Polar Derivatives | Math Coach Amy

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Key Ideas Easy Medium Hard

\(\dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos\theta-r\sin\theta\)

\(\dfrac{dy}{d\theta}=\dfrac{dr}{d\theta}\sin\theta+r\cos\theta\)

What to Remember

Convert, then use a parametric slope

\(x=r\cos\theta\) and \(y=r\sin\theta\).

\(\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}\)

The product rule on each coordinate

\(\dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos\theta-r\sin\theta\)

\(\dfrac{dy}{d\theta}=\dfrac{dr}{d\theta}\sin\theta+r\cos\theta\)

Horizontal and vertical in polar form

Horizontal when \(dy/d\theta=0\) and \(dx/d\theta\neq 0\). Vertical when \(dx/d\theta=0\) and \(dy/d\theta\neq 0\).

At the pole, r is zero

A point where \(r=0\) is the origin. The slope formula can still have a value there. Compute the two \(\theta\)-derivatives instead of plugging \((0,0)\) into an \(x\)-\(y\) formula you do not have.

Level 1

Easy

r is constant, or the angle is a special value.

Example

For the circle \(r=2\), find \(dy/dx\) at \(\theta=\pi/3\).

\(x=2\cos\theta\) and \(y=2\sin\theta\).

\(\dfrac{dy}{dx}=\dfrac{2\cos\theta}{-2\sin\theta}=-\cot\theta\)

At \(\pi/3\): \(-\dfrac{1}{\sqrt{3}}\).

Answer: \(-\dfrac{1}{\sqrt{3}}\)

For \(r=4\), where is the tangent horizontal?

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\(dy/d\theta=4\cos\theta=0\) when \(\cos\theta=0\).

In \([0,2\pi)\), \(\theta=\pi/2\) and \(\theta=3\pi/2\).

There \(dx/d\theta=-4\sin\theta\) is \(-4\) or \(4\), not zero.

Answer: \(\theta=\pi/2\) and \(3\pi/2\)

For the ray \(\theta=\pi/6\) in rectangular language, the line through the origin has slope \(\tan(\pi/6)\). Find that slope.

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\(\dfrac{1}{\sqrt{3}}\)

Answer: \(\dfrac{1}{\sqrt{3}}\)

Convert \(r=6\cos\theta\) into a rectangular equation by multiplying by \(r\) is a known circle trick. First find \(r\) at \(\theta=0\) and the rectangular point.

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\(r=6\)

\(x=6\) and \(y=0\).

Answer: \((6,0)\)

For \(r=\theta\), find \(dy/d\theta\) and \(dx/d\theta\) in general.

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\(x=\theta\cos\theta\) and \(y=\theta\sin\theta\).

\(dx/d\theta=\cos\theta-\theta\sin\theta\)

\(dy/d\theta=\sin\theta+\theta\cos\theta\)

Answer: \(\cos\theta-\theta\sin\theta\) and \(\sin\theta+\theta\cos\theta\)

Use those derivatives to find the slope of \(r=\theta\) at \(\theta=0\).

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\(dx/d\theta=1\) and \(dy/d\theta=0\).

Slope 0.

Answer: \(0\)

Level 2

Medium

A specific angle on a cardioid or a cosine circle.

Example

For \(r=2\cos\theta\), find \(dy/dx\) at \(\theta=\pi/3\).

\(dr/d\theta=-2\sin\theta\) and \(r=1\) at this angle.

\(dx/d\theta=-2\sin\theta\cos\theta-r\sin\theta\)

At \(\pi/3\): \(-2\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}-1\cdot\dfrac{\sqrt{3}}{2}=-\sqrt{3}\).

\(dy/d\theta=-2\sin^2\theta+r\cos\theta\)

At \(\pi/3\): \(-2\cdot\dfrac{3}{4}+1\cdot\dfrac{1}{2}=-1\).

\(\dfrac{dy}{dx}=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}\)

Answer: \(\dfrac{1}{\sqrt{3}}\)

For \(r=1+\cos\theta\), find the slope at \(\theta=\pi/2\) and the rectangular point.

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\(r=1\) and \(dr/d\theta=-1\).

\(dx/d\theta=(-1)(0)-1\cdot 1=-1\)

\(dy/d\theta=(-1)(1)+1\cdot 0=-1\)

Slope \(1\).

Point: \(x=0\), \(y=1\).

Answer: slope 1 at \((0,1)\)

For \(r=\theta\), find \(dy/dx\) at \(\theta=\pi/4\).

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\(\dfrac{dy}{dx}=\dfrac{1+\theta}{1-\theta}\) after canceling \(\sqrt{2}/2\) at this angle.

At \(\pi/4\): \(\dfrac{1+\pi/4}{1-\pi/4}\).

Answer: \(\dfrac{1+\pi/4}{1-\pi/4}\)

Show the cancellation that produces that quotient.

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\(dx/d\theta=\dfrac{\sqrt{2}}{2}\left(1-\dfrac{\pi}{4}\right)\)

\(dy/d\theta=\dfrac{\sqrt{2}}{2}\left(1+\dfrac{\pi}{4}\right)\)

The common factor divides out.

Answer: \(\dfrac{1+\pi/4}{1-\pi/4}\)

For \(r=2\cos\theta\), \(r=0\) at \(\theta=\pi/2\). Find \(dx/d\theta\) and \(dy/d\theta\) there.

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\(dr/d\theta=-2\) and \(r=0\).

\(dx/d\theta=-2\cdot 0-0\cdot 1=0\)

\(dy/d\theta=-2\cdot 1+0\cdot 0=-2\)

The tangent is vertical at the origin.

Answer: vertical tangent at the origin

Which pair gives a horizontal tangent?

A. \(dy/d\theta=0\) and \(dx/d\theta\neq 0\)

B. \(dx/d\theta=0\) and \(dy/d\theta\neq 0\)

C. both zero

D. \(r=0\)

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Horizontal means the numerator of \(dy/dx\) is zero and the denominator is not.

Answer: A. \(dy/d\theta=0\) and \(dx/d\theta\neq 0\)

Level 3

Hard

A tangent line in the plane, using the polar slope.

Example

For \(r=1+\cos\theta\) at \(\theta=\pi/2\), the point is \((0,1)\) and the slope is 1. Write the tangent line.

\(y-1=1(x-0)\)

\(y=x+1\)

Answer: \(y=x+1\)

Find \(dy/dx\) for \(r=1+\cos\theta\) at \(\theta=\pi/3\).

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\(r=1+\dfrac{1}{2}=\dfrac{3}{2}\) and \(dr/d\theta=-\dfrac{\sqrt{3}}{2}\).

\(dx/d\theta=\left(-\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)-\dfrac{3}{2}\cdot\dfrac{\sqrt{3}}{2}\)

\(-\dfrac{\sqrt{3}}{4}-\dfrac{3\sqrt{3}}{4}=-\sqrt{3}\)

\(dy/d\theta=\left(-\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)+\dfrac{3}{2}\cdot\dfrac{1}{2}\)

\(-\dfrac{3}{4}+\dfrac{3}{4}=0\)

The tangent is horizontal.

Answer: horizontal, slope 0

Find the rectangular point for the previous tangent.

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\(x=\dfrac{3}{2}\cdot \dfrac{1}{2}=\dfrac{3}{4}\)

\(y=\dfrac{3}{2}\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{4}\)

Answer: \(\left(\dfrac{3}{4},\dfrac{3\sqrt{3}}{4}\right)\)

For \(r=2\sin\theta\), find the slope at \(\theta=\pi/6\).

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\(r=1\) and \(dr/d\theta=\sqrt{3}\).

\(dx/d\theta=\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}-1\cdot\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{1}{2}=1\)

\(dy/d\theta=\sqrt{3}\cdot\dfrac{1}{2}+1\cdot\dfrac{\sqrt{3}}{2}=\sqrt{3}\)

Slope \(\sqrt{3}\).

Answer: \(\sqrt{3}\)

Explain why the slope of \(r=2\) at \(\theta=0\) is undefined.

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At \(\theta=0\), \(dy/d\theta=2\) and \(dx/d\theta=0\).

The tangent is vertical. The point is \((2,0)\).

Answer: vertical tangent at \((2,0)\)

A polar curve passes through the origin when \(\theta=\alpha\). Why might \(dy/dx\) still have a value?

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\(x=r\cos\theta\) and \(y=r\sin\theta\) are parametric equations in \(\theta\).

The slope is the ratio of \(dy/d\theta\) and \(dx/d\theta\), which can be nonzero when \(r=0\).

Answer: use the derivatives with respect to \(\theta\)

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