Find the point(s) on the curve where the tangent is horizontal. $x = t ^ { 3 } - 3 t + 4 , \quad y = t ^ { 3 } - 3 t ^ { 2 } + 4$
A) $( 4,4 ) , ( 6,0 )$
B) $( 0,0 ) , ( 4 , - 4 )$
C) $( - 1,1 ) , ( - 4 , - 4 )$
D) $( 4 , - 4 )$
E) None of these

Question

Quizplus · Accepted Answer

The tangent is horizontal when the derivative of y with respect to x is 0. So, we need to find the derivative of y with respect to x and set it equal to 0.$$\frac{dy}{dx} = \frac{d}{dt}(t^3 - 3t^2 + 4) \frac{dt}{dx}$$$$= (3t^2 - 6t) \frac{dt}{dx}$$$$= 3t(t - 2) \frac{dt}{dx}$$Setting this equal to 0, we get:$$3t(t - 2) = 0$$$$t = 0, 2$$Substituting these values of t back into the original equations for x and y, we get the points (4, 4) and (6, 0).

Find the Point(s) on the Curve Where the Tangent Is x=t3−3t+4,y=t3−3t2+4x = t ^ { 3 } - 3 t + 4 , \quad y = t ^ { 3 } - 3 t ^ { 2 } + 4x=t3−3t+4,y=t3−3t2+4

Find the Point(s) on the Curve Where the Tangent Is $x = t ^ { 3 } - 3 t + 4 , \quad y = t ^ { 3 } - 3 t ^ { 2 } + 4$