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If `x= 2costheta-cos 2theta` and `y=2 sintheta - sin 2theta`.Find `(d^2y)/(dx^2)` at `theta=pi/2` |
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Answer» We have `x=(2cos theta-cos 2 theta)and y=(2 sin theta-sin 2 theta)` `rArr(dx)/(d theta)=(-2sin theta+2 sin 2 theta)and (dy)/(d theta)=(2cos theta-2 cos 2 theta)` `rArr(dy)/(dx)=((dy//d theta))/((dx//d theta))=((2cos theta-2 cos 2theta))/((-2sin theta+2sin 2 theta))=((cos theta-cos 2 theta))/((sin 2 theta-sin theta))` `=(2sin((3 theta)/(2))sin.(theta)/(2))/(2cos((3theta)/(2))sin.(theta)/(2))=tan.(3theta)/(2)` `rArr(d^(2)y)/(dx^(2))=(d)/(dx)(tan.(3theta)/(2))=(3)/(2)sec^(2).(3theta)/(2).(d theta)/(dx)=(3)/(2)sec^(2).(3 theta)/(2).(1)/(2(sin 2 theta-sin theta))` `rArr((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(3)/(2).sec^(2)((3pi)/(4)).(1)/((2sinpi-sin.(pi)/(2)))` `=(-3)/(4)sec^(2).(pi)/(4)=(-3)/(4)xx(sqrt2)^(2)=(-3)/(2)` `" "[becausesec.(3pi)/(4)=sec(pi-(pi)/(4))=-sec.(pi)/(4)].` |
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