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Chapter 8: Problem 66

Sketch a graph of the polar equation and find the tangents at the pole. $$ r=3 \cos 2 \theta $$

Short Answer

Expert verified
The graph of the polar equation \(r=3 \cos 2 \theta\) is a rose with 4 petals. The tangent lines at the pole (origin) of this graph can be found by first converting it to cartesian coordinates, differentiating to find the derivative (slope at any point), then evaluating this derivative at the pole.

Step by step solution

01

Plotting the Polar Graph

We can interpret the equation \(r=3 \cos 2 \theta\) as a rose with 4 petals (twice the coefficient of theta). Also, each petal extends from the origin (pole) to \(r=3\) (amplitude of the cosine function). Plot points of this function starting from angle theta = 0 and moving in increments (like 蟺/4) till we cover a full circle. Since the cosine function oscillates, we will end up tracing all the petals in one complete rotation.
02

Conversion to Cartesian Coordinates

To find the tangent lines, it will be helpful to translate the equation to cartesian coordinates. We use the relationships: \(x = r\cos \theta\), \(y = r\sin \theta\), and \(r^2 = x^2 + y^2\). Therefore, substituting for \(r\) into \(x = r\cos \theta\) and \(y = r\sin \theta\) gives \(x = 3\cos 2\theta \cdot \cos \theta\) and \(y= 3\cos 2\theta \cdot \sin \theta\). These can be further reduced using double-angle formulae to \(x = 3(1 - 2\sin^2 \theta)\) and \(y = 3(2\sin \theta \cos \theta)\) respectively.
03

Differentiating and Finding Tangent Lines

First, find the derivative of \(y\) with respect to \(x\) by implicit differentiation, this will give us the slope of the tangent at any point. At the pole (origin), \(x = y = 0\). Therefore, substitute \(\theta=0\) (note we set \(\theta\), not \(x\) or \(y\), to zero because this sets both \(x\) and \(y\) to zero) into the derivative to get the slopes for the tangent lines at the pole. Intersect these lines with the graph for the visual understanding.

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Most popular questions from this chapter

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=a \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \theta=\frac{\pi}{2} \end{array} $$

Show that the polar equation for \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta} \cdot \quad\) Ellipse

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=t^{2}+t, \quad y=t^{2}-t $$

Give the integral formulas for the area of the surface of revolution formed when the graph of \(r=f(\theta)\) is revolved about (a) the \(x\) -axis and (b) the \(y\) -axis.

Use the integration capabilities of a graphing utility to approximate to two decimal places the area of the surface formed by revolving the curve about the polar axis. $$ r=\theta, \quad 0 \leq \theta \leq \pi $$
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