Parametric Equations- Calculus BC Practice Problems

What Are Parametric Equations?

Parametric equations define a curve by expressing both x and y as functions of a third variable, usually t. Instead of y = f(x), you get:

x = f(t) and y = g(t)

In Calculus BC, you'll need to find derivatives, tangent lines, and arc length for these curves. The good news? The process is straightforward once you know the steps.

Eliminating the Parameter

Before you can do much with parametric curves, you often need to eliminate t to get a standard Cartesian equation. Here's how:

Method: Solve One Equation for t, Substitute into the Other

Example: Given x = t + 1 and y = t², eliminate the parameter.

Solve for t: t = x - 1

Substitute: y = (x - 1)²

That's it. You now have the Cartesian form.

Derivatives of Parametric Equations

This is where most students struggle. The derivative dy/dx for a parametric curve is NOT dy/dt divided by dx/dt as separate fractions. You must combine them:

dy/dx = (dy/dt) / (dx/dt)

You divide dy/dt by dx/dt. The t's cancel out, leaving you with dy/dx in terms of t.

Key Rule

If dx/dt = 0 at a point, the tangent is vertical, not horizontal. If dy/dt = 0, the tangent is horizontal.

Never set dy/dx = 0 to find horizontal tangents. Check dy/dt = 0 instead.

Calculus BC Practice Problems

Problem 1

Given x = 3t - 2 and y = 2t² + 1, find dy/dx at t = 1.

Solution:

dx/dt = 3

dy/dt = 4t

dy/dx = (4t) / 3

At t = 1: dy/dx = 4/3

Problem 2

Find the equation of the tangent line to x = cos(t), y = sin(t) at t = π/4.

Solution:

dx/dt = -sin(t)

dy/dt = cos(t)

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

At t = π/4: dy/dx = -cot(π/4) = -1

Point: x = cos(π/4) = √2/2, y = sin(π/4) = √2/2

Point-slope form: y - √2/2 = -1(x - √2/2)

Problem 3

Find dy/dx and d²y/dx² for x = t³ - 2t and y = t² + 3t.

Solution:

dx/dt = 3t² - 2

dy/dt = 2t + 3

First derivative: dy/dx = (2t + 3) / (3t² - 2)

For the second derivative, use:

d²y/dx² = d/dt [dy/dx] ÷ dx/dt

d²y/dx² = [d/dt ((2t + 3)/(3t² - 2))] / (3t² - 2)

This requires quotient rule on the numerator. The algebra gets messy—don't panic. Work through it step by step.

Arc Length of Parametric Curves

The arc length formula for parametric curves is:

L = ∫ from a to b √[(dx/dt)² + (dy/dt)²] dt

Notice this is not the same as the arc length formula in rectangular coordinates. The dy/dx term doesn't appear.

Problem 4

Find the arc length of x = t², y = t³ from t = 0 to t = 2.

Solution:

dx/dt = 2t

dy/dt = 3t²

√[(2t)² + (3t²)²] = √[4t² + 9t⁴] = t√(4 + 9t²)

L = ∫₀² t√(4 + 9t²) dt

Use substitution: u = 4 + 9t², du = 18t dt, so t dt = du/18

When t = 0, u = 4. When t = 2, u = 4 + 36 = 40

L = ∫₄⁴⁰ √u · (1/18) du = (1/18) ∫₄⁴⁰ u^(1/2) du

L = (1/18) · (2/3) [u^(3/2)]₄⁴⁰

L = (1/27) [40^(3/2) - 4^(3/2)]

L = (1/27) [40√40 - 8]

Area Bounded by Parametric Curves

For area under a parametric curve:

A = ∫ y dx = ∫ y (dx/dt) dt

The limits are in terms of t, not x.

Problem 5

Find the area under x = t², y = t from t = 0 to t = 2.

Solution:

A = ∫₀² (t)(2t) dt = ∫₀² 2t² dt = [2t³/3]₀² = 16/3

Parametric Equations Cheat Sheet

Common Mistakes to Avoid

• Forgetting to divide when finding dy/dx—just multiplying dy/dt and dx/dt together
• Using the wrong arc length formula—parametric uses √[(dx/dt)² + (dy/dt)²], not √[1 + (dy/dx)²]
• Setting dy/dx = 0 to find horizontal tangents—check dy/dt = 0 instead
• Forgetting to convert limits when switching between x and t
• Not checking if dx/dt = 0 when finding tangent slopes—vertical tangents exist

Getting Started

Work through these steps for any parametric problem:

1. Find dx/dt and dy/dt first—write them down
2. Decide what you're solving for: derivative, area, or arc length
3. Use the corresponding formula from the table above
4. Substitute and integrate or evaluate as needed
5. Check your work: are units consistent? Does your answer make sense?

Parametric equations test your ability to switch between coordinate systems. Master the conversions and the rest follows.