 # 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 ka1leE 2021-01-02 Answered

Find $\frac{dw}{dt}$ using the appropriate Chain Rule. Evaluate $\frac{dw}{dt}$ at the given value of t. Function: Value: $t=0$

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