Step by step
The question
Question: find the slope at t = 1, when x = t² + t and y = t² − t.
x = t2 + t
y = t2 − t
1 dx/dt and dy/dt
The power rule, with t: dx/dt = 2t + 1.
dxdt = 2t + 1
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)/(2t + 1).
dydx = 2t − 12t + 1
3 At the point
Put in t = 1: dy/dt is 1 and dx/dt is 3.
The slope is 1/3.
dydx = 13
dydt: 2(1) − 1 = 2 − 1 = 1
dxdt: 2(1) + 1 = 2 + 1 = 3
Careful!
dy goes on top.
Upside down gives 3: wrong.
Careful: not slope = 31 ✗
So the slope at t = 1 is 1/3.
= slope = 13
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 0.333: the slope 0.333 our rule gives.
Right!
t = 1 ± 0.001: ΔyΔx ≈ 0.333
Let's look back.
First: dx/dt and dy/dt.
Then: dy/dx.
Then: at the point.
Last: check.
Answer
= slope = 13