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

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
asked 2021-01-02
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

Answers (1)

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
\(e^t(\sin t+\cos t),\ 1\)
0

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