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 don't think that's 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 don't 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 that's 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-

start with TanTheta=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?

Explain Bot Explain Bot answered
9 months ago

I apologize for the confusion. Let's go through the calculations again.

For the first problem, we have the equation tan(theta) = h/2, where theta is the angle and h is the height of the rocket. We want to find dh/dt when theta = 1/200 radians/sec.

To find dh/dt, we'll take the derivative of both sides of the equation with respect to time (t):

sec^2(theta) * d(theta)/dt = (1/2) * (dh/dt)
d(theta)/dt is given as 1/200 radians/sec, so let's substitute that in:
sec^2(theta) * (1/200) = (1/2) * (dh/dt)

Now, we need to find the value of sec^2(theta), which is equal to 1 + tan^2(theta). Since tan(theta) = h/2, we can substitute that in as well:
sec^2(theta) = 1 + (h/2)^2

Now let's plug in the values:
1 + (h/2)^2 * (1/200) = (1/2) * (dh/dt)

The value of h is given as 20 km. Plugging that in, we get:
1 + (20/2)^2 * (1/200) = (1/2) * (dh/dt)

Simplifying further:
1 + 100/40000 = (1/2) * (dh/dt)
1 + 1/400 = (1/2) * (dh/dt)
801/400 = (1/2) * (dh/dt)

Now we can solve for dh/dt:
dh/dt = (2 * 801)/400 = 1602/400 = 4.005 km/sec (approx)

So the speed of the rocket at that instant is approximately 4.005 km/sec.

Now let's move on to the second problem.

We have the equation 16/x = 6/(x - D), where x is the distance from the lamp to the end of the shadow, and D is the distance from the lamp to the girl. We want to find dD/dt, the rate at which D is changing.

To solve for dD/dt, we'll take the derivative of both sides of the equation with respect to time (t):

d(16/x)/dt = d(6/(x - D))/dt

Let's simplify the equation first:
16/x = 6/(x - D)

Now let's differentiate both sides:
d(16/x)/dt = d(6/(x - D))/dt
-16/x^2 * dx/dt = -6/(x - D)^2 * (dx/dt - dD/dt)

We are given dx/dt as 5 ft/sec. Substituting that in:
-16/x^2 * 5 = -6/(x - D)^2 * (5 - dD/dt)

Simplifying further:
-80/x^2 = -6/(x - D)^2 * (5 - dD/dt)

Now let's solve for dD/dt:
dD/dt = 5 - (80/x^2) * (x - D)^2 / 6

Since we don't have information about x and D, we cannot find a specific numerical value for dD/dt. However, this equation gives us the general formula to calculate the rate at which the girl is walking, given the values of x and D.

I hope this explanation clarifies the process for solving these related rates problems. If you have further questions, feel free to ask!