Worked examples
The question
Question: find dy/dx, when x = t² and y = t³.
x = t2
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
Careful!
dy goes on top: dy over dt, divided by dx over dt.
Not the other way round.
Careful: not dx/dtdy/dt ✗
So dy/dx = 3/2 × t.
dydx = 32t
3 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.
Last: check.
▶ Watch it solved
The question
Question: find the tangent line at t = 2, when x = t² and y = t³ − 3t.
x = t2
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.
Upside down gives 4/9: wrong.
Careful: not slope = 49 ✗
4 The tangent line
The point at t = 2: x = 4, y = 2.
point = (4, 2)
Point-slope form: y minus y one equals the slope times x minus x one.
y − 2 = 94(x − 4)
Open the bracket and get y alone: y = 9/4 × x − 7.
= y = 94x − 7
Draw the curve: for each t, the point (x, y).
The point at t = 2.
And the tangent line just touches the curve there.
5 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.
Then: the tangent line.
Last: check.
▶ Watch it solved
The question
Question: find the horizontal tangents, when x = t² and y = t³ − 12t.
x = t2
y = t3 − 12t
1 dx/dt and dy/dt
The power rule, with t: dx/dt = 2t.
dxdt = 2t
And dy/dt = 3t² − 12.
dydt = 3t2 − 12
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 − 12)/(2t).
dydx = 3t2 − 122t
3 Where dy/dt = 0
A horizontal tangent has slope 0: the top, dy/dt, is 0, and the bottom is not.
dydt = 0
3t2 − 12 = 0
Factor it.
t² − 4 is a difference of two squares: t² minus 2 squared.
So it is (t − 2)(t + 2).
3(t + 2)(t − 2) = 0
A product is 0 when one of its parts is 0.
So t = −2 or t = 2.
t = −2 or t = 2
Check dx/dt is not 0 there, and find each point.
t = −2: dx/dt = −4 ≠ 0, point (4, 16)
t = 2: dx/dt = 4 ≠ 0, point (4, −16)
So the horizontal tangents touch at (4, 16) and (4, −16).
points : (4, 16) and (4, −16)
Careful!
Horizontal means the top, dy/dt, is 0.
dx/dt = 0 gives vertical tangents.
Careful: not dxdt = 0 ✗
Draw the curve: for each t, the point (x, y).
The curve is flat at these points.
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 0: flat.
Right!
t = −2 ± 0.001: ΔyΔx ≈ 0
Let's look back.
First: dx/dt and dy/dt.
Then: dy/dx.
Then: where dy/dt = 0.
Last: check.
Answer points : (4, 16) and (4, −16)
▶ Watch it solved
