What points on the given curve x = 4t^3, y = 3 + 32t − 20t^2

Question

does the tangent line have slope 1?

Answer 1

The curve equations are x = 4t³ and y = 3 + 32t - 20t² and the slope of tangent line is 1.

Differentiate with respect to x.

dy/dx = (dy/dt)/(dx/dt)

= [(d/dt)(3 + 32t - 20t²)]/[(d/dt)(4t³)]

= (32 - 40t)/(12t²)

= (8 - 10t)/3t²

Therefore, (8 - 10t)/3t² = 1

8 - 10t = 3t²

3t² + 10t - 8 = 0

3t² + 12t - 2t - 8 = 0

3t(t + 4) - 2(t + 4) = 0

(3t - 2)(t + 4) = 0

3t - 2 = 0 and t + 4 = 0

t = 2/3 and t = -4

If t = 2/3 then

x = 4(2/3)³ = 1.185 and

y = 3 + 32(2/3) - 20(2/3)² = 15.4.

If t = -4 then

x = 4(-4)³ = -256 and

y = 3 + 32(-4) - 20(-4)² = -445.

The points are (1.19, 15.4) and (-256, -445).