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Question

If F(x,y)=x3xy2y4,x=2cos3tF(x, y)=x^3-x y^2-y^4, x=2 \cos 3 t, and y=3sinty=3 \sin t, find dF/dt at t=0.

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Given F(x,y)=x3xy2y4F\left(x,y \right)=x^{3}-xy^{2}-y^{4}, x=2cos3tx=2\cos 3t and y=3sinty=3\sin t

dFdt=dFdxdxdt+dFdydydt=3x2(6)sin3t+(2xy4y3)(3cost)=18x2sin3t+3cost(12cos3tsint12sint3)\begin{align*} \dfrac{dF}{dt}&=\dfrac{dF}{dx}\dfrac{dx}{dt}+\dfrac{dF}{dy}\dfrac{dy}{dt}\\ &=3x^{2}\cdot \left( -6\right)\sin 3t+\left(2xy-4y^{3} \right)\cdot \left(3\cos t\right)\\ &=-18x^{2}\sin 3t+3\cos t\left(12\cos 3t\sin t-12\sin t^{3} \right)\end{align*}

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