Step by step
The question
Question: find the slope at t = 1, when x = 3t and y = t² − 4.
x = 3t
y = t2 − 4
1 dx/dt and dy/dt
The power rule, with t: dx/dt = 3.
dxdt = 3
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/3.
dydx = 2t3
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
Careful!
dy goes on top.
Upside down gives 3/2: wrong.
Careful: not slope = 32 ✗
So the slope at t = 1 is 2/3.
= slope = 23
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 0.667: 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.
Last: check.
Answer
= slope = 23