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.
I got 84.155 for A. Correct?
(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.
what about -11/175 is not exact enough? Do you need a decimal?
That would be 0.062857
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.
To find the maximum height of the rocket during its flight, we need to determine the vertex of the function y = 60ln(x+1) - 6x. The vertex represents the highest point of the rocket's trajectory.
The vertex of a logarithmic function of the form y = a ln(bx+c) + d can be found using the following formula:
Vertex(x, y) = (-c/b, d)
In our case, the equation y = 60ln(x+1) - 6x can be rewritten as y = 60ln(x+1) - 6(x+1) - 6, which simplifies to y = 60ln(x+1) - 6x - 6.
Comparing this form with the standard logarithmic form y = a ln(bx+c) + d, we find:
a = 60
b = 1
c = 1
d = -6
Using the formula for the vertex, we can now find the x-coordinate of the vertex:
x = -c/b = -1/1 = -1
Substituting this value back into the equation, we get:
y = 60ln(-1+1) - 6(-1) - 6
y = -6
Therefore, the maximum height of the rocket during its flight is 6 feet.
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To find the rocket's vertical velocity (dy/dt) when its horizontal velocity (dx/dt) is 50ft/sec and it is directly over a point 20 feet from the launch site, we need to differentiate the equation of the rocket's trajectory with respect to time.
The equation of the rocket's trajectory is y = 60ln(x+1) - 6x. Differentiating with respect to time (t) using the chain rule, we get:
dy/dt = (dy/dx) * (dx/dt)
We are given that dx/dt = 50 ft/sec.
To find dy/dx, we need to differentiate y = 60ln(x+1) - 6x with respect to x:
dy/dx = d/dx (60ln(x+1) - 6x) = 60(1/(x+1)) - 6 = 60/(x+1) - 6
Now, we can substitute the given value of x = 20 into dy/dx to find the vertical velocity at that point:
dy/dt = (dy/dx) * (dx/dt)
dy/dt = (60/(20+1) - 6) * 50
dy/dt = (60/21 - 6) * 50
dy/dt = (20/7 - 6) * 50
dy/dt = (20 - 42) * 50/7
dy/dt = -22 * 50/7
dy/dt = -157.14 ft/sec (rounded to two decimal places)
Therefore, the rocket's vertical velocity at a point 20 feet from the launch site when its horizontal velocity is 50 ft/sec is approximately -157.14 ft/sec.