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Workbook for Stewart’s "Calculus"

Cory Wilson

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Section 10.3 Polar Coordinates

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Subsection 10.3.1 Before Class

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Figure 74. Pre-Class Video 1
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Figure 75. Pre-Class Video 2
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Subsubsection 10.3.1.1 The Polar Coordinate System

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Question 10.3.1.
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The Cartesian coordinate system carries two pieces of information: a horizontal distance, x, and a vertical distance, y. If polar coordinates deal with circular objects, what two pieces of information should a polar coordinate carry?
Solution.
Radius and angle
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Example 10.3.2.
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Use the polar grid below to plot the points. Be sure to label the points.
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  1. 🔗
    (2,−π6)
  2. 🔗
    (−3,π)
  3. 🔗
    (1,π4)
  4. 🔗
    (4,−2π3)
  5. 🔗
    (−1,5π6)
  6. 🔗
    (2,0)
A blank polar grid
Solution.
Done in video, need to fix after "image in solution" issue is figured out.
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Example 10.3.3.
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Consider the polar point (1,π4) plotted above. This is not the only way of describing the point.
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  1. 🔗
    Give two or three different ways of expressing this point in polar coordinates.
  2. 🔗
    Do you see a pattern in the expressions? What is it?
Solution.
  1. Answers vary
  2. Answers vary
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A Point in Polar Coordinates.
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Any point (r,θ) in polar coordinates can also be represented by the points
(r,θ+2nπ)
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and
(−r,θ+(2n+1)π)
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for any n∈Z
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There is a connection between Cartesian coordinates and polar coordinates; we can exploit trigonometry to find the relationship.
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Example 10.3.4.
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  1. 🔗
    Below is a circle of radius r, with an inscribed right triangle. Label as much as you can on the figure.
  2. 🔗
    Use the triangle and trigonometric functions to obtain a conversion from polar coordinates to Cartesian coordinates.
A circle of radius \(r\text{,}\) with an inscribed right triangle
Solution.
To be filled out when the image issue is resolved
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Polar-to-Cartesian Conversion.
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If a point P has the polar coordinate expression (r,θ), then its Cartesian coordinate expression is given by x=rcos⁡θ and y=rsin⁡θ
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Cartesian-to-Polar Conversion.
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If a point P has the Cartesian coordinate expression (x,y), then its polar coordinate expression is given by r2=x2+y2 and θ=tan−1⁡(yx)
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Example 10.3.5.
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Convert the following points from polar to Cartesian coordinates.
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  1. 🔗
    (2,3π2)
  2. 🔗
    (2,π4)
  3. 🔗
    (−3,−π3)
Solution.
Technically there are an infinite number of representations, but here are perhaps the most straightforward.
  1. (0,−2)
  2. (1,1)
  3. (−32,−332)
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Example 10.3.6.
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Convert the following points from Cartesian to polar coordinates.
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  1. 🔗
    (−4,4)
  2. 🔗
    (3,1)
  3. 🔗
    (3,33)
Solution.
Same comment as above
  1. (42,3π4)
  2. (2,π6)
  3. (6,π3)
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Example 10.3.7.
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Convert each equation from polar to rectangular coordinates. If possible, identify the curve.
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  1. 🔗
    r2=16
  2. 🔗
    r=4sin⁡θ
  3. 🔗
    θ=π3
  4. 🔗
    r2cos⁡2θ=1
Solution.
  1. x2+y2=16, a circle of radius 4 centered at the origin.
  2. x2+(y−2)2=4, a circle of radius 2 centered at (0,−2)
  3. y=3x, a line of slope 3
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Example 10.3.8.
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Convert each equation from rectangular coordinates to polar coordinates.
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  1. 🔗
    y=2
  2. 🔗
    4y2=x
  3. 🔗
    y=x
Solution.
  1. r=2csc⁡θ
  2. 4r2sin2⁡θ=rcos⁡θ
  3. θ=π4
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Subsection 10.3.2 Pre-Class Activities

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Example 10.3.9.

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Use this space to write any questions you might have from the videos
Solution.
Answers vary
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Example 10.3.10.

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Plot each point below on a polar grid. Then, convert each from polar coordinates to Cartesian coordinates.
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  1. 🔗
    (2,5π6)
  2. 🔗
    (1,−2π3)
  3. 🔗
    (−1,5π4)
Solution.
Again, the plots will have to wait, but the conversions are below.
  1. (−3,1)
  2. Missing or unrecognized delimiter for \right
  3. (22,22)
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Example 10.3.11.

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Convert r=4csc⁡θ into rectangular coordinates, and describe/identify the graph.
Solution.
This was conveniently done earlier in the before class: y=4
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Example 10.3.12.

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Express the ellipse x29+y24=1 in polar coordinates.
Solution.
4r2cos2⁡θ+9r2sin2⁡θ=1, so this simplifies as r2(4+5sin2⁡θ)=1
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Subsection 10.3.3 In Class

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Subsubsection 10.3.3.1 Polar Curves

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Example 10.3.13.
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Graph the polar equation θ=π4
Solution.
This ends up the line y=x, but again with the graphing issue.
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Example 10.3.14.
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Let r=2sin⁡θ
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  1. 🔗
    Make a table of values for the curve, for 0≤θ≤π
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    Use the table to sketch the curve
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    Find the Cartesian equation for the curve
Solution.
  1. ,
    θ r=2sin⁡θ
    0 0
    π6 1
    π4 2
    π3 3
    π2 2
    2π3 3
    3π4 2
    5π6 1
    π 0
  2. Graphing issues, but you should see a circle of radius 1 centered at (0,1)
  3. x2+(y−1)2=1
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Example 10.3.15.
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Sketch the polar curve r=1+cos⁡θ. This shape is called a cardioid.
Solution.
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Example 10.3.16.
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Sketch the polar curve r=sin⁡2θ. This shape is called a rose of four leaves.
Solution.
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Symmetry in Polar Graphs.
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Consider the polar curve r=f(θ)
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    The curve is symmetric about the polar axis if f(θ)=f(−θ)
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    The curve is symmetric about the pole if −f(θ)=f(θ) or f(θ)=f(θ+π)
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    The curve is symmetric about θ=π2 if f(θ)=f(π−θ)
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Example 10.3.17.
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What kind of symmetries does the cardioid r=1+cos⁡θ possess? What about the cardioid r=1+sin⁡θ?
Solution.
Let f(θ)=1+cos⁡θ and g(θ)=1+sin⁡θ
f(−θ)=f(θ), so the first cardioid is symmetric about the polar axis. It is clear that −f(θ)≠f(θ), so there is no symmetry about the pole. f(π−θ)=1+cos⁡πcos⁡θ−sin⁡θsin⁡π=1−cos⁡θ≠f(θ), so there is no symmetry about θ=π2
g(−θ)=−g(θ), so the second cardioid is not symmetric about the polar axis. −f(θ)≠f(θ), so there is no symmetry about the pole. f(π−θ)=1+sin⁡πcos⁡θ+cos⁡πsin⁡θ=1−sin⁡θ≠f(θ), so there is no symmetry about θ=π2
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Example 10.3.18.
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Use symmetries to help you graph r2=cos⁡4θ.
Solution.
Graphing issues
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Subsubsection 10.3.3.2 Tangents to Polar Curves

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Derivative of a Polar Curve.
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Given the polar curve r=f(θ), consider the parametric equations x=rcos⁡θ=f(θ)cos⁡θ and y=rsin⁡θ=f(θ)sin⁡θ. Then,
dydx=dy/dθdx/dθ=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ
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Example 10.3.19.
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Consider the cardioid r=1+sin⁡θ.
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  1. 🔗
    Find the slope of the tangent line when θ=π6.
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    Find the points where the tangent line is horizontal or vertical.
Solution.
  1. dydx=cos⁡θ(1+2sin⁡θ)(1+sin⁡θ)(1−2sin⁡θ), so
    dydx|θ=π/6=30
    which is undefined.
  2. The tangent line is horizontal when cos⁡θ(1+2sin⁡θ)=0⟺θ=π2,7π6,11π6.
    The tangent line is vertical when (1+sin⁡θ)(1−2sin⁡θ)=0⟺θ=3π2,π6,5π6
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Example 10.3.20.
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Find the slope of the tangent line to the polar curve r=2cos⁡θ at θ=π3.
Solution.
drdθ|π/3=−2sin⁡θ|π/3=−3, sin⁡(π3)=32 and cos⁡(π3=12), so
Missing or unrecognized delimiter for \right
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Example 10.3.21.
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Find the points on the curve r=1−sin⁡θ where the tangent line is horizontal or vertical.
Solution.
drdθ=−cos⁡θ, so the tangent line is horizontal when
−cos2⁡θ−sin⁡θ−sin2⁡θ=0
which gives θ=\3pi2. The tangent line is vertical when
−cos⁡θsin⁡θ−cos⁡θ+sin⁡θcos⁡θ=0
i.e. when cos⁡θ=0⟺θ=π2,3π2.
This means that the tangent line is never horizontal.
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Subsection 10.3.4 After Class Activities

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Example 10.3.22.

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Sketch the curve r=2(1+cos⁡θ). Then, convert the equation to Cartesian coordinates.
Solution.
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Example 10.3.23.

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The graph below shows r as a function of θ, in Cartesian coordinates. Use the graph to sketch the corresponding polar curve.
Solution.
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Example 10.3.24.

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Find the slope of the tangent line to the polar curve r=2+sin⁡3θ at θ=π4.
Solution.
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