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I'm trying to understand how to deal with related rates and here is my problem.

When a rocket is 2 km high it is moving vertically at 300 km/hr. At this time, how fast is the angle of elevation increasing as seen by a person on the ground 5km from the launchpad?

la picture

So we have

$$\frac{dy}{dt}=300 km/hr$$ $$y=2 km$$ $$x=5km$$

And we have to find $$\frac{d\Theta}{dt}=?$$

My solution is next:

$$\sin(\Theta)=\frac{y}{\sqrt{29}}$$ where $\sqrt{29}$ is the hypotenuse

Then i take derivative of this thing $$\cos(\Theta)*\frac{d\Theta}{dt} = \frac{\frac{dy}{dt}}{\sqrt{29}}$$ We know, that $\cos(\Theta)$ is $\frac{5}{\sqrt{29}}$ and $\frac{dy}{dt}$ is $300$ so that we have $$\frac{5}{\sqrt{29}}*\frac{d\Theta}{dt}=\frac{300}{\sqrt{29}}$$ so that $$\frac{d\Theta}{dt}=\frac{300}{\sqrt{29}}*\frac{\sqrt{29}}{5}$$ $$\frac{d\Theta}{dt}=60$$

But it doesn't look like the right answer. Could you explain what did I do wrong, please?

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    As a quick Mathjax hint, use `\sqrt{29}` instead of `\sqrt(29)`. – 2017-01-02
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    in my version it is, but i have the true answer and it's another, but my version seems legit for me, that's the problem – 2017-01-02
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    Another Question: How do we know if the answer is in radians or degrees? – 2017-01-02
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    i don't know to be honest – 2017-01-02
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    the answer is supposed to be $\frac{1500}{29}$ – 2017-01-02
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    The length of the hyp is not constant. – 2017-01-02
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    this makes sense – 2017-01-02
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    @DavidQuinn why is it only valid for radians? – 2017-01-02

2 Answers 2

1

A modification on your solution that works:

$\sin(\theta)=\frac{y}{\sqrt{5^2+y^2}} \\\ $

Differentiating both sides gives

$ \theta' cos(\theta)=\frac{y' \sqrt{5^2+y^2}-y \cdot \frac{1}{2} \cdot \frac{2y y'}{\sqrt{5^2+y^2}}}{(\sqrt{5^2+y^2})^2} \\\ $

So now you can replace the expression that represents the hyp with $\sqrt{29}$

So you can write

$\theta' \cos(\theta)=\frac{y' \sqrt{29}-y \cdot \frac{1}{2} \frac{2 y y'}{\sqrt{29}}}{(\sqrt{29})^2}$

We can also make all other substitutions:

$\theta' \frac{5}{\sqrt{29}}=\frac{300 \sqrt{29}-2 \cdot \frac{1}{2} \frac{2 (2) (300)}{\sqrt{29}}}{(\sqrt{29})^2}$

Solving for $\theta'$ gives:

$\theta'=\frac{\sqrt{29}}{5} \cdot \frac{300 \sqrt{29}-\frac{1200}{\sqrt{29}}}{29} $

Distribute the $\sqrt{29}$ on top there and the $5$ there on bottom:

$\theta'=\frac{300(29)-1200}{5(29)}$

So now the simplified answer is:

$\theta'=\frac{1500}{29} \frac{rad}{hr}$

And yes I know this is not the easiest way... I just wanted to show what the op's way would look like if it was done correctly.

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    I hate spelling out the word that I abbreviate as hyp because I forget how to spell it :p. And thanks. – 2017-01-02
2

We have

$$\tan(\theta)=\frac{y}{5}$$ and by differentiation

$$(1+\tan^2(\theta))\frac{d \theta}{dt}=\frac{1}{5}\frac{dy}{dt}$$

thus

$$\frac{d\theta}{dt}=\frac{300}{5(1+(\frac{2}{5})^2)}$$

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    i know the answer but i don't know why my version is wrong ;( – 2017-01-02
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    It is because you assumed the hyp to be constant in your solution. The height of the triangle is increasing so the point from observation to the rocket is increasing. Does this make any sense @user2686299 ? – 2017-01-02
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    this makes sense #2 – 2017-01-02
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    thank you Salah Fatima and @randomgril, now I feel myself much better – 2017-01-02