x = t^2, y = t^3 - 3t, find the horizontal tangents Step by Step

Step by step

The question

Question: find the horizontal tangents, when x = t² and y = t³ − 3t.

x = t²

y = t³ − 3t

1 dx/dt and dy/dt

The power rule, with t: dx/dt = 2t.

dxdt = 2t

And dy/dt = 3t² − 3.

dydt = 3t² − 3

2 dy/dx

The slope dy over dx is dy over dt, divided by dx over dt: the dt's cancel.

dydx = dydtdxdt

So dy/dx is (3t² − 3)/(2t).

dydx = 3t² − 32t

3 Where dy/dt = 0

A horizontal tangent has slope 0: the top, dy/dt, is 0, and the bottom is not.

dydt = 0

3t² − 3 = 0

Factor it.

t² − 1 is a difference of two squares: t² minus 1 squared.

So it is (t − 1)(t + 1).

3(t + 1)(t − 1) = 0

A product is 0 when one of its parts is 0.

So t = −1 or t = 1.

t = −1 or t = 1

Check dx/dt is not 0 there, and find each point.

t = −1: dx/dt = −2 ≠ 0, point (1, 2)

t = 1: dx/dt = 2 ≠ 0, point (1, −2)

So the horizontal tangents touch at (1, 2) and (1, −2).

points : (1, 2) and (1, −2)

Careful!

Horizontal means the top, dy/dt, is 0.

dx/dt = 0 gives vertical tangents.

Careful: not dxdt = 0 ✗

Draw the curve: for each t, the point (x, y).

The curve is flat at these points.

4 Check

Let's check with a calculator: move t a tiny bit around −1, and divide the change in y by the change in x.

About 0: flat.

Right!

t = −1 ± 0.001: ΔyΔx ≈ 0

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: where dy/dt = 0.

Last: check.

Answer