ka1leE

2021-01-02

, value $t=0$, find $\frac{dw}{dt}$ using the appropriate Chain Rule and evaluate $\frac{dw}{dt}$ at the given value of t.

curwyrm

If
$\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}$
$\frac{dw}{dt}=\frac{d}{dx}\left[x\mathrm{sin}y\right]\frac{d}{dt}\left[{e}^{t}\right]+\frac{d}{dy}\left[x\mathrm{sin}y\right]\frac{d}{dt}\left[\pi -t\right]$
$\frac{dw}{dt}=\mathrm{sin}y\left({e}^{t}\right)+\left(x\mathrm{cos}y\right)\left(-1\right)$
$\frac{dw}{dt}={e}^{t}\mathrm{sin}y-x\mathrm{cos}y$
so
$\frac{dw}{dt}={e}^{t}\mathrm{sin}\left(\pi -t\right)-{e}^{t}\mathrm{cos}\left(\pi -t\right)$
$\mathrm{sin}\left(\pi -t\right)=\mathrm{sin}t$ 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
$\frac{dw}{dt}={e}^{0}\left(\mathrm{sin}0+\mathrm{cos}0\right)$
$\frac{dw}{dt}=1\left(0+1\right)$
$\frac{dw}{dt}=1$