You CAN Ace Calculus

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

On this page we cover the most common calculus problems involving polar coordinates. The two main topics are differentiation and integration.
As mentioned on the main polar coordinates page, polar coordinates are just parametric equations. If you are familiar with parametric equations, this material should be very intuitive.

Differentiation - Slope and Tangent Lines

In order to find slope and, subsequently, a tangent line to a smooth graph, we need to be able to calculate the derivative of a parametric equation.

We have the graph represented by the parametric equations \( x = r(\theta)\cos(\theta) \) and \( y = r(\theta)\sin(\theta) \). Here we are emphasizing that r is a a function of \(\theta\) by writing \(r(\theta)\). The derivative is given by

\(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} }\)     where \( r' = dr/d\theta \).

Polar Slope Proof

Theorem: Polar Slope

For a set of polar equations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) where \(r\) is a function of \(\theta\), the slope \(dy/dx\) of a smooth function is given by

\(\displaystyle{\frac{dy}{dx} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)}}\)

where \( r' = dr/d\theta \)

To prove this, we use the theorem of parametrics derivative, the result of which is \(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }\)

In our case, the parameter is \(\theta\) and our parametric equations are \( x(\theta) = r(\theta)\cos(\theta) \) and \( y(\theta) = r(\theta)\sin(\theta) \). So our parametric derivative is \(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} }\)

Using the chain rule, we calculate
\(\displaystyle{\frac{dy}{d\theta} = r'\sin(\theta) + r \cos(\theta)}\) and \(\displaystyle{ \frac{dx}{d\theta} = r'\cos(\theta) - r \sin(\theta)}\)

Putting these results in the parametric derivative equation, we have

\(\displaystyle{ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{r'\sin(\theta) + r \cos(\theta)}{r'\cos(\theta) - r \sin(\theta)} }\) [qed]

This equation gives us the slope of the graph at every point. Once we have the slope, we can calculate the equation of a tangent line using the technique found on the main tangent page. Before we go on, let's work some practice problems.

Basic Problems

Calculate the slope of the tangent line to the curve \(r=4\) at \(\theta=\pi/4\).

Problem Statement

Calculate the slope of the tangent line to the curve \(r=4\) at \(\theta=\pi/4\).

Solution

1269 solution video

video by PatrickJMT

close solution

Calculate the slope of the tangent line to the curve \(r=1+5\cos\theta\) at \(\theta=\pi/4\).

Problem Statement

Calculate the slope of the tangent line to the curve \(r=1+5\cos\theta\) at \(\theta=\pi/4\).

Solution

1270 solution video

video by PatrickJMT

close solution

Calculate the slope of the tangent line to the curve \(r=\sin(5\theta)\) at \(\theta=\pi/5\).

Problem Statement

Calculate the slope of the tangent line to the curve \(r=\sin(5\theta)\) at \(\theta=\pi/5\).

Solution

1271 solution video

video by PatrickJMT

close solution

Find the tangent line to the curve \(r=\cos(2\theta)\) at \(\theta=\pi/4\).

Problem Statement

Find the tangent line to the curve \(r=\cos(2\theta)\) at \(\theta=\pi/4\).

Solution

1279 solution video

video by Krista King Math

close solution

Intermediate Problems

Find the points on the polar curve \(r=3\cos(\theta)\) where the graph of the tangent line is vertical or horizontal.

Problem Statement

Find the points on the polar curve \(r=3\cos(\theta)\) where the graph of the tangent line is vertical or horizontal.

Solution

1280 solution video

video by Krista King Math

close solution

Integration - Area

To calculate the area defined by a polar curve r, which is a function of \(\theta\), between the rays \(\theta = \theta_0\) and \(\theta = \theta_1\), we use the integral

\(\displaystyle{A = \frac{1}{2} \int_{\theta_0}^{\theta_1}{ r^2 d\theta}}\)

Here is a good in-depth video discussing area using polar coordinates.

MIT OCW - Lec 33 | MIT 18.01 Single Variable Calculus, Fall 2007 [35min]

video by MIT OCW

Basic Problems

Calculate the area enclosed by the polar curve \(r=2\cos\theta\).

Problem Statement

Calculate the area enclosed by the polar curve \(r=2\cos\theta\).

Solution

1383 solution video

video by Krista King Math

close solution

Calculate the area bounded by one loop of \(r=2\cos(4\theta)\).

Problem Statement

Calculate the area bounded by one loop of \(r=2\cos(4\theta)\).

Solution

1384 solution video

video by Krista King Math

close solution

Calculate the area enclosed by one loop of \(r=\sin(4\theta)\).

Problem Statement

Calculate the area enclosed by one loop of \(r=\sin(4\theta)\).

Solution

1272 solution video

video by PatrickJMT

close solution

Calculate the area enclosed by \(r=3+\sin\theta\).

Problem Statement

Calculate the area enclosed by \(r=3+\sin\theta\).

Solution

1276 solution video

video by PatrickJMT

close solution

Intermediate Problems

Calculate the area inside \(r=1+2\cos(\theta)\) and outside \(r=2\).

Problem Statement

Calculate the area inside \(r=1+2\cos(\theta)\) and outside \(r=2\).

Solution

1382 solution video

video by Krista King Math

close solution

Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).

Problem Statement

Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).

Final Answer

\(\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }\)

Problem Statement

Calculate the area inside both \(r=\cos\theta\) and \(r=\sin(2\theta)\).

Solution

The final answer at the end of the video is incorrect. At about the 11:26 mark he notes his mistake.

1273 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \frac{\pi}{4} - \frac{3\sqrt{3}}{16} }\)

close solution

Calculate the region that lies inside \(r=1+\sin\theta\) and outside \(r=1\).

Problem Statement

Calculate the region that lies inside \(r=1+\sin\theta\) and outside \(r=1\).

Solution

1274 solution video

video by PatrickJMT

close solution

Calculate the area of the region inside both \(r=9\cos\theta\) and \(r=9\sin\theta\).

Problem Statement

Calculate the area of the region inside both \(r=9\cos\theta\) and \(r=9\sin\theta\).

Solution

1275 solution video

video by PatrickJMT

close solution

Calculate the area outside \(r=\sin\theta\) and inside \(r=2\sin\theta\).

Problem Statement

Calculate the area outside \(r=\sin\theta\) and inside \(r=2\sin\theta\).

Solution

1277 solution video

video by PatrickJMT

close solution

Calculate the area of the region inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\).

Problem Statement

Calculate the area of the region inside both \(r=\sin(2\theta)\) and \(r=\cos(2\theta)\).

Solution

1278 solution video

video by Krista King Math

close solution

Integration - Arc Length

To calculate the arc length of a smooth curve, we can use this integral.

\(\displaystyle{s = \int_{\theta_0}^{\theta_1}{\sqrt{r^2 + [r']^2} ~ d\theta}}\)
where r is a function of \(\theta\), \( r' = dr/d\theta \) and we are looking at the arc from \( \theta = \theta_0 \) to \( \theta = \theta_1 \).

Basic Problems

Calculate the arc length of the polar curve \( r=e^{\theta/2} \), \(0\leq\theta\leq4\pi\).

Problem Statement

Calculate the arc length of the polar curve \( r=e^{\theta/2} \), \(0\leq\theta\leq4\pi\).

Solution

1385 solution video

video by Krista King Math

close solution

Intermediate Problems

Calculate the arc length of the polar curve \(r=\sin^2(\theta/2)\), \(0\leq\theta\leq\pi\).

Problem Statement

Calculate the arc length of the polar curve \(r=\sin^2(\theta/2)\), \(0\leq\theta\leq\pi\).

Solution

1386 solution video

video by Krista King Math

close solution

Integration - Surface Area

To calculate the surface area of a polar curve revolved about an axis, we use these integrals.

The equation of the polar curve is in the form \( x=r(\theta) \cos(\theta) \) and \( y = r(\theta) \sin(\theta) \) and the curve that is being revolved is defined to be from \(\theta_0\) to \(\theta_1\). We also define \( ds = \sqrt{r^2 + [dr/d\theta]^2}~dt \).

surface area

rotation about the x-axis
(the polar axis)

\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{y~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \sin(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\)

rotation about the y-axis

\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{x~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \cos(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\)

So, why the \(ds\) term? If you look at the previous section on arc length, you will notice that this term appears in that integral. This is called a differential length and is just a convenient way of writing these integrals.

Intermediate Problems

Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the x-axis.

Problem Statement

Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the x-axis.

Solution

1268 solution video

video by Krista King Math

close solution
Real Time Web Analytics