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A First Course in Linear Algebra, 2017a

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8.1. Polar Coord<strong>in</strong>ates and Polar Graphs 437<br />

θ 2<br />

θ 1<br />

To f<strong>in</strong>d the po<strong>in</strong>t <strong>in</strong> the plane correspond<strong>in</strong>g to the ordered pair ( f (θ),θ), we follow the same process<br />

as when f<strong>in</strong>d<strong>in</strong>g the po<strong>in</strong>t correspond<strong>in</strong>g to (r,θ).<br />

Consider the follow<strong>in</strong>g example of this procedure, <strong>in</strong>corporat<strong>in</strong>g computer software.<br />

Example 8.5: Graph<strong>in</strong>g a Polar Equation<br />

Graph the polar equation r = 1 + cosθ.<br />

Solution. We will use the computer software Maple to complete this example. The command which<br />

produces the polar graph of the above equation is: > plot(1+cos(t),t= 0..2*Pi,coords=polar). Here we use<br />

t to represent the variable θ for convenience. The command tells Maple that r is given by 1 + cos(t) and<br />

that t ∈ [0,2π].<br />

y<br />

x<br />

The above graph makes sense when considered <strong>in</strong> terms of trigonometric functions. Suppose θ =<br />

0,r = 2andletθ <strong>in</strong>crease to π/2. As θ <strong>in</strong>creases, cosθ decreases to 0. Thus the l<strong>in</strong>e from the orig<strong>in</strong> to the<br />

po<strong>in</strong>t on the curve should get shorter as θ goes from 0 to π/2. As θ goes from π/2toπ, cosθ decreases,<br />

eventually equal<strong>in</strong>g −1 atθ = π. Thus r = 0 at this po<strong>in</strong>t. This scenario is depicted <strong>in</strong> the above graph,<br />

which shows a function called a cardioid.<br />

The follow<strong>in</strong>g picture illustrates the above procedure for obta<strong>in</strong><strong>in</strong>g the polar graph of r = 1 + cos(θ).<br />

In this picture, the concentric circles correspond to values of r while the rays from the orig<strong>in</strong> correspond<br />

to the angles which are shown on the picture. The dot on the ray correspond<strong>in</strong>g to the angle π/6 is located<br />

at a distance of r = 1 + cos(π/6) from the orig<strong>in</strong>. The dot on the ray correspond<strong>in</strong>g to the angle π/3 is<br />

located at a distance of r = 1 + cos(π/3) from the orig<strong>in</strong> and so forth. The polar graph is obta<strong>in</strong>ed by<br />

connect<strong>in</strong>g such po<strong>in</strong>ts with a smooth curve, with the result be<strong>in</strong>g the figure shown above.

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