Question

Camille launches a model rocket in an open field near her house. The rocket has a bit of a problem, being slightly off balance. Its trajectory is described by the function y=60ln(x+1)-6x for 0 ≤ x ≤ 36.15, where y is is the rocket's height (in feet) above the ground, and x is the horizontal distance (in feet) between the launch site and a point directly below the rocket. Below is a graph of the trajectory of the rocket: (pls search this question, to see the visual on Chegg's website).

A. What is the maximum height of the rocket during its flight?

B. Given that the rocket's horizontal velocity (dx/dt) is 50ft/sec when it's directly over a point 20 (horizontal) feet from the launch site, determine the rocket's vertical velocity (dy/dt) at that point.

Answer 1

(A) dy/dx = 60/(x+1) - 6
so, where is that zero? Evaluate y there.

(B) dy/dx = (dy/dt) / (dx/dt) so
dy/dt = y'(20) / 50 = (60/21 - 6)/50 = -11/175
makes sense, since it's past the point of maximum height.

Answer 2

dude shut the h.e.ll up please

Answer 3

I got 84.155 for A. Correct?

Answer 4

That's what I got.

Answer 5

What was your exact answer for B?

Answer 6

what about -11/175 is not exact enough? Do you need a decimal?
That would be 0.062857

Answer 7

Yep. Can you also show your work for A?

Answer 8

oobleck I think you wrote the formula wrong. You said dy/dx = (dy/dt) / (dx/dt) which is right but then you said dy/dt = (dy/dx) / (dx/dt). It should have been dy/dt = (dy/dx) * (dx*dt) which would have given the correct answer for B.