x = t^2 + 2, y = t^3 - t, find the tangent line at t = 1 Step by Step

Step by step

The question

Question: find the tangent line at t = 1, when x = t² + 2 and y = t³ − t.

x = t² + 2

y = t³ − t

1 dx/dt and dy/dt

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

dxdt = 2t

And dy/dt = 3t² − 1.

dydt = 3t² − 1

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 − 1)/(2t).

dydx = 3t² − 12t

3 At the point

Put in t = 1: dy/dt is 2 and dx/dt is 2.

The slope is 1.

dydx = 22 = 1

dydt: 3(1)² − 1 = 3 − 1 = 2

dxdt: 2(1) = 2

Careful!

dy goes on top.

Upside down gives 1: wrong.

Careful: not slope = 22 ✗

4 The tangent line

The point at t = 1: x = 3, y = 0.

point = (3, 0)

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

y − 0 = 1(x − 3)

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

= y = x − 3

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

The point at t = 1.

And the tangent line just touches the curve there.

5 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 1: the slope 1 our rule gives.

Right!

t = 1 ± 0.001: ΔyΔx ≈ 1

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Then: the tangent line.

Last: check.

Answer