Related Rates Problem

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onhcetum

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A missile is fired vertically from a point that is 5 miles from a tracking station and at the same elevation as the tracking station. For the first 20 seconds of flight, the missile's angle of elevation changes at a constant rate of 2 degrees per second. Find the velocity of the missile when the angle of elevation is 30 degrees.

I'm totally lost. I have no clue where to begin. I know how to differentiable so that's not the problem. The problem is that I don't know what the original function is and what this actually looks like drawn. Perhaps, someone can interpret what this actually looks like?

Thank you, it is much appreciated.
 
Hello, onhcetum!

Did you make a sketch?

A missile is fired vertically from a point that is 5 miles from a tracking station
and at the same elevation as the tracking station.
For the first 20 seconds of flight, the missile's angle of elevation changes
at a constant rate of 2 degrees per second.
Find the velocity of the missile when the angle of elevation is 30 degrees.
Click to expand...
Code: 
    M *
      |   *
      |       *
    h |           *
      |               *
      |                @  *
    P *-----------------------* S
      : - - - - - 5 - - - - - :

The missile is fired from \(\displaystyle P\)
and is located at \(\displaystyle M\!:\;h = MP.\)

The tracking station is at \(\displaystyle S\!:\;PS = 5\) miles.

Let \(\displaystyle \theta \,=\,\angle MSP\)
\(\displaystyle \text{We are told that: }\:\frac{d\theta}{dt} \:=\:2\text{ deg/sec} \:=\:\frac{\pi}{90}\text{ rad/sec}\)

\(\displaystyle \text{We want }\frac{dh}{dt}\text{ when }\theta \,=\, 30^o \,=\,\frac{\pi}{6}\text{ radians.}\)


\(\displaystyle \text{In the right triangle: }\:\tan\theta \:=\:\frac{h}{5} \quad\Rightarrow\quad h \:=\:5\tan\theta\)

\(\displaystyle \text{Differentiate with respect to time: }\:\frac{dh}{dt} \:=\:5\sec^2\!\theta\,\frac{d\theta}{dt} \;\;\;\bf{[1]}\)

\(\displaystyle \text{We have: }\:\sec\tfrac{\pi}{6} \,=\,\frac{2}{\sqrt{3}} \quad\Rightarrow\quad \sec^2\!\tfrac{\pi}{6} \,=\,\frac{4}{3}\)
. . \(\displaystyle \text{and: }\:\frac{d\theta}{dt} \,=\,\frac{\pi}{90}\)

\(\displaystyle \text{Substitute into \bf{[1]}: }\;\frac{dh}{dt} \:=\:5\left(\frac{4}{3}\right)\left(\frac{\pi}{90}\right) \;=\;\frac{2\pi}{27}\)

\(\displaystyle \text{The missile is rising at }\frac{2\pi}{27}\text{ miles/sec} \;\approx\;1229\text{ ft/sec}\)

 
Hey, thank you. I understood everything you did.

I followed your steps, but I did everything in degree.

(5) x sec(30)^2 x 2 = 40/3 = 13.333 miles/sec.

Is this acceptable?

Thank you.
 
Can you tell me why I would need to do that considering how all my numbers are given in degrees?
 
13.33 miles per second translates to almost 48,000 mph. That is fast, even for a rocket.

It is a good idea to use radians anytime you are dealing with calculus.

Degrees are arbitrary units of measure and should not be used in mathematics, in general. Radians are absolute. Degrees is OK for surveying, but in calculus, use radians.
 
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