Super Hard Challenge Calculus Problem. PLEASE HELP ME!!?

Question

So the question goes like this:
A TV camera at ground level is filming the liftoff of a space shuttle that is rising vertically according to the position eqn: s(t)=50t^2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from the launch pad.

Find the velocity of the space shuttle when it is 5000ft high.
[[I did this already, and I got 1000 ft/s. I know that it is right.]]

Find the rate of change in the angle of elevation of the camera when the space shuttle is 5000 ft. high.
[[I can't figure this one out for the life of me. I think it has something to do with inverse tangent, and with the previous anwser (1000 ft/sec).]]

If you can solve this, you're my hero and a freaking genius!!

Answer 1

first, draw a diagram of the situation so you can see that tan(theta) = s(t)/2000=50t^2/2000=t^2/40

now take the arc tan so that theta=arctan[t^2/40]

you want to find d(theta)/dt, so take d[arctan(t^2)/40]/dt

recall that d/dt{arctan[t]}=1/(1+t^2), in our case we have:

d[arctan[t^2/40]/dt = {1/[1+(t^4/1600)]}2t

where you have to remember to square (t^2/40) as shown and use the chain rule to get the factor of 2t

From here you can simplify and use the value of t when s is 5000 ft high (t=10 secs)

Answer 2

Let the distance from the camera to launch pad be d, the height of the shuttle is s = 50t^2.

The right triangle formed by the two sides, s and d, has a hypotoneus :

r = sqrt(d^2 + 2500t^4)

Now the angle the camera makes with the ground q is

q = arctan(s/d) =arctan(50t^2/d)

or tan(q) = 50t^2/d

differentiate by t:

sec^2(q) dq/dt = 100t/d

But sec(q) = 1/cos(q) and cos(q) = d/r so

dq/dt = 100 t cos^2(q)/d = 100t/d *(d/r)^2

dq/dt = 100 d t /(d^2+2500t^4)

You can plug in values from here

Answer 3

v=ds/dt = 100 t 5000 = 50t ^2so t= 10 and v=1000f/s
tan @ = s/2000 so
@ =act tan s/2000)
d@/dt= d@/ds*ds/dt = 1/[1+(s/2000)^2] *1/2000 * 100t
again t=10 s and s= 5000.
You can do the substitutions

Answer 4

The angle, call it a, is described by tan(a)=s/2000.
So d/dt (that eq) will give you the relation between a, da/dt (the desired number) and ds/dt (which you have).

Answer 5

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