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Find the equation of the tangent to the curve `x=sin3t ,y=cos2t`at `t=pi/4dot` |
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Answer» `x_1=sin(3/4pi),x_1=1/G` `y_1=cos(pi/2)=y_1=0` `dy/dx=(dy/dt)/(dx/dx)=(-2sin2t)/(3cos3t)=(-2sin(3/4pi)/(3cos(3/4pi)))` `=(-2)/(3*(1/sqrt2))` `m=(2sqrt2)/3` `y-0=(2sqrt2)/3(x-1/sqrt2)` `y=(2sqrt2x)/3-2/3` `(2sqrt2)/3x-y=2/3`. |
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