The position vector of an object that is moving in the xy-plane is (t+cost, t^3-75t) for t on [0, 2pi]. At which value of y does the object's path have a horizontal tangent?
I think this is the value at which dy/dt=0, or at t=5. At t=5, y= -250.
x = t + cos t
y = t^3 - 75 t
I think I want dy/dx = 0 not dy/dt = 0 but it amounts to the same here
one way
dy = (3 t^2- 75) dt
dx = (1 - sin t)
when is dy = 0 ?
3 t^2 = 75
t^2 = 25
t = +/- 5, yes use t = 5
then y = t^3-75t = 125 - 375 = - 250 agree
x = t + cos t
y = t^3 - 75 t
dy/dx = (dy/dt)/(dx/dt) = (3t^2-75)/(1-sint)
As Damon said, finding where dy/dx=0 is the same as finding where dy/dt=0, since the denominator is only zero when t=pi/2.
To find the value of y at which the object's path has a horizontal tangent, we need to find when the derivative of y with respect to t, dy/dt, is equal to zero.
Given that the position vector of the object is (t + cos(t), t^3 - 75t), we can calculate the derivative of y by taking the derivative of the second component of the position vector with respect to t:
dy/dt = d/dt (t^3 - 75t)
Taking the derivative of each term separately, we get:
dy/dt = 3t^2 - 75
Setting dy/dt equal to zero and solving for t:
3t^2 - 75 = 0
Dividing both sides of the equation by 3:
t^2 - 25 = 0
Factoring the left-hand side of the equation:
(t - 5)(t + 5) = 0
Setting each factor equal to zero and solving for t:
t - 5 = 0 or t + 5 = 0
t = 5 or t = -5
Since the range for t is [0, 2π], we can disregard the value t = -5, as it is outside the given range.
Therefore, at t = 5, the object's path has a horizontal tangent.
To find the corresponding value of y, substitute t = 5 into the equation for y:
y = t^3 - 75t
y = 5^3 - 75(5)
y = 125 - 375
y = -250
So the value of y at which the object's path has a horizontal tangent is y = -250.