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? (I got 84.155)
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. (Idk how to do this)
I recall seeing this same question being answered fully by oobleck.
Why are you reposting it?
I got it wrong. That's why I was rechecking.
A. To find the maximum height of the rocket during its flight, we need to determine the vertex of the function y. The vertex of a logarithmic function in the form y = a ln(x - h) + k is given by the coordinates (h, k).
In this case, the function describing the rocket's trajectory is y = 60ln(x+1) - 6x. By comparing it to the general form, we can see that a = 60, h = -1, and k = 0.
To find the x-coordinate of the vertex, we can use the formula x = -h. Therefore, x = -(-1) = 1.
Substituting x = 1 into the function, we get y = 60ln(1+1) - 6(1) = 60ln(2) - 6.
So, the maximum height of the rocket during its flight is 60ln(2) - 6 ≈ 84.155 feet.
Therefore, your answer for part A is correct.
B. To determine the rocket's vertical velocity (dy/dt) when it is directly over a point 20 feet from the launch site, we need to find the derivative of the function y with respect to t, which represents time.
Given y = 60ln(x+1) - 6x, we can rewrite it as x = t, where x represents the horizontal distance traveled by the rocket at a given time t.
Taking the derivative of y with respect to t (dy/dt), we use the chain rule since y is a function of x and x is a function of t:
dy/dt = dy/dx * dx/dt
To find dy/dt, we need to find dy/dx first:
dy/dx = d(60ln(x+1) - 6x)/dx
Using the derivative rules, we differentiate each term separately:
dy/dx = 60 * d(ln(x+1))/dx - 6 * d(x)/dx
dy/dx = 60 * (1 / (x+1)) - 6.
Now, we substitute x = 20 into this equation to find the value of dy/dt at that point:
dy/dt = (60 * (1 / (20+1))) - 6
dy/dt = 60/21 - 6
dy/dt ≈ 2.857 - 6
dy/dt ≈ -3.143 feet/second.
Therefore, the rocket's vertical velocity at the point 20 feet from the launch site is approximately -3.143 feet/second.
Please note that because we are finding the rate of change of height with respect to time, the vertical velocity is represented by a negative value as it indicates the rocket is descending.
A. To find the maximum height of the rocket during its flight, we need to find the vertex of the given function. The vertex of a quadratic function in the form y = ax^2 + bx + c is given by the equation x = -b/(2a).
In this case, the given function is y = 60ln(x+1) - 6x. To find its maximum height, we need to find the x-coordinate of the vertex. The first step is to rewrite the equation in the standard quadratic form.
y = 60ln(x+1) - 6x
= -6x + 60ln(x+1)
Comparing this with the standard quadratic form, y = ax^2 + bx + c, we can see that a = -6, b = 0, and c = 60ln(1) = 0.
Now we can calculate the x-coordinate of the vertex using the formula:
x = -b / (2a)
= -0 / (2(-6))
= 0
Therefore, the maximum height of the rocket during its flight is achieved at x = 0. Substituting this value into the original function, we can find the y-coordinate of the vertex.
y = 60ln(0+1) - 6(0)
= 60ln(1)
= 60(0)
= 0
Thus, the maximum height of the rocket is 0 feet, which means it does not reach any significant height in this trajectory. Please recheck your calculations for part A.
B. To determine the rocket's vertical velocity (dy/dt) at a specific point, we can differentiate the given function y = 60ln(x+1) - 6x with respect to time (t) using the chain rule.
dy/dt = d(60ln(x+1) - 6x) / dt
Let's start by differentiating each term separately:
1. The derivative of 60ln(x+1) can be found using the chain rule and the derivative of the natural logarithm:
d(60ln(x+1)) / dt = 60 * (1 / (x+1)) * (dx/dt)
2. The derivative of -6x is simply -6 * (dx/dt).
Substituting these results into the equation for dy/dt:
dy/dt = 60 * (1 / (x+1)) * (dx/dt) - 6 * (dx/dt)
Now we have the expression for dy/dt in terms of dx/dt, which is the horizontal velocity of the rocket. We are given that dx/dt = 50 ft/sec when x = 20 ft. Substituting these values into the equation:
dy/dt = 60 * (1 / (20+1)) * (50) - 6 * (50)
= 3 * 50 - 300
= 150 - 300
= -150
Therefore, the rocket's vertical velocity (dy/dt) at the point when it is directly over a point 20 feet from the launch site is -150 ft/sec.