009C Sample Final 2, Problem 9 Detailed Solution

A curve is given in polar coordinates by

$r=\sin(2\theta ).$

(a) Sketch the curve.

(b) Compute  $y'={\frac {dy}{dx}}.$

(c) Compute  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{d^2y}{dx^2}.}

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Background Information:
How do you calculate  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'}   for a polar curve  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=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{(\frac{dr}{d\theta})\sin\theta+r\cos\theta}{(\frac{dr}{d\theta})\cos\theta-r\sin\theta}.}

Solution:

(a)
400px

(b)

Step 1:
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=\sin(2\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}=2\cos(2\theta).}

Step 2:
Since
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {dy}{dx}}={\frac {({\frac {dr}{d\theta }})\sin \theta +r\cos \theta }{({\frac {dr}{d\theta }})\cos \theta -r\sin \theta }},}
we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {y'}&=&\displaystyle {\frac {2\cos(2\theta )\sin \theta +\sin(2\theta )\cos \theta }{2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta }}\\&&\\&=&\displaystyle {\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}\end{array}}}
since
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin(2\theta )=2\sin \theta \cos \theta ,~\cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta .}

(c)

Step 1:
We have   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {({\frac {dy'}{d\theta }})}{({\frac {dr}{d\theta }})\cos \theta -r\sin \theta }}.}
So, first we need to find   ${\frac {dy'}{d\theta }}.$
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy'}{d\theta }}&=&\displaystyle {{\frac {d}{d\theta }}{\bigg (}{\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}}}.}\end{array}}}

Step 2:
Now, using the resulting formula for   ${\frac {dy'}{d\theta }},$  we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}}.}

Final Answer:
(a)    See above
(b)    Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta }{\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta }}}
(c)
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )(-4\cos \theta \sin ^{2}\theta +2\cos ^{3}\theta -3\sin ^{2}\theta \cos \theta )-(2\cos ^{2}\theta \sin \theta -\sin ^{3}\theta )(-3\cos ^{2}\theta \sin \theta -4\sin \theta \cos ^{2}\theta +2\sin ^{3}\theta )}{(\cos ^{3}\theta -2\sin ^{2}\theta \cos \theta )^{2}(2\cos(2\theta )\cos \theta -\sin(2\theta )\sin \theta )}}}

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