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3 years ago

13

A point is chosen at random in the larger circle. What percent of the time will the point be in the shaded region? Round to the

nearest tenth of a percent. Please help me!!

1 answer:

Andru [333]3 years ago

6 0

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image for the step by step explanation of the answer to the question

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You are given the polar curve r = e^θ

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The tangent to $r(\theta)=e^\theta$ has slope $\frac{\mathrm dy}{\mathrm dx}$ , where

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By the chain rule, we have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

and by the product rule,

$\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta$

$\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta$

so that with $\frac{\mathrm dr}{\mathrm d\theta}=e^\theta$ , we get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}$

The tangent line is horizontal when the slope is 0; this happens for

$-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi$

where $n$ is any integer. In the interval $0\le\theta\le2\pi$ , this happens for $n=1,2$ , or

$\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4$

i.e at the points

$(r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)$

and

$(r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)$

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Step-by-step explanation:

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Answer:

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