Related Rates

[fbellman]

Hi there, I'm a first time poster.

I'm lost on these two problems (the first especially) and would appreciate any help!

1. A rocket is rising vertically at a rate of 5,400 miles per hour. An observer on the ground is standing 20 miles from the rocket's launch point. How fast (in radians per second) is the angle of elevation between the ground and the observer's line of sight of the rocket increasing when the rocket is at an elevation of 40 miles?

2. Let y=2e^cos(x).
A) Calculate dy/dx and (d^2)y/dx^2.
B) If x and y both vary with time in such a way that y increases at a steady rate of 5 units per second, at what rate is x changing when x= pi/2?

For the second problem, I THINK I determined the first and second derivatives:
I got that the first derivative is -2sin(x)*e^cos(x).
And the second derivative is 2sin^2 (x) * e^cos(x) - 2cos(x) * e
Is this all correct?

Again, thanks a lot for any help!

 

[ChaoticLlama]

The first thing that would help is if you have not already done so is draw a diagram.

with the seperation between the observer and the launch site at 20 miles, and the vertical representing the rockets height.

This is what is known:
dy/dt = 5400 mi/h (speed of rocket)
x = 20 mi
y = 40 mi (at the desired instant)
--> Therefore the hypotaneuse equals 20√5 mi at that same desired instant

\(\displaystyle \tan (\Theta ) = \frac{y}{{20}}\)

Now differentiate (implicitly wrt time [t]!)

\(\displaystyle \sec ^2 (\Theta )\frac{{d\Theta }}{{dt}} = \left( {\frac{1}{{20}}} \right)\frac{{dy}}{{dt}}\)

Now the dy/dt is easy enough to deal with.. but the sec²(θ) takes some more manipulation.

First: know that sec(θ) = 1/cos(θ) **1**

Here, as this specific instant, the adjacent side to the angle is 20 mi and the hyp is 20√5

Therefore
cos(θ) = 20 / (20√5)
cos(θ) = 1 / √5 **2**

Combine lines **1** and **2** to find that
sec²(θ) = 5

Now it's really easy from here.

\(\displaystyle \begin{array}{l}
5\frac{{d\Theta }}{{dt}} = \left( {\frac{1}{{20}}} \right)5400 \\
\frac{{d\Theta }}{{dt}} = 54 \\
\end{array}\)

 

[soroban]

Hello, fbellman!

Welcome aboard!

ChaoticLlama beat me to #1 . . . except for the punchline . . . *snicker*

\(\displaystyle \;\;\;54\) radians/hour\(\displaystyle = \:0.015\) radians/sec.

2. Let \(\displaystyle y\:=\:2\cdot e^{\cos(x)}\)

A) Calculate \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{d^2y}{dx^1}\)

B) If \(\displaystyle x\) and \(\displaystyle y\) both vary with time in such a way that
\(\displaystyle y\) increases at a steady rate of 5 units per second,
at what rate is \(\displaystyle x\) changing when \(\displaystyle x\,=\,\frac{\pi}{2}\) ?

I THINK I determined the first and second derivatives:

I got that the first derivative is: \(\displaystyle \,-2\cdot\sin(x)\cdot e^{\cos(x)}\)

And the second derivative is: \(\displaystyle \,2\cdot\sin^2(x)\cdot e^{cos(x){\,-\,2\cdot \cos(x)\cdot e^{\cos(x)}\)

Is this all correct?\(\displaystyle \;\;\) . . . Yes! Nice work!

B) Differentiate with respect to time: \(\displaystyle \L\:\frac{dy}{dt}\;=\;-2\cdot\sin(x)\cdot e^{\cos(x)}\cdot\left(\frac{dx}{dt}\right)\)

We are given \(\displaystyle \frac{dy}{dt}\,=\,5\) and when \(\displaystyle x\,=\,\frac{\pi}{2}\) we have:

\(\displaystyle \L\;\;\;5\;= \;-2\cdot\sin\left(\frac{\pi}{2}\right)\cdot e^^{\cos(\frac{\pi}{2})}\cdot\left(\frac{dx}{dt}\right)\;=\;-2(1)(e^0)\left(\frac{dx}{dt}\right)\)

Therefore: \(\displaystyle \L\;\frac{dx}{dt}\:=\:-2.5\) units/sec

 

[ChaoticLlama]

*smack*

*buries Soroban*

No one ever has to know... :twisted:

 

[fbellman]

Hi, ChaoticLlama and soroban!!

Thank you both very much for the help and the warm welcomes! The work above makes perfect sense and helped a lot, thank you so much!