Parametric differentiation, and the second derivative

x = t², y = t³ − 3t

• t = 0 is the origin.
• From t = −2 to 2 it swings out and back.
• It crosses itself at (3, 0).

Where it crosses itself

• Two values of t, one point.
• t and −t give the same x.

1. x(−t) = (−t)² = x(t)same x
2. need y(t) = y(−t)
3. t³ − 3t = −t³ + 3t
4. 2t³ − 6t = 0collect terms
5. 2t(t² − 3) = 0factorise
6. t = 0t and −t are equal
7. t = ±√3
8. x = 3, y = 3√3 − 3√3 = 0
9. crosses itself at (3, 0)

The first derivative

• Differentiate x and y in t.
• dy/dx = dy/dt ÷ dx/dt.
• In the formula booklet.

1. x = t², y = t³ − 3t
2. dx/dt = 2t
3. dy/dt = 3t² − 3
4. dy/dx = (dy/dt) ÷ (dx/dt)chain rule
5. = (3t² − 3)/(2t)

Stationary points and vertical tangents

• Stationary: dy/dt = 0.
• Vertical: dx/dt = 0.

1. 3t² − 3 = 0dy/dt = 0
2. t² = 1, so t = ±1
3. t = 1: x = 1, y = 1 − 3 = −2
4. t = −1: x = 1, y = −1 + 3 = 2
5. (1, −2) and (1, 2)stationary points
6. 2t = 0, so t = 0dx/dt = 0
7. vertical tangent at (0, 0)

The tangent at t = 2

• Point and gradient: same t.
• Finish in y = mx + c.

1. t = 2: x = 2² = 4
2. y = 2³ − 3(2) = 2
3. point (4, 2)
4. m = (3(2)² − 3)/(2(2))into dy/dx
5. = 9/4
6. y − 2 = (9/4)(x − 4)point-gradient form
7. y = (9/4)x − 9 + 2expand
8. y = (9/4)x − 7

The normal at t = 2

• Normal: perpendicular.
• Gradients multiply to −1.
• End in y = mx + c.

1. point (4, 2), tangent m = 9/4
2. m = −1 ÷ (9/4) = −4/9negative reciprocal
3. y − 2 = −(4/9)(x − 4)same point
4. y = −(4/9)x + 16/9 + 2expand
5. y = −(4/9)x + 34/9
6. (9/4) × (−4/9) = −1

Where candidates stopped

• d/dt alone is not finished.
• Then ÷ dx/dt = 2t.

1. dy/dx = (3t² − 3)/(2t)
2. = (3/2)t − 3/(2t)split the fraction
3. d/dt: 3/2 + 3/(2t²)not d²y/dx² yet
4. d²y/dx² = (3/2 + 3/(2t²)) ÷ 2t
5. = ((3t² + 3)/(2t²)) ÷ 2tcommon denominator
6. = (3t² + 3)/(4t³)

Using it to classify

• Positive: a minimum.
• Negative: a maximum.

1. t = 1: (3 + 3)/4 = 3/2
2. 3/2 > 0: (1, −2) is a minimum
3. t = −1: (3 + 3)/(−4) = −3/2
4. −3/2 < 0: (1, 2) is a maximum
5. stopped early: 3/2 + 3/2 = 3
6. 3 > 0 at both: two minima?impossible

Trig: x = 2cos t, y = 3sin t

• Work through t again.
• Then ÷ dx/dt = −2sin t.

1. dx/dt = −2sin t
2. dy/dt = 3cos t
3. dy/dx = 3cos t/(−2sin t)
4. = −(3/2)cot t
5. d/dt: (3/2)cosec² td/dt cot t = −cosec² t
6. ÷ (−2sin t): −3/(4sin³ t)÷ dx/dt
7. t = π/2: −3/(4(1)³) = −3/4
8. negative: concave down

The answers: question 1

1. 1. dx/dt = 2t, dy/dt = 3t² − 12
2. dy/dx = (3t² − 12)/(2t)
3. 3t² − 12 = 0, so t = ±2top = 0
4. t = 2: x = 5, y = 8 − 24 = −16
5. t = −2: x = 5, y = −8 + 24 = 16
6. (5, −16) and (5, 16)

The answers: question 2

1. 2. dy/dx = (3/2)t − 6/tsplit the fraction
2. d/dt: 3/2 + 6/t²
3. ÷ 2t: (3t² + 12)/(4t³)÷ dx/dt
4. t = 2: (12 + 12)/32 = 3/4
5. 3/4 > 0: (5, −16) is a minimum

The answers: questions 3 and 4

1. 3. x = e⁰ = 1, y = e⁰ − 0 = 1t = 0
2. dx/dt = eᵗ, dy/dt = 2e²ᵗ − 4
3. m = (2 − 4)/1 = −2at t = 0
4. y − 1 = −2(x − 1), so y = −2x + 3
5. 4. dx/dt = 3cos t
6. dy/dt = −2sin t
7. dy/dx = −2sin t/(3cos t)
8. = −(2/3)tan t
9. t = π/4: −(2/3)(1) = −2/3