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

Step by step

The question

Question: find the tangent line at t = 2, when x = t² and y = t³ − 3t.

x = t²

y = t³ − 3t

1 dx/dt and dy/dt

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

dxdt = 2t

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)/(2t).

dydx = 3t² − 32t

3 At the point

Put in t = 2: dy/dt is 9 and dx/dt is 4.

The slope is 9/4.

dydx = 94

dydt: 3(2)² − 3 = 12 − 3 = 9

dxdt: 2(2) = 4

Careful!

dy goes on top.

Upside down gives 4/9: wrong.

Careful: not slope = 49 ✗

4 The tangent line

The point at t = 2: x = 4, y = 2.

point = (4, 2)

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

y − 2 = 94(x − 4)

Open the bracket and get y alone: y = 9/4 × x − 7.

= y = 94x − 7

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

The point at t = 2.

And the tangent line just touches the curve there.

5 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 2.25: the slope 2.25 our rule gives.

Right!

t = 2 ± 0.001: ΔyΔx ≈ 2.25

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Then: the tangent line.

Last: check.

Answer