x = 2t + 1, y = t^2 - 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 = 2t + 1 and y = t² − t.

x = 2t + 1

y = t² − t

1 dx/dt and dy/dt

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

dxdt = 2

And dy/dt = 2t − 1.

dydt = 2t − 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 (2t − 1)/2.

dydx = 2t − 12

3 At the point

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

The slope is 1/2.

dydx = 12

dydt: 2(1) − 1 = 2 − 1 = 1

dxdt: 2

Careful!

dy goes on top.

Upside down gives 2: wrong.

Careful: not slope = 21 ✗

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 = 12(x − 3)

Open the bracket and get y alone: y = 1/2 × x − 3/2.

= y = 12x − 32

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 0.5: the slope 0.5 our rule gives.

Right!

t = 1 ± 0.001: ΔyΔx ≈ 0.5

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Then: the tangent line.

Last: check.

Answer