Related rates problem

**Exotique** · April 21st 2010, 05:19 PM

I just want to check if my reasoning and answer are correct. The question is:
A rocket is launched straight up, and it's altitude is given by $h=10t^2$ after $t$ seconds. You are on the ground 300m from the launch site watching the rocket going up. The line of sight from you to the rocket makes and angle $\theta$ with the horizontal. By how many radians per second is $\theta$ changing 10 seconds after the launch?

My reasoning is as follows:
Need to find $\frac{d\theta}{dt}$

$h=10t^2$ , so $\frac{dh}{dt}=20t$ .

$\tan\theta=\frac{h}{300}$ , so $\frac{d\theta}{dh}=\frac{300}{h^2+90000}$ .

Remembering $t=10$ , $\frac{dh}{dt}=20t=200$

Also, since $h=10t^2$ , $h=1000$ when $t=10$ .

Therefore, $\frac{d\theta}{dh}=\frac{300}{h^2+90000}=\frac{300 }{91000}$

So using Chain Rule:
$\frac{d\theta}{dt}=\frac{d\theta}{dh}\cdot\frac{dh }{dt}$

$=\frac{300}{91000}\cdot200=\frac{60}{91}$

Any help will be much appreciated.

**CaptainBlack** · April 21st 2010, 11:02 PM

Originally Posted by Exotique

I just want to check if my reasoning and answer are correct. The question is:
A rocket is launched straight up, and it's altitude is given by $h=10t^2$ after $t$ seconds. You are on the ground 300m from the launch site watching the rocket going up. The line of sight from you to the rocket makes and angle $\theta$ with the horizontal. By how many radians per second is $\theta$ changing 10 seconds after the launch?

My reasoning is as follows:
Need to find $\frac{d\theta}{dt}$

$h=10t^2$ , so $\frac{dh}{dt}=20t$ .

$\tan\theta=\frac{h}{300}$ , so $\frac{d\theta}{dh}=\frac{300}{h^2+90000}$ .

Remembering $t=10$ , $\frac{dh}{dt}=20t=200$

Also, since $h=10t^2$ , $h=1000$ when $t=10$ .

Therefore, $\frac{d\theta}{dh}=\frac{300}{h^2+90000}=\frac{300 }{91000}$

So using Chain Rule:
$\frac{d\theta}{dt}=\frac{d\theta}{dh}\cdot\frac{dh }{dt}$

$=\frac{300}{91000}\cdot200=\frac{60}{91}$

Any help will be much appreciated.

$\tan(\theta)=\frac{h}{300}$

Now differentiate with respect to $t$ :

$<br /> \sec^2(\theta) \frac{d\theta}{dt}=\frac{1}{300}\frac{dh}{dt}<br />$

Now at $t=10$ , $h=1000$ , $\tan(\theta)=10/3$ so $\cos(\theta)=3/\sqrt{109}$ and $\frac{dh}{dt}=200$

CB

**eddie2042** · April 21st 2010, 11:20 PM

Originally Posted by Exotique

Any help will be much appreciated.

Not what I got. It's basically a right triange with the the adjacent side to $\theta$ being 300m and the opposite side is the height, $h = 10t^2$ .

To relate all of them, you use TOA.

$tan {\theta} = \frac{opposite}{adjacent}$

$tan {\theta} = \frac{10t^2}{300}$

$tan {\theta} = \frac{t^2}{30}$ << simplifying it a bit..

now we take the derivative of both sides with respect to time. Since $\theta$ is a variable, we have to use the chain rule.

$\frac{d}{dt}\left[tan {\theta}\right] =\frac{d}{dt}\left[\frac{t^2}{30}\right]$

$sec^2{\theta} \frac{d\theta }{dt} = \frac{t}{15}$

$\frac{d\theta }{dt} = \frac{t}{15sec^2{\theta}}$

now we plug in our numbers; make sure to use radians on your calculator; not degrees.

At $t = 10$ , $h = 1000 m$ and using some trigonometry we find that $\theta \approx 1.279339532$ . Use the memory function on your calculator!

$\frac{d\theta }{dt} = \frac{10}{15sec^2(1.279339532)}$

$\frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s$

**Exotique** · April 22nd 2010, 12:48 AM

Originally Posted by eddie2042

Not what I got. It's basically a right triange with the the adjacent side to $\theta$ being 300m and the opposite side is the height, $h = 10t^2$ .

To relate all of them, you use TOA.

$tan {\theta} = \frac{opposite}{adjacent}$

$tan {\theta} = \frac{10t^2}{300}$

$tan {\theta} = \frac{t^2}{30}$ << simplifying it a bit..

now we take the derivative of both sides with respect to time. Since $\theta$ is a variable, we have to use the chain rule.

$\frac{d}{dt}\left[tan {\theta}\right] =\frac{d}{dt}\left[\frac{t^2}{30}\right]$

$sec^2{\theta} \frac{d\theta }{dt} = \frac{t}{15}$

$\frac{d\theta }{dt} = \frac{t}{15sec^2{\theta}}$

now we plug in our numbers; make sure to use radians on your calculator; not degrees.

At $t = 10$ , $h = 1000 m$ and using some trigonometry we find that $\theta \approx 1.279339532$ . Use the memory function on your calculator!

$\frac{d\theta }{dt} = \frac{10}{15sec^2(1.279339532)}$

$\frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s$

Can you please explain what might be wrong with my reasoning?

**CaptainBlack** · April 22nd 2010, 02:03 AM

Originally Posted by Exotique

Can you please explain what might be wrong with my reasoning?

if $h=1000$ then $h^2=1000000$

CB

**Exotique** · April 22nd 2010, 03:08 AM

Seems I failed to spot that. Peer review works wonders, thank you! But if I had squared $h$ after substituting, we would have arrived at the same answer, so my logic is correct, I assume.

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