Asked by questioner

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.

Answers

Answered by oobleck
(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.
Answered by nathan
dude shut the h.e.ll up please
Answered by questioner
I got 84.155 for A. Correct?
Answered by oobleck
That's what I got.
Answered by questioner
What was your exact answer for B?
Answered by oobleck
what about -11/175 is not exact enough? Do you need a decimal?
That would be 0.062857
Answered by questioner
Yep. Can you also show your work for A?
Answered by Joe
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.
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