Step by step
The question
Question: find the slope at t = −1, when x = t² − 1 and y = t³.
x = t2 − 1
y = t3
1 dx/dt and dy/dt
The power rule, with t: dx/dt = 2t.
dxdt = 2t
And dy/dt = 3t².
dydt = 3t2
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 over (2t).
dydx = 3t22t
Simplify: divide the numbers, take away the powers of t.
= 32t
3 At the point
Put in t = −1: dy/dt is 3 and dx/dt is −2.
The slope is −3/2.
dydx = 3−2 = −32
dydt: 3(−1)2 = 3
dxdt: 2(−1) = −2
Careful!
dy goes on top.
Upside down gives −2/3: wrong.
Careful: not slope = −23 ✗
So the slope at t = −1 is −3/2.
= slope = −32
Draw the curve: for each t, the point (x, y).
The point at t = −1.
4 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 −1.5: the slope −1.5 our rule gives.
Right!
t = −1 ± 0.001: ΔyΔx ≈ −1.5
Let's look back.
First: dx/dt and dy/dt.
Then: dy/dx.
Then: at the point.
Last: check.
Answer
= slope = −32