Parametric Equation Derivatives Worksheet | Calculus

📚 Essential Parametric Derivative Formulas:

First Derivative: dy/dx = (dy/dt) ÷ (dx/dt)

Second Derivative: d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)

Tangent Line Slope at t=t₀: m = (dy/dt)/(dx/dt) evaluated at t₀

Critical Condition: dx/dt ≠ 0 for derivative to exist

Problem 1: First Derivative (Polynomial) Beginner

x(t) = t², y(t) = t³ - 3t

Find dy/dx in terms of t. Then evaluate at t = 2.

1 Find dx/dt:

dx/dt = d(t²)/dt = 2t

2 Find dy/dt:

dy/dt = d(t³ - 3t)/dt = 3t² - 3

3 Apply formula:

dy/dx = (dy/dt) ÷ (dx/dt) = (3t² - 3)/(2t)

4 Simplify:

dy/dx = (3t)/2 - 3/(2t)

5 Evaluate at t = 2:

dy/dx|_{t=2} = (3·2)/2 - 3/(2·2) = 3 - 3/4 = 9/4 = 2.25

💡 Interpretation: When t = 2, the curve is at point (4, 2) with slope 2.25.

Problem 2: Second Derivative (Trigonometric) Intermediate

x(t) = cos(t), y(t) = sin(2t)

Find d²y/dx². What is the concavity at t = π/4?

1 First derivatives:

dx/dt = -sin(t)

dy/dt = 2cos(2t)

2 First derivative dy/dx:

dy/dx = (2cos(2t))/(-sin(t)) = -2cos(2t)/sin(t)

3 Derivative of dy/dx with respect to t:

d/dt(dy/dx) = d/dt[-2cos(2t)/sin(t)]

Using quotient rule: [4sin(2t)·sin(t) + 2cos(2t)·cos(t)]/sin²(t)

4 Second derivative formula:

d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)

= [4sin(2t)sin(t) + 2cos(2t)cos(t)]/sin²(t) ÷ (-sin(t))

= -[4sin(2t)sin(t) + 2cos(2t)cos(t)]/sin³(t)

5 Evaluate at t = π/4:

sin(π/4) = √2/2, sin(π/2) = 1, cos(π/2) = 0

d²y/dx²|_{t=π/4} = -[4·1·(√2/2) + 2·0·(√2/2)]/((√2/2)³)

= -[2√2]/(√2/4) = -8

💡 Interpretation: Negative second derivative means concave down at t = π/4.

Problem 3: Tangent Line Equation Intermediate

x(t) = eᵗ, y(t) = t·eᵗ

Find the equation of the tangent line at t = 0.

1 Point on curve at t = 0:

x(0) = e⁰ = 1

y(0) = 0·e⁰ = 0

Point: (1, 0)

2 Derivatives:

dx/dt = eᵗ

dy/dt = eᵗ + t·eᵗ (product rule)

3 Slope at t = 0:

dy/dx = (dy/dt)/(dx/dt) = (eᵗ + t·eᵗ)/eᵗ = 1 + t

m = 1 + 0 = 1

4 Tangent line equation:

y - y₀ = m(x - x₀)

y - 0 = 1(x - 1)

y = x - 1

💡 Verification: The line y = x - 1 passes through (1,0) with slope 1.

Problem 4: Vertical & Horizontal Tangents Advanced

x(t) = t³ - 3t, y(t) = t² - 4

Find all points where the curve has (a) horizontal tangents (b) vertical tangents.

1 Derivatives:

dx/dt = 3t² - 3 = 3(t² - 1)

dy/dt = 2t

2 Horizontal tangents occur when dy/dt = 0 (provided dx/dt ≠ 0):

2t = 0 ⇒ t = 0

dx/dt|_{t=0} = -3 ≠ 0 ✓

Point: (x(0), y(0)) = (0, -4)

3 Vertical tangents occur when dx/dt = 0 (provided dy/dt ≠ 0):

3(t² - 1) = 0 ⇒ t = ±1

Check t = 1: dy/dt = 2 ≠ 0 ✓

Point: (x(1), y(1)) = (-2, -3)

Check t = -1: dy/dt = -2 ≠ 0 ✓

Point: (x(-1), y(-1)) = (2, -3)

4 Summary:

Horizontal tangent at (0, -4)

Vertical tangents at (-2, -3) and (2, -3)

💡 Graphical Insight: This is a parametric representation of a curve with two cusps and one flat point.

📊 Parametric Derivatives Quick Reference

Type	Condition	Interpretation
Horizontal Tangent	dy/dt = 0, dx/dt ≠ 0	Slope = 0
Vertical Tangent	dx/dt = 0, dy/dt ≠ 0	Slope = ∞
Cusp	dx/dt = dy/dt = 0	Sharp point
Regular Point	dx/dt ≠ 0	Smooth curve