A rocket follows a path given by y=x-1/91x^3 where distances are in miles. If the horizontal velocity is given by vx=1.4x, find the direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.
So far I've only taken the derivative for y and don't know what to do afterwards
Vx=1.4x
Vy=1-(3/91)x^2
Maybe you mean
y=x - (1/91) x^3
if so
y = 0 at ground, start and finish
0 = x - (1/91) x^3
x = 0 works of course
x^2 = 91
x = 9.54
Vx = 1.4 x = 13.4 miles/min at ground
what is Vy ?
dy/dt = dy/dx * dx/dt = [1-(3/91)x^2 ] Vx = [ 1-(3/91)9.54^2 ] 13.4 miles/min
To find the direction of the velocity when the rocket hits the ground, you can use calculus to analyze the rate of change of y with respect to x.
Given that the horizontal velocity is Vx = 1.4x, we know that the derivative of y with respect to x, dy/dx, represents the vertical velocity component (Vy).
Let's find dy/dx by taking the derivative of y = x - (1/91)x^3 with respect to x:
dy/dx = d/dx (x - (1/91)x^3)
= 1 - (3/91)(3x^2)
= 1 - (9/91)x^2
Now we have the expression for Vy, the vertical velocity component.
When the rocket hits the ground, y = 0. So, let's find the value of x when y = 0:
0 = x - (1/91)x^3
x(1 - (1/91)x^2) = 0
This gives us two solutions: x = 0 and x = ∛91.
Since we are assuming level terrain, the rocket won't hit the ground at x = 0 because that would mean it starts underground. Therefore, the rocket will hit the ground at x = ∛91.
Now, let's substitute this value of x into Vy to find its direction:
Vy = 1 - (9/91)(∛91)^2
= 1 - (9/91)(91^(2/3))
= 1 - (9/91)(~4.497)
≈ 1 - 0.3981
= 0.6019
The vertical velocity component (Vy) when the rocket hits the ground is approximately 0.6019 miles per minute.
To determine the direction of the velocity, we need to consider the sign of Vy. Since Vy is positive (greater than 0), the rocket is moving upwards when it hits the ground.
To find the direction of the velocity when the rocket hits the ground, we need to consider the y-component of the velocity, Vy, when y = 0 (since the rocket hits the ground at y = 0).
By differentiating the equation for y with respect to x, we find the derivative:
dy/dx = 1 - (3/91) * 3x^2
Next, we can find the value of x when y = 0. Substituting y = 0 into the equation for y, we have:
0 = x - (1/91) * x^3
0 = x * (1 - (1/91) * x^2)
Therefore, either x = 0 or (1 - (1/91) * x^2) = 0.
If x = 0, then the rocket would already be on the ground and we cannot determine the direction of velocity.
For (1 - (1/91) * x^2) = 0, we solve for x:
1 - (1/91) * x^2 = 0
x^2 = 91
x = ± √91
Now, we can find the corresponding Vy value for each x using the given vy equation:
Vy = 1 - (3/91) * x^2
For x = √91, Vy = 1 - (3/91) * (√91)^2 = 1 - 3 = -2
For x = -√91, Vy = 1 - (3/91) * (-√91)^2 = 1 - 3 = -2
Hence, the direction of the velocity when the rocket hits the ground is downward (negative y-direction) with a Vy value of -2 miles per minute.