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`"If "x=(2costheta-cos 2theta)and y=(2sin theta-sin 2theta)," find "((d^(2))/(dx^(2)))_(theta=(pi)/(2))` |
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Answer» Correct Answer - `(-3)/(2)` `(dy)/(dx)=((dy//d theta))/((dx//d theta))=tan.(3 theta)/(2)` `rArr(d^(2)y)/(dx^(2))=(3)/(2)sec^(2)(3theta)/(2).(d theta)/(dx)=(3)/(2)sec^(2).(3theta)/(2).(1)/(2(sin 2theta-sin theta))` `rArr((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(3)/(4)sec^(2).(3pi)/(4).(1)/((sin pi-sin.(pi)/(2)))=(3)/(4)xx2xx(1)/((0-1))=(-3)/(2).` |
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