For the parametric curve defined x=2t^3-12t^2-30t+9 and y=t^2-4t+6 ..find dy/dx and then where is the tangent to the curve vertical (give the cartesian coordiantes of the points. and find the tang. to the curve horizontal
dx/dt= 6t^2-24t-30
dy/dt= 2t-4
dy/dx= (t-2)/(3t^2-12t-15)
Now, if dy/dx= inf, then
3t^2-12t-15=0 or
(t-5)(t+1)=0 check that.
solve for t. Put that value of t in the original equations to get x,y for the point where the tangent is vertical.
To find the tangent to the curve where it is vertical, we need to find the values of t where dy/dx becomes infinitely large (dy/dx = ∞).
To do so, we set the denominator of dy/dx equal to zero, since division by zero results in infinity.
3t^2 - 12t - 15 = 0
We can factor this quadratic equation:
(t - 5)(t + 1) = 0
Solving for t:
t = 5 or t = -1
Now, we substitute these values of t back into the original equations to find the corresponding x and y values for the points where the tangent is vertical.
For t = 5:
x = 2(5^3) - 12(5^2) - 30(5) + 9
x = 0
y = 5^2 - 4(5) + 6
y = 16
So, the Cartesian coordinates for the point where the tangent is vertical are (0, 16).
For t = -1:
x = 2(-1^3) - 12(-1^2) - 30(-1) + 9
x = 56
y = (-1)^2 - 4(-1) + 6
y = 9
Thus, the Cartesian coordinates for the second point where the tangent is vertical are (56, 9).
To find the tangent to the curve where it is horizontal, we need to find the values of t where dy/dx equals zero (dy/dx = 0).
The expression for dy/dx is:
dy/dx = (t - 2)/(3t^2 - 12t - 15)
Setting dy/dx equal to zero, we have:
(t - 2)/(3t^2 - 12t - 15) = 0
Since the numerator can't be zero, we need to solve for the values of t where the denominator is zero.
Simplifying the denominator to (t - 5)(t + 1), we have:
3t^2 - 12t - 15 = 0
Factorizing this quadratic equation, we get:
(t - 5)(t + 1) = 0
From this, we find two values of t:
t = 5 or t = -1
Therefore, the tangent to the curve is horizontal at the points where t = 5 and t = -1. To find the corresponding Cartesian coordinates, substitute these values of t into the original equations for x and y:
For t = 5:
x = 2(5^3) - 12(5^2) - 30(5) + 9
x = 0
y = 5^2 - 4(5) + 6
y = 16
So, one point where the tangent is horizontal is (0, 16).
For t = -1:
x = 2(-1^3) - 12(-1^2) - 30(-1) + 9
x = 56
y = (-1)^2 - 4(-1) + 6
y = 9
Thus, the other point where the tangent is horizontal is (56, 9).