Steply

x = 4t, y = t^2 - 2t, find the horizontal tangents

Step by step, with the reason for every step · Parametric curves · Calculus 2

Step by step

The question

Question: find the horizontal tangents, when x = 4t and y = t² − 2t.

x = 4t
y = t2 − 2t

1 dx/dt and dy/dt

The power rule, with t: dx/dt = 4.

dxdt = 4

And dy/dt = 2t − 2.

dydt = 2t − 2

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 (2t − 2)/4.

dydx = 2t − 24

3 Where dy/dt = 0

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

dydt = 0
2t − 2 = 0

Factor it.

2(t − 1) = 0

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

So t = 1.

t = 1

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

t = 1: dx/dt = 4 ≠ 0, point (4, −1)

So the horizontal tangents touch at (4, −1).

points : (4, −1)

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
points : (4, −1)

Your turn

Try this one yourself, then check:

Find the horizontal tangents, when x = 3t
Show the answer
(3, −1)

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