Parametric and Polar Practice Problems with Answers

These parametric and polar practice problems cover slopes, tangent lines, second derivatives and arc length for parametric curves, then converting, slopes and area in polar coordinates. Try each one first. The answers come first, then full solutions.

The key formulas:

Problems

Parametric curves

1. For x = t^2, y = t^3, find \frac{dy}{dx} and its value at t = 2.
2. Find the tangent line to the curve in problem 1 at t = 2.
3. Eliminate the parameter: x = 2\cos t, y = 3\sin t.
4. For x = t^2, y = t^3, find \frac{d^2y}{dx^2}.
5. Find the points where x = t^2 - 1, y = t^3 - 3t has a horizontal tangent.
6. Find the arc length of x = t^2, y = \frac{2}{3}t^3 for 0 \le t \le 1.

Polar coordinates

7. Convert the polar point (r, \theta) = \left(4, \frac{\pi}{3}\right) to Cartesian coordinates.
8. Convert r = 4\sin\theta to a Cartesian equation and describe the curve.
9. Find the slope \frac{dy}{dx} of r = 1 + \cos\theta at \theta = \frac{\pi}{2}.

Polar area

10. Find the area inside r = 2\cos\theta.
11. Find the area of one petal of r = \cos(2\theta).
12. Find the area inside the cardioid r = 1 + \cos\theta.

Answers

Step-by-step solutions

1. \frac{dx}{dt} = 2t and \frac{dy}{dt} = 3t^2:

2. At t = 2 the point is (4, 8) and the slope is 3: y - 8 = 3(x - 4), so y = 3x - 4.

3. \cos t = \frac{x}{2} and \sin t = \frac{y}{3}. Use \cos^2 t + \sin^2 t = 1:

This is an ellipse.

4. Differentiate \frac{dy}{dx} = \frac{3t}{2} with respect to t to get \frac{3}{2}, then divide by \frac{dx}{dt} = 2t:

5. \frac{dy}{dt} = 3t^2 - 3 = 0 at t = \pm 1, and \frac{dx}{dt} = 2t \neq 0 there. At t = 1: (0, -2). At t = -1: (0, 2).

6. \frac{dx}{dt} = 2t and \frac{dy}{dt} = 2t^2, so the speed is \sqrt{4t^2 + 4t^4} = 2t\sqrt{1 + t^2} for t \ge 0. With u = 1 + t^2:

7. x = 4\cos\frac{\pi}{3} = 2 and y = 4\sin\frac{\pi}{3} = 2\sqrt{3}.

8. Multiply both sides by r: r^2 = 4r\sin\theta, so x^2 + y^2 = 4y. Complete the square:

A circle of radius 2 centered at (0, 2).

9. Use x = r\cos\theta, y = r\sin\theta with r' = -\sin\theta. At \theta = \frac{\pi}{2}: r = 1 and r' = -1.

So \frac{dy}{dx} = \frac{-1}{-1} = 1.

10. The circle is traced once for -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}:

This matches a circle of radius 1.

11. One petal runs from \theta = -\frac{\pi}{4} to \frac{\pi}{4}, where \cos 2\theta \ge 0. Use \cos^2 2\theta = \frac{1 + \cos 4\theta}{2}:

12. Expand (1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta. Over [0, 2\pi], these integrate to 2\pi, 0 and \pi:

Common mistakes on these problems

• Dividing second derivatives. In problem 4, \frac{d^2y/dt^2}{d^2x/dt^2} = \frac{6t}{2} = 3t is wrong.
• Using \frac{dr}{d\theta} as the slope. In problem 9, r' = -1 but the slope of the curve is 1.
• Limits that trace a curve twice. r = 2\cos\theta on [0, 2\pi] goes around twice and gives 2\pi.
• Forgetting the \frac{1}{2}. Polar area is \frac{1}{2}\int r^2\,d\theta, not \int r\,d\theta.

Keep practicing

For unlimited problems with instant feedback, try the parametric curves lesson, the parametric arc length lesson, the polar slopes lesson and the polar area lesson.