x = t^3 + 1, y = t^2 + 3, 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 = t³ + 1 and y = t² + 3.

x = t³ + 1

y = t² + 3

1 dx/dt and dy/dt

The power rule, with t: dx/dt = 3t².

dxdt = 3t²

And dy/dt = 2t.

dydt = 2t

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/(3t²).

dydx = 2t3t²

Simplify: divide the numbers, take away the powers of t.

= 23t

3 At the point

Put in t = −1: dy/dt is −2 and dx/dt is 3.

The slope is −2/3.

dydx = −23

dydt: 2(−1) = −2

dxdt: 3(−1)² = 3

Careful!

dy goes on top.

Upside down gives −3/2: wrong.

Careful: not slope = 3−2 ✗

4 The tangent line

The point at t = −1: x = 0, y = 4.

point = (0, 4)

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

y − 4 = −23(x − 0)

Open the bracket and get y alone: y = −2/3 × x + 4.

= y = −23x + 4

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 −0.667: the slope −0.667 our rule gives.

Right!

t = −1 ± 0.001: ΔyΔx ≈ −0.667

Let's look back.

First: dx/dt and dy/dt.

Then: dy/dx.

Then: at the point.

Then: the tangent line.

Last: check.

Answer