009C Sample Final 1, Problem 7 Detailed Solution

A curve is given in polar coordinates by

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r=1+\sin \theta }

(a) Sketch the curve.

(b) Compute  $y'={\frac {dy}{dx}}$.

(c) Compute  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y''={\frac {d^{2}y}{dx^{2}}}} .

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Background Information:
How do you calculate  $y'$  for a polar curve  $r=f(\theta )?$
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=r\cos(\theta)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=r\sin(\theta),}   we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\frac{dy}{dx}=\frac{\big(\frac{dr}{d\theta}\big)\sin\theta+r\cos\theta}{\big(\frac{dr}{d\theta}\big)\cos\theta-r\sin\theta}.}

Solution:

(a)
400px

(b)

Step 1:
First, recall we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\frac{dy}{dx}=\frac{\big(\frac{dr}{d\theta}\big)\sin\theta+r\cos\theta}{\big(\frac{dr}{d\theta}\big)\cos\theta-r\sin\theta}.}
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=1+\sin\theta,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dr}{d\theta}=\cos\theta.}
Hence,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\frac{\cos\theta\sin\theta+(1+\sin\theta)\cos\theta}{\cos^2\theta-(1+\sin\theta)\sin\theta}.}

Step 2:

Thus, we have        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{y'} & = & \displaystyle{\frac{2\cos\theta\sin\theta+\cos\theta}{\cos^2\theta-\sin^2\theta-\sin\theta}}\\ &&\\ & = & \displaystyle{\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}.}\\ \end{array}}

(c)

Step 1:
We have  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2y}{dx^2}=\frac{\big(\frac{dy'}{d\theta}\big)}{\big(\frac{dr}{d\theta}\big)\cos\theta-r\sin\theta}.}
So, first we need to find  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy'}{d\theta}.}
We have
${\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}{\bigg )}}\\&&\\&=&\displaystyle {\frac {(\cos(2\theta )-\sin \theta )(2\cos(2\theta )-\sin \theta )-(\sin(2\theta )+\cos \theta )(-2\sin(2\theta )-\cos \theta )}{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta +\sin ^{2}\theta +\cos ^{2}\theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\&&\\&=&\displaystyle {\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{2}}}\\\end{array}}$
since  $\sin ^{2}\theta +\cos ^{2}\theta =1$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 2\cos ^{2}(2\theta )+2\sin ^{2}(2\theta )=2.}

Step 2:
Now, using the resulting formula for  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy'}{d\theta }},}   we get
${\frac {d^{2}y}{dx^{2}}}={\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}.$

Final Answer:
(a)     See above.
(b)     ${\frac {\sin(2\theta )+\cos \theta }{\cos(2\theta )-\sin \theta }}$
(c)     ${\frac {3-3\sin \theta \cos(2\theta )+3\sin(2\theta )\cos \theta }{(\cos(2\theta )-\sin \theta )^{3}}}$

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