Step by step
The question
Question: find the slope at t = 2, when x = t² + 1 and y = t³ − 3t.
x = t2 + 1
Both x and y move with a third letter, t, like time.
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: dy over dt, divided by dx over dt.
Upside down gives 4/9: wrong.
Careful: not slope = 49 ✗
So the slope at t = 2 is 9/4.
slope = 94
Draw the curve: for each t, the point (x, y).
The point at t = 2.
4 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.
Last: check.
Answer
slope = 94