Saturday
April 30, 2016

# Homework Help: More related rates.

Posted by Tracy on Friday, November 3, 2006 at 4:02pm.

Heyy, its me again. So last night I asked a related rate question, and I think I still don't get it. I tried it a few differnt ways and I think I'm just missing something...I don't think a speed of a rocket can be 0.018km/hr:S. So heres what happened:
A camera, located 2 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 20 km, the camera is rotating at the rate of 1/200 radians per second. What is the speed of the rocket at that instant? Give your answer in km/sec.
I drew a picture which is a triangle. And I know that tan theata =20/2 (I think), and we know dtheata/dt = 1/200 radians. We are looking for dh/dt. So tan theata =h/2 so the derviative is sec^2 theata dtheata/dt =1/2dh/dt which means that 2(1/cos^2theata)(dtheata/dt)=dh/dt
So when I do this out, I get 0.018 and I dont think thats right. I think I'm missing something important and I can't figure out what it is.

And also, I have one more question. Me and my friends worked on this for liek 2 hours, we all did the same thing and got the smae answer no matter waht we did but it says its wrong so I dont know what is going wrong.
A girl 6 ft tall is walking away from a streetlight. The streetlight is attached to a pole that is 16 feet tall, and is generating a shadow. If the tip of the girl's shadow is moving at 5 ft/sec, how fast is she walking? Give your answer in ft/sec.

The answer I got was -3, using similar triangles. And we also did it a differnt way and got 8.3 but thats also wrong.

Thanks ahead of time, I know its alot, I'm sooo frsutrated.

Here you know the angle, and you are given dTheta/dt. You are looking for dh/dt

You know TanTheta=h/2, and you know dtheta/dt when h=2-

take the derivative with respect to time. It will come out rather simply.

sec^2theta * dTheta/dt = 1/2 dh/dt

dh/dt= 2*sec^2Theta dT/dt
= 2* (h^2+ 3)^1/2 /2 * 1/200
= 2* (400+ 3)^1/2 /2 * 1/200
=appx .1 km/sec check my work.

let the distance x be the distance from the tall lamp to the end of the shadow.
16/x= 6/(x-D) where D is the distance from the lamp to the girl. This is from similar triangles.

16(x-D)= 6x
16 dx/dt - 16dD/dt= 6 dx/dt
given dx/dt is 5ft/sec, dD/dt is the unknown.

Okay, so heres where I get lost:
= 2* (h^2+ 3)^1/2 /2 * 1/200
= 2* (400+ 3)^1/2 /2 * 1/200
=appx .1 km/sec check my work.
Where did you get the h^2 +3 ^1/2
And what value is h to make it 400?