frac{d\theta}{dt} = \frac{1}{12}(cos^2\theta)$
$\frac{d\theta}{dt} = \frac{25}{408}$

$csc\theta = \frac{r}{300}$
$-csc\theta cot\theta (\frac{d\theta}{dt}) = \frac{1}{300} (\frac{dr}{dt})$
$\frac{-sqrt{340000}}{12} (\frac{5x100}{408} = \frac{dr}{dt}$
$\frac{dr}{dt} = -59.54811...$

$csc\theta = \frac{r}{300}$
$-csc\theta(cot\theta) (\frac{d\theta}{dt}) = \frac{1}{300} (\frac{dr}{dt})$
$\frac{-\sqrt{340000}}{12} (\frac{500}{408}) = \frac{dr}{dt}$
$\frac{dr}{dt} = -59.54811...$

Still not getting the answer

4. ## Re: Related Rates

Originally Posted by rofredtan
Still not getting the answer. My teacher teaches us as if we were algebra dummies or something like that, so probably he explained to us in a very complicated way, thinking that we got it?

$\frac{dr}{dt} = ?$
$\frac{ds}{dt} = \frac{25ft}{s}$
$r = \sqrt{300^2+s^2}$
$= \sqrt{340000}$

$tan\theta=\frac{s}{300}$
$(sec^2\theta)(\frac{d\theta}{dt}) = \frac{1}{300} (\frac{ds}{dt})$
$\frac{d\theta}{dt} = \frac{1}{12}(cos^2\theta)$
$\frac{d\theta}{dt} = \frac{25}{408}$

$csc\theta = \frac{r}{300}$
$-csc\theta cot\theta (\frac{d\theta}{dt}) = \frac{1}{300} (\frac{dr}{dt})$
$\frac{-sqrt{340000}}{12} (\frac{5x100}{408} = \frac{dr}{dt}$
$\frac{dr}{dt} = -59.54811...$

$csc\theta = \frac{r}{300}$
$-csc\theta(cot\theta) (\frac{d\theta}{dt}) = \frac{1}{300} (\frac{dr}{dt})$
$\frac{-\sqrt{340000}}{12} (\frac{500}{408}) = \frac{dr}{dt}$
$\frac{dr}{dt} = -59.54811...$

Still not getting the answer
I have told you exactly how to do it but you have chosen to ignore what I posted.

Furthermore, you said "How fast must she let out the string when the kite is 500 ft away from her?"

when clearly the given answer requires "... the kite is [400] ft away from her?"

Using 400 ft and what I posted, the correct answer falls out in a couple of lines.

5. ## Re: Related Rates

Hello.
Just to post the solution...

You were right mr fantastic, it was indeed using that formula. For a moment I thought all related rates problems could only be solved using trig relationships, that's why I kind of "ignored" your solution. My apologies for that.

The problem can be solved with the values in this picture:

And taking the derivative with respect to t of both sides of the following equation:
$r = \sqrt{300^2 + s^2}$

Can be solved in an easier way by taking the derivative of both sides from the following equation and solving:
$r^2 = 300^2 + s^2$

Even though I didn't use your solution because by that time I DIDN'T KNOW about the usage of the quadratic formula for solving these problems, I thank you for having tried to help. If I did know about it, then I would probably solved the problem using your solution.

,

### content

Click on a term to search for related topics.