# Parametric and Polar Practice Problems with Answers

> 12 parametric and polar practice problems with answers: dy/dx, tangent lines, arc length, converting coordinates, polar slopes and polar area.

Source: https://practicecalculus.com/practice-problems/parametric-and-polar
Updated: 2026-11-06

$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus BC, Unit 9
- **You need:** The [chain rule](/rules/chain-rule), trig values and [trig integrals](/practice-problems/trig-integrals)
- **Main idea:** Parametric: divide the $t$-derivatives. Polar: use $x = r\cos\theta$, $y = r\sin\theta$
- **Practice with new numbers:** [Polar area lesson](/learn/calculus-2/power-series-parametric-polar/polar-area)

## Key takeaways

- For parametric curves, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, and $\frac{d^2y}{dx^2}$ is the $t$-derivative of $\frac{dy}{dx}$ divided by $\frac{dx}{dt}$ again.
- Convert with $x = r\cos\theta$, $y = r\sin\theta$ and $r^2 = x^2 + y^2$.
- Polar area is $\frac{1}{2}\int r^2\,d\theta$. Choose limits that trace the region exactly once.

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:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad L = \int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt, \qquad A = \frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta$$

## 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

| Problem | Answer |
| --- | --- |
| 1 | $\frac{dy}{dx} = \frac{3t}{2}$; at $t = 2$, slope $3$ |
| 2 | $y = 3x - 4$ |
| 3 | $\frac{x^2}{4} + \frac{y^2}{9} = 1$ |
| 4 | $\frac{3}{4t}$ |
| 5 | $(0, -2)$ and $(0, 2)$ |
| 6 | $\frac{2}{3}\left(2\sqrt{2} - 1\right) \approx 1.219$ |
| 7 | $\left(2, 2\sqrt{3}\right)$ |
| 8 | $x^2 + (y - 2)^2 = 4$, a circle |
| 9 | $1$ |
| 10 | $\pi$ |
| 11 | $\frac{\pi}{8}$ |
| 12 | $\frac{3\pi}{2}$ |

## Step-by-step solutions

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

$$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}, \qquad \text{at } t = 2: \ 3$$

**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 $\cos^2 t + \sin^2 t = 1$:

$$\frac{x^2}{4} + \frac{y^2}{9} = 1$$

This is an ellipse.

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

$$\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$$

**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 $u = 1 + t^2$:

$$\int_0^1 2t\sqrt{1 + t^2}\,dt = \left[\frac{2}{3}(1 + t^2)^{3/2}\right]_0^1 = \frac{2}{3}\left(2\sqrt{2} - 1\right)$$

**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:

$$x^2 + (y - 2)^2 = 4$$

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 $r' = -1$.

$$\frac{dy}{d\theta} = r'\sin\theta + r\cos\theta = -1, \qquad \frac{dx}{d\theta} = r'\cos\theta - r\sin\theta = -1$$

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

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

$$\frac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta = 2\cdot\frac{\pi}{2} = \pi$$

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 $\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}$:

$$\frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)\,d\theta = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$$

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

$$\frac{1}{2}(2\pi + 0 + \pi) = \frac{3\pi}{2}$$

## 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](/learn/calculus-2/power-series-parametric-polar/parametric-curves), the [parametric arc length lesson](/learn/calculus-2/power-series-parametric-polar/parametric-arc-length), the [polar slopes lesson](/learn/calculus-2/power-series-parametric-polar/polar-slopes) and the [polar area lesson](/learn/calculus-2/power-series-parametric-polar/polar-area).

## Frequently asked questions

### How do you find the second derivative of a parametric curve?

Differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$. It is not $\frac{d^2y/dt^2}{d^2x/dt^2}$. Problem 4 shows the steps.

### Why is polar area one half r squared?

A thin slice of a polar region is almost a sector of a circle with radius $r$ and angle $d\theta$. A sector has area $\frac{1}{2}r^2\,d\theta$, and integrating adds up the slices.

### Where does a parametric curve have a horizontal tangent?

Where $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. A vertical tangent is where $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$. If both are zero, you need more work.