Question

4 OE 6 4 12 4 -4.
a- find equation in terms of polar coordinates that represent the figure
b - determine the area contained within the circle given in part (a) by using tbe definite integral
c- use def of the derivitive in polar coordinates to find points (r, (theta)) on the circle where vertical tangent line exists
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Answer #1

\mathbf{a.}$From the graph we see that the radius of the circle is 2 and centre is (0,0),therefore the equation of the circle is $ (x-0)^2+(y-0)^2=2^2\\$In Cartesian co-ordinate system we have $ x=rcos\theta ,y=rsin\theta,$ then $ \\x^2+y^2=r^2cos^2\theta+r^2sin^2\theta =2^2\Rightarrow r=2,$this is the polar equation the circle $.\\ \mathbf{b.}$The area bounded by the circle is $ A=\int_{0}^{2\pi}\frac{1}{2}r^2d\theta\\=\frac{1}{2}\int_{0}^{2\pi}2^2d\theta =[2\theta]_{0}^{2\pi}=4\pi\\ \mathbf{c.}$We know $ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{f'(\theta)sin\theta-f(\theta)cos\theta}{f'(\theta)cos\theta-f(\theta)sin\theta}\\$The circle has verticle tangent if $\frac{dx}{d\theta }=0\Rightarrow f'(\theta)cos\theta-f(\theta)sin\theta=0\\$Here $ r=f(\theta)=2,$ then $ f'(\theta)=0\\ $Therefore $ \frac{dx}{d\theta }=0\Rightarrow 2sin\theta=0\Rightarrow sin\theta=0=sin0$ or $ sin\pi\Rightarrow \theta =0,\pi \\$Therefore the required points are (2,0) and $(2,\pi).

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