Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Polar Coordinates

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Tangents of Polar Curves (Derivatives)
First Derivative
(a.k.a slope of the tangent to the curve)
dydx=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ   or   y′(θ)x′(θ)\displaystyle \frac{dy}{dx}=\frac{\displaystyle \frac{dr}{d\theta}\sin\theta+r\cos\theta}{\displaystyle \frac{dr}{d\theta}\cos\theta-r\sin\theta}\ \ \ \text{or}\ \ \ \frac{y'\left(\theta\right)}{x'\left(\theta\right)}dxdy​=dθdr​cosθ−rsinθdθdr​sinθ+rcosθ​   or   x′(θ)y′(θ)​ 
*Recall that y=rsin⁡θy=r\sin\thetay=rsinθ, x=rcos⁡θx=r\cos\thetax=rcosθ

The polar curve has
A Horizontal Tangent when
drdθsin⁡θ+rcos⁡θ   or   y′(θ)=0\displaystyle \frac{dr}{d\theta}\sin\theta+r\cos\theta\ \ \ \text{or}\ \ \ y'\left(\theta\right)=0dθdr​sinθ+rcosθ   or   y′(θ)=0 and
drdθcos⁡θ−rsin⁡θ   or   x′(θ)≠0\displaystyle \frac{dr}{d\theta}\cos\theta-r\sin\theta\ \ \ \text{or}\ \ \ x'\left(\theta\right)\ne0dθdr​cosθ−rsinθ   or   x′(θ)=0
A Vertical Tangent when
drdθcos⁡θ−rsin⁡θ   or   x′(θ)=0\displaystyle \frac{dr}{d\theta}\cos\theta-r\sin\theta\ \ \ \text{or}\ \ \ x'\left(\theta\right)=0dθdr​cosθ−rsinθ   or   x′(θ)=0 and
drdθsin⁡θ+rcos⁡θ   or   y′(θ)≠0\displaystyle \frac{dr}{d\theta}\sin\theta+r\cos\theta\ \ \ \text{or}\ \ \ y'\left(\theta\right)\ne0dθdr​sinθ+rcosθ   or   y′(θ)=0
*If both y′(θ)=0y'\left(\theta\right)=0y′(θ)=0 and x′(θ)=0x'\left(\theta\right)=0x′(θ)=0, use L'Hosputal's rule

Practice Question
Find the slope of the tangent line to the curve r=sec⁡2θcsc⁡θr = \sec^2 \theta \csc\theta 
  r=sec2θcscθ at θ=π/6.\theta = \pi/6. θ=π/6.

Practice: Tangents
Consider the polar curve r=eθr=e^{\theta}r=eθ where θ∈[0,2π]\theta\in\left[0,2\pi\right]θ∈[0,2π].
a) Find the slope of the tangent line to the curve at θ=π\theta=\piθ=π.
b) Find all points at which the tangent line to the curve is horizontal.
c) Find all points at which the tangent line to the curve is vertical.

a.) Slope is 0
b.) θ=π4, 5π4\theta=\frac{\pi}{4},\ \frac{5\pi}{4}θ=4π​, 45π​
c.) θ=3π4, 7π4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4}θ=43π​, 47π​

a.) Slope is 0
b.) θ=3π4, 7π4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4}θ=43π​, 47π​
c.) θ=π4, 5π4\theta=\frac{\pi}{4},\ \frac{5\pi}{4}θ=4π​, 45π​

a.) Slope is 1
b.) θ=π4, 5π4\theta=\frac{\pi}{4},\ \frac{5\pi}{4}θ=4π​, 45π​
c.) θ=3π4, 7π4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4}θ=43π​, 47π​

a.) Slope is 1
b.) θ=3π4, 7π4\theta=\frac{3\pi}{4},\ \frac{7\pi}{4}θ=43π​, 47π​
c.) θ=π4, 5π4\theta=\frac{\pi}{4},\ \frac{5\pi}{4}θ=4π​, 45π​

a.) Slope is 1
b.) θ=π4\theta=\frac{\pi}{4}θ=4π​
c.) θ=3π4, 5π4,7π4\theta=\frac{3\pi}{4},\ \frac{5\pi}{4},\frac{7\pi}{4}θ=43π​, 45π​,47π​

I don't know

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Applications of Polar Curves
Consider the polar curve r=r(θ)r=r\left(\theta\right)r=r(θ) where a≤θ≤ba\le\theta\le ba≤θ≤b. Suppose that f′(θ)f'\left(\theta\right)f′(θ) is continuous on [a,b]\left[a,b\right][a,b] and the curve is traversed exactly once on [a,b]\left[a,b\right][a,b].
Arc Length
L=∫ab(r(θ))2+r′(θ)2  ⁣dθL={\displaystyle\int}^b_a
\sqrt{\left(r(\theta)\right)^2+r'(\theta )^2}\de{\theta}L=∫ab​(r(θ))2+r′(θ)2​ dθ

Areas Under the Curve
A=∫ab12[r(θ)]2  ⁣dθA={\displaystyle\int}^b_a\frac{1}{2}[r(\theta)]^2\de{\theta}A=∫ab​21​[r(θ)]2 dθ

Area Bounded Between Two Polar Curves
A=12∫ab[(router)2−(rinner)2]  ⁣dθA=\frac{1}{2}{\displaystyle\int}^b_a[(r_{outer})^2-(r_{inner})^2]\de{\theta}A=21​∫ab​[(router​)2−(rinner​)2] dθ

Wize Tip
Always graph the curves before using the formula!

Practice: Arc Length
Compute the total length of the curve r=2acos⁡θr = 2a \cos \theta r=2acosθ for a>0a>0a>0 and θ∈[0,π]\theta\in\left[0,\pi\right]θ∈[0,π]

Practice Question
Write the integral representing the area of the inner loop of r=2+4sin⁡θ.r = 2 + 4 \sin \theta. 
  r=2+4sinθ.

A=12∫π/65π/6(4sin⁡θ)2  ⁣dθA=\frac{1}{2}{\displaystyle\int}^{5\pi/6}_{\pi/6}(4\sin\theta)^2\de{\theta}A=21​∫π/65π/6​(4sinθ)2 dθ

A=2∫π/65π/6(4sin⁡θ)2  ⁣dθA=2{\displaystyle\int}^{5\pi/6}_{\pi/6}(4\sin\theta)^2\de{\theta}A=2∫π/65π/6​(4sinθ)2 dθ

A=12∫π/65π/6(2+4sin⁡θ)2  ⁣dθA=\frac{1}{2}{\displaystyle\int}^{5\pi/6}_{\pi/6}(2+4\sin\theta)^2\de{\theta}A=21​∫π/65π/6​(2+4sinθ)2 dθ

A=12∫7π/611π/6(2+4sin⁡θ)2  ⁣dθA=\frac{1}{2}{\displaystyle\int}^{11\pi/6}_{7\pi/6}(2+4\sin\theta)^2\de{\theta}A=21​∫7π/611π/6​(2+4sinθ)2 dθ

A=2∫7π/611π/6(2+4sin⁡θ)2  ⁣dθA=2{\displaystyle\int}^{11\pi/6}_{7\pi/6}(2+4\sin\theta)^2\de{\theta}A=2∫7π/611π/6​(2+4sinθ)2 dθ

I don't know

Practice: Area Between Polar Curves
Find the area of the region lying inside the limacon curve of equation r=1+2cosθr = 1 + 2 cos \thetar=1+2cosθ, but outside the circle rrr = 2. 

53+π2\frac{5\sqrt{3}+\pi}{2}253​+π​

53−π2\frac{5\sqrt{3}-\pi}{2}253​−π​

53−π3\frac{5\sqrt{3}-\pi}{3}353​−π​

53+π3\frac{5\sqrt{3}+\pi}{3}353​+π​

532−π3\frac{5\sqrt{3}}{2}-\frac{\pi}{3}253​​−3π​

I don't know

Extra Practice

Practice: Polar Curve Tangent

Find the slope of the tangent line to the curve r=sec⁡2θcsc⁡θr = \sec^2 \theta \csc\theta 
  r=sec2θcscθ at θ=π/6.\theta = \pi/6. θ=π/6.

Practice: Area inside Polar Curves

Practice Question
Write the integral representing the area of the inner loop of r=2+4sin⁡θ.r = 2 + 4 \sin \theta. 
  r=2+4sinθ.

Practice: Polar Curve Tangent

Practice Question
Find the slope of the tangent line to the curve r=sec⁡2θcsc⁡θr = \sec^2 \theta \csc\theta 
  r=sec2θcscθ at θ=π/6.\theta = \pi/6. θ=π/6.

Practice: Polar Curve Tangent

Practice Question
Find the slope of the tangent line to the curve r=sec⁡2θcsc⁡θr = \sec^2 \theta \csc\theta 
  r=sec2θcscθ at θ=π/6.\theta = \pi/6. θ=π/6.

Practice: Area inside Polar Curves

Practice Question
Write the integral representing the area of the inner loop of r=2+4sin⁡θ.r = 2 + 4 \sin \theta. 
  r=2+4sinθ.

Practice: Area inside Polar Curves

Practice Question
Write the integral representing the area of the inner loop of r=2+4sin⁡θ.r = 2 + 4 \sin \theta. 
  r=2+4sinθ.

Practice: Area inside Polar Curves

Practice Question
Write the integral representing the area of the inner loop of r=2+4sin⁡θ.r = 2 + 4 \sin \theta. 
  r=2+4sinθ.

Practice: Polar Curve Tangent

Practice Question
Find the slope of the tangent line to the curve r=sec⁡2θcsc⁡θr = \sec^2 \theta \csc\theta 
  r=sec2θcscθ at θ=π/6.\theta = \pi/6. θ=π/6.

Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Polar Coordinates

Popular Courses

Tangents of Polar Curves (Derivatives)

First Derivative

Practice Question

Practice: Tangents

Applications of Polar Curves

Arc Length $L={\displaystyle\int}^b_a \sqrt{\left(r(\theta)\right)^2+r'(\theta )^2}\de{\theta}$

Areas Under the Curve $A={\displaystyle\int}^b_a\frac{1}{2}[r(\theta)]^2\de{\theta}$

Area Bounded Between Two Polar Curves

$A=\frac{1}{2}{\displaystyle\int}^b_a[(r_{outer})^2-(r_{inner})^2]\de{\theta}$

Practice: Arc Length

Practice: Area Under Polar Curve

Practice Question

Practice: Area Between Polar Curves

Extra Practice

Practice: Polar Curve Tangent

Practice: Area inside Polar Curves

Practice: Polar Curve Tangent

Practice: Polar Curve Tangent

Practice: Area inside Polar Curves

Practice: Area inside Polar Curves

Practice: Area inside Polar Curves

Practice: Polar Curve Tangent

Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Polar Coordinates

Popular Courses

Tangents of Polar Curves (Derivatives)

First Derivative

Practice Question

Practice: Tangents

Applications of Polar Curves

Arc Length L=∫ab(r(θ))2+r′(θ)2 ⁣dθL={\displaystyle\int}^b_a \sqrt{\left(r(\theta)\right)^2+r'(\theta )^2}\de{\theta}L=∫ab​(r(θ))2+r′(θ)2​ dθ

Areas Under the Curve A=∫ab12[r(θ)]2 ⁣dθA={\displaystyle\int}^b_a\frac{1}{2}[r(\theta)]^2\de{\theta}A=∫ab​21​[r(θ)]2 dθ

Area Bounded Between Two Polar Curves

A=12∫ab[(router)2−(rinner)2] ⁣dθA=\frac{1}{2}{\displaystyle\int}^b_a[(r_{outer})^2-(r_{inner})^2]\de{\theta}A=21​∫ab​[(router​)2−(rinner​)2] dθ

Practice: Arc Length

Practice: Area Under Polar Curve

Practice Question

Practice: Area Between Polar Curves

Extra Practice

Practice: Polar Curve Tangent

Practice: Area inside Polar Curves

Practice: Polar Curve Tangent

Practice: Polar Curve Tangent

Practice: Area inside Polar Curves

Practice: Area inside Polar Curves

Practice: Area inside Polar Curves

Practice: Polar Curve Tangent

Arc Length $L={\displaystyle\int}^b_a \sqrt{\left(r(\theta)\right)^2+r'(\theta )^2}\de{\theta}$

Areas Under the Curve $A={\displaystyle\int}^b_a\frac{1}{2}[r(\theta)]^2\de{\theta}$

$A=\frac{1}{2}{\displaystyle\int}^b_a[(r_{outer})^2-(r_{inner})^2]\de{\theta}$