Calculus with polar curves

05 Theory - Polar tangent lines, arclength

Theory 1

Polar arclength formula

The arclength of the polar graph of , for :

To derive this formula, convert to Cartesian with parameter :

From here you can apply the familiar arclength formula with in the place of .

Extra - Derivation of polar arclength formula

Let and convert to parametric Cartesian, so:

Then:

Therefore:

Therefore:

Link to original

06 Illustration

Example - Length of the inner loop

Length of the inner loop

Consider the limaçon given by .

How long is the inner loop? Set up an integral for this quantity.

Solution

The inner loop is traced by the moving point when . This can be seen from the graph:

center

Therefore the length of the inner loop is given by this integral:

Link to original

07 Theory - Polar area

Theory 2

Sectorial area from polar curve

The “area under the curve” concept for graphs of functions in Cartesian coordinates translates to a “sectorial area” concept for polar graphs. To compute this area using an integral, we divide the region into Riemann sums of small sector slices.

center

center

To obtain a formula for the whole area, we need a formula for the area of each sector slice.

Area of sector slice

Let us verify that the area of a sector slice is .

center

Take the angle in radians and divide by to get the fraction of the whole disk.

Then multiply this fraction by (whole disk area) to get the area of the sector slice.

Now use and for an infinitesimal sector slice, and integrate these to get the total area formula:


One easily verifies this formula for a circle.

Let be a constant. Then:


The sectorial area between curves:

Sectorial area between polar curves

Subtract after squaring, not before!

This aspect is not similar to the Cartesian version:

Link to original

08 Illustration

Area between circle and limaçon

Area between circle and limaçon

Find the area of the region enclosed between the circle and the limaçon .

Solution

First draw the region:

center

The two curves intersect at . Therefore the area enclosed is given by integrating over :

Link to original

Area of small loops

Area of small loops

Consider the following polar graph of :

center

Find the area of the shaded region.

Solution

Find bounds for one small loop. Lower left loop occurs first. This loop is when .

Now set up area integral:

Power-to-frequency conversion: with :

Link to original

Overlap area of circles

Overlap area of circles

Compute the area of the overlap between crossing circles. For concreteness, suppose one of the circles is given by and the other is given by .

Solution

Drawing of the overlap:

center

Notice: total overlap area = area of red region. Bounds for red region: .

Area formula applied to :

Power-to-frequency: :

Link to original

Complex algebra

Videos

Review Videos

Videos, Organic Chemistry Tutor

Link to original

01 Theory - Complex arithmetic

Theory 1

The complex numbers are sums of real and imaginary numbers. Every complex number can be written uniquely in ‘Cartesian’ form:

To add, subtract, scale, and multiply complex numbers, treat ‘’ like a constant.

Simplify the result using .

For example:


Complex conjugate

Every complex number has a complex conjugate:

For example:

In general, .

Conjugates are useful mainly because they eliminate imaginary parts:

In general:


Complex division

To divide complex numbers, use the conjugate to eliminate the imaginary part in the denominator.

For example, reciprocals:

More general fractions:

Multiplication preserves conjugation

For any :

Therefore, one can take products or conjugates in either order.

Link to original

02 Illustration

Example - Complex multiplication

Complex multiplication

Compute the products:

(a) (b)

Solution

(a)

Expand:

Simplify :


(b)

Expand:

Simplify :

Link to original

Example - Complex division

Complex division

Compute the following divisions of complex numbers:

(a) (b) (c) (d)

Solution

(a)

Conjugate is :

Simplify:


(b)

Conjugate is :


(c)

Factor out the :

Use :


(d)

Denominator conjugate is :

Simplify:

Link to original