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.
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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 
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 

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 

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 

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 

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 

video by Krista King Math 

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You CAN Ace Calculus
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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