\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Polar Coordinates - Slope and Tangent Lines

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On this page we cover a common calculus problem involving polar coordinates, determining slope and equation of tangent lines.
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.

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 \(r(\theta)\). Let's look at the derivative equation and the proof. Do not skip this proof. It is easy to understand and will help you understand when you can use this equation.

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. Let's work some practice problems.

Practice

Basic

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 video

video by PatrickJMT

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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 video

video by PatrickJMT

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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 video

video by PatrickJMT

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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 video

video by Krista King Math

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Intermediate

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 video

video by Krista King Math

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You CAN Ace Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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