Question

At what points on the given curve x = 4t3, y = 3 + 40t − 2t2 does the tangent line have slope 1?

Answer 1

ANSWER

The points are:

(32,-85)

and

$( \frac{500}{27} , \frac{577}{9} )$

EXPLANATION

The given parametric equation that defines the curve is:

$x = 4 {t}^{3}$

$y = 3 + 40t - 2 {t}^{2}$

We differentiate to get,

$\frac{dx}{dt} = 12 {t}^{2}$

$\frac{dy}{dt} = 40 - 4t$

The slope of the tangent is given by;

$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$

$\frac{dy}{dx} = \frac{12 {t}^{2} }{40 - 4t}$

For the slope to be 1, then we must solve the following equation for t.

$\frac{12 {t}^{2} }{40 - 4t} = 1$

$12 {t}^{2} = 40 - 4t$

$12 {t}^{2} + 4t - 4 0 = 0$

$3{t}^{2} + t - 10 = 0$

$(t + 2)(3t - 5) = 0$

$t = - 2 \: \: or \: \: t = \frac{5}{3}$

When t=-2, we put -2 into the expression for x and y to get

x=4(-2)³ and y=3+40(-2)-2(-2)²

This implies that,

x=32, y=-85

Hence one point is (32,-85)

Similarly, when

$t = \frac{5}{3}$

$x = \frac{500}{27} ,y = \frac{577}{9}$

Another point is

$( \frac{500}{27} , \frac{577}{9} )$