At what point on the curve x = 3t^2 + 5, y = t^3 − 7 does the tangent line have slope 1/2 ?

Answer 1

(A) x=3t² + 5

(B) y=t³ - 7

Add 7 and get the cube root of both sides of equation (B)

t=(y+7)^1/3

Substitute:

x = 3((y+7)^1/3)² +5

x = 3(y+7)^2/3 + 5

Get the derivative of both sides with respect to x

1= 2(y+7)^-1/3 (dy/dx)

The slope of a tangent line is dy/dx = 1/2.

Substitute:

1= 2(y+7)^-1/3 (1/2)

1 = (y+7)^-1/3

Raise to -3 on both sides:

1 = y + 7

y= -6

Use x = 3(y+7)^2/3 + 5 to solve for x:

x = 3(-6+7)^2/3 + 5

x = 3(1) +5

x= 8

At the point (8,-6)