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A television camera is positioned $ 4000 ft $ from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the speed is $ 600 ft/s $ when it has risen $ 3000 ft. $(a) How fast is the distance from the television camera to the rocket changing at that moment?(b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

a) 360 $\mathrm{ft} / \mathrm{s}$b) 0.096 $\mathrm{rad} / \mathrm{s}$

03:57

Wen Z.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

03:34

A television camera is pos…

0:00

A television camera at gro…

02:39

An observer on the ground …

As shown in the accompanyi…

02:40

For the camera and rocket …

01:36

Angular Rate of Change A t…

05:39

A video camera is focused …

04:07

Height of a Rocket A rocke…

02:56

09:14

A rocket travels verticall…

So here's what we have is a camera and we have its distance to a rocket and we could try this. And this is a height right here that we have off the ground h and we'll shorten that up. Then we know that this distance right here is 4000 ft. That's 4000 ft from the launch pad. And the launch pad is right here. This right here is gonna be our X distance. So we know some initial conditions of 3000 ft being R H value this height. And we know that the change in height with respect to time his 600 ft per second and we want to know the rate of change in X. So how far it's going away from the camera. So what we do is we find a relationship. We find this relationship between X and H that's Pythagorean theorem, because what we have is a right triangle. So X squared equals 4000 squared plus h square. We differentiate that so we get the two x dx DT equals two HD HD t. Then we saw for DX DT and we get h over x d h d t now we know D h d t we know h The only thing we don't know which we need to know is X So the way we find X, the specific moment in time Yes, we have The X equals the square root of our 4000 squared plus 3000 square. That's 3000 square because we know that at that moment in time, that is our, um each distance because we have the 3000 and 4000 that ends up giving us an X value 5000. So now we have our X. We have our d h d t, and we have our h. This can all get plugged into our equation and we end up getting that are DX DT is going to end up going 360 ft per second. So that tells us that the change in acts with respect to time it's changing at a rate of 360 ft per second from the camera to the rocket. Then for part B, we want to find the rate of the angle change. So we know that the tangent of theta is equal to H over 4000, taking the derivative. We get sequence squared Theta equals one over 4000 D h d t. So therefore, our this is defeated ET so then defeated et equals, um a co sign squared beta over 4000 D h d t Because we multiply both sides by coastline data CoSine squared data. Somebody take the we need to find the specific moment in time because we already have a d h d t, and we already have, um we need to find our data value. So we know that data is going to equal the inverse tangent of H over 4000, and that's gonna give us and are tangent. Since H is 3000, we know that this is going to be the inverse tangent of 0.75 So that gets plugged in because now we have our theta, And now we have our Now we have our r d HDTV from before, which is 600 ft per second. So we plug all that in, we get defeated, DT equals 0.96 And remember the units. That's radiance per second

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