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Exercise 8.6 · Q7

Q.Let U(x,y)=exsin⁡yU(x,y)=e^x\sin y, where x=st2, y=s2t, s,t∈Rx=st^2,\ y=s^2t,\ s,t\in\mathbb R. Find ∂U∂s,∂U∂t\dfrac{\partial U}{\partial s},\dfrac{\partial U}{\partial t} and evaluate them at s=t=1s=t=1.

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Apply the two-parameter chain rule (17)/(18) with x=st2, y=s2tx=st^2,\,y=s^2t, then substitute s=t=1s=t=1.

Step 1. Partial derivatives of U=exsin⁡yU=e^x\sin y. Ux=exsin⁡yU_x=e^x\sin y. Uy=excos⁡y\quad U_y=e^x\cos y.

Step 2. Partial derivatives of x=st2, y=s2tx=st^2,\,y=s^2t. ∂x∂s=t2\dfrac{\partial x}{\partial s}=t^2. ∂x∂t=2st\dfrac{\partial x}{\partial t}=2st. ∂y∂s=2st\dfrac{\partial y}{\partial s}=2st. ∂y∂t=s2\dfrac{\partial y}{\partial t}=s^2.

Step 3. Combine via (17)/(18).

∂U∂s=Ux∂x∂s+Uy∂y∂s=exsin⁡y⋅t2+excos⁡y⋅2st=ex[t2sin⁡y+2stcos⁡y],\frac{\partial U}{\partial s} = U_x\frac{\partial x}{\partial s}+U_y\frac{\partial y}{\partial s} = e^x\sin y\cdot t^2 + e^x\cos y\cdot2st = e^x\big[t^2\sin y+2st\cos y\big],

∂U∂t=Ux∂x∂t+Uy∂y∂t=exsin⁡y⋅2st+excos⁡y⋅s2=ex[2stsin⁡y+s2cos⁡y].\frac{\partial U}{\partial t} = U_x\frac{\partial x}{\partial t}+U_y\frac{\partial y}{\partial t} = e^x\sin y\cdot2st + e^x\cos y\cdot s^2 = e^x\big[2st\sin y+s^2\cos y\big].

Step 4. Evaluate at s=t=1s=t=1. x=st2=1, y=s2t=1x=st^2=1,\ y=s^2t=1, so ex=ee^x=e.

∂U∂s∣1,1=e[1⋅sin⁡1+2(1)(1)cos⁡1]=e(sin⁡1+2cos⁡1),\frac{\partial U}{\partial s}\bigg|_{1,1} = e\big[1\cdot\sin1+2(1)(1)\cos1\big] = e(\sin1+2\cos1), …

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