# Find dw/dt using the appropriate Chain Rule. Evaluate frac{dw}{dt} at the given value of t. Function: w=xsin y, x=e^t, y=pi-t Value: t = 0

Question
Differential equations
Find dw/dt using the appropriate Chain Rule. Evaluate $$\frac{dw}{dt}$$ at the given value of t. Function: $$w=x\sin y,\ x=e^t,\ y=\pi-t$$ Value: t = 0

2021-01-03
Let $$w=x\sin y,\ x=e^t,\ y=\pi-t$$
Apply the chain Rule for One independet Variable
$$\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}$$
So
$$\frac{dw}{dt}=\frac{d}{dx}[x\sin y]\frac{d}{dt}[e^t]+\frac{d}{dy}[x\sin y]\frac{d}{dt}[\pi-t]$$
Calculate the Partial derivatives, you get
$$\frac{dw}{dt}=\sin y(e^t)+(x\cos y)(-1)$$
Simplify
$$\frac{dw}{dt}=e^t\sin y-x\cos y$$
Substitute $$x=e^t,\ y=\pi-t,$$ so
$$\frac{dw}{dt}=e^t\sin(\pi-t)-e^t\cos(\pi-t)$$
Remember that $$\sin(\pi-t)=\sin t$$ and $$\cos(\pi-t)=-\cos t ,$$ so
$$\frac{dw}{dt}=e^t\sin t-e^t(-\cos t)$$
$$\frac{dw}{dt}=e^t(\sin t+\cos t)$$
Evaluate $$\frac{dw}{dt}$$ when t =0
$$\frac{dw}{dt}=e^0(\sin0+\cos0)$$
$$\frac{dw}{dt}=1(0+1)$$
$$\frac{dw}{dt}=1$$
Results
$$e^t(\sin t+\cos t),\ 1$$

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