At what point on the curve x = 3t^2 + 2, y=t^3-8 does the tangent line have slope 1/2?

Answer 1

x = 3 t² + 2 , y = t³ - 8

dx / dt = 3 ∙ 2 t = 6 t

dy / dt = 3 t²

( dy / dt ) / ( dx / dt ) = dy / dx = y´ = 3 t² / 6 t = t / 2

The first derivative of a function is the slope of the tangent line for any point on the function, so you must find where y' = 1 / 2

t / 2 = 1 / 2

t = 1

So :

x = 3 t² + 2 = x = 3 ∙ 1² + 2 = 3 ∙ 1 + 2 = 3 + 2 = 5

y = t³ - 8 = y = 1³ - 8 = 1 - 8 = - 7

The tangent line have slope 1 / 2 at point ( 5 , - 7 )

Answer 2

dy/dx = dy/dt / dx/dt = 3t^2/6t = t/2
So, dy/dx = 1/2 when t=1, at (5,-7)

Answer 3

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