x = t^2, y = t^3 - 12t, find the horizontal tangents Step by Step

Step by step

The question

Question: find the horizontal tangents, when x = t² and y = t³ − 12t.

x = t²

y = t³ − 12t

1 dx/dt and dy/dt

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

dxdt = 2t

And dy/dt = 3t² − 12.

dydt = 3t² − 12

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 − 12)/(2t).

dydx = 3t² − 122t

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 − 12 = 0

Factor it.

t² − 4 is a difference of two squares: t² minus 2 squared.

So it is (t − 2)(t + 2).

3(t + 2)(t − 2) = 0

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

So t = −2 or t = 2.

t = −2 or t = 2

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

t = −2: dx/dt = −4 ≠ 0, point (4, 16)

t = 2: dx/dt = 4 ≠ 0, point (4, −16)

So the horizontal tangents touch at (4, 16) and (4, −16).

points : (4, 16) and (4, −16)

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 −2, and divide the change in y by the change in x.

About 0: flat.

Right!

t = −2 ± 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