A point moves along the curve that has the parametric equations:
x=9t^3- 〖6t〗^2 and y= 〖9t〗^3+27
x and y being in yards and t being in minutes. Find the magnitude and direction of the acceleration when t= 5/2.
Round answers to 3 significant figures.
To find the magnitude and direction of acceleration, we need to determine the derivatives of x and y with respect to t, and then find the second derivative of each. Let's start by finding dx/dt and dy/dt.
Given:
x = 9t^3 - 6t^2
y = 9t^3 + 27
Find dx/dt:
Taking the derivative of x with respect to t:
dx/dt = d/dt(9t^3 - 6t^2)
= 27t^2 - 12t
Find dy/dt:
Taking the derivative of y with respect to t:
dy/dt = d/dt(9t^3 + 27)
= 27t^2
Now, let's find the second derivatives, d^2x/dt^2 and d^2y/dt^2.
Find d^2x/dt^2:
Taking the derivative of dx/dt with respect to t:
d^2x/dt^2 = d/dt(27t^2 - 12t)
= 54t - 12
= 54(t - 2/9)
Find d^2y/dt^2:
Taking the derivative of dy/dt with respect to t:
d^2y/dt^2 = d/dt(27t^2)
= 54t
Now, we have the expressions for the second derivatives of x and y with respect to t. To find the acceleration, we'll substitute t = 5/2 into these equations.
Substituting t = 5/2 in d^2x/dt^2:
d^2x/dt^2 = 54(5/2 - 2/9)
= 54(45/18 - 4/18)
= 54(41/18)
= 123
Substituting t = 5/2 in d^2y/dt^2:
d^2y/dt^2 = 54(5/2)
= 135
The magnitude of acceleration: |a| = sqrt((d^2x/dt^2)^2 + (d^2y/dt^2)^2)
= sqrt(123^2 + 135^2)
≈ sqrt(15129 + 18225)
≈ sqrt(33354)
≈ 182.698
The direction of acceleration can be obtained by finding the angle θ between the acceleration vector (d^2x/dt^2, d^2y/dt^2) and the positive x-axis. The direction is given by:
θ = atan(d^2y/dt^2 / d^2x/dt^2)
Substituting the values:
θ = atan(135 / 123)
≈ atan(1.09756)
≈ 47.905 degrees
Therefore, the magnitude of acceleration is approximately 182.698 yards per minute squared, and the direction is approximately 47.905 degrees from the positive x-axis.