x = t + 2, y = t^3 - 3t, find the tangent line at t = 0 Step by Step

Step by step

The question

Question: find the tangent line at t = 0, when x = t + 2 and y = t³ − 3t.

x = t + 2

y = t³ − 3t

1 dx/dt and dy/dt

The power rule, with t: dx/dt = 1.

dxdt = 1

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)/1.

dydx = 3t² − 31

3 At the point

Put in t = 0: dy/dt is −3 and dx/dt is 1.

The slope is −3.

dydx = −31

dydt: 3(0)² − 3 = 0 − 3 = −3

dxdt: 1

Careful!

dy goes on top.

Upside down gives −1/3: wrong.

Careful: not slope = 1−3 ✗

4 The tangent line

The point at t = 0: x = 2, y = 0.

point = (2, 0)

Point-slope form: y minus y one equals the slope times x minus x one.

y − 0 = −3(x − 2)

Open the bracket and get y alone: y = −3x + 6.

= y = −3x + 6

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

The point at t = 0.

And the tangent line just touches the curve there.

5 Check

Let's check with a calculator: move t a tiny bit around 0, and divide the change in y by the change in x.

About −3: the slope −3 our rule gives.

Right!

t = 0 ± 0.001: ΔyΔx ≈ −3

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Then: the tangent line.

Last: check.

Answer