Q.(a) If $w = x+2y+z^2$ and $x=\cos t$; $y=\sin t$; $z=t$ find $\dfrac{dw}{dt}$ by using chain rule. Also find $\dfrac{dw}{dt}$ by substitution of $x, y$ and $z$ in $w$ and hence verify the result. OR

(a) computes $dw/dt$ for a composite function both via the multivariable chain rule and by direct substitution, and confirms they agree; (b) applies Newton's law of cooling twice in succession to find the temperature 10 minutes after the cup was placed in the room.

(a) $\dfrac{dw}{dt}$ for $w=x+2y+z^2$, $x=\cos t,\ y=\sin t,\ z=t$

1. Chain rule: $\dfrac{\partial w}{\partial x}=1,\ \dfrac{\partial w}{\partial y}=2,\ \dfrac{\partial w}{\partial z}=2z$, and $\dfrac{dx}{dt}=-\sin t,\ \dfrac{dy}{dt}=\cos t,\ \dfrac{dz}{dt}=1$.
2. $\dfrac{dw}{dt}=1\cdot(-\sin t)+2\cdot(\cos t)+2z\cdot(1)=-\sin t+2\cos t+2z$.
3. Substitute $z=t$: $\dfrac{dw}{dt}=2t+2\cos t-\sin t$.
4. Direct substitution: $w=x+2y+z^2=\cos t+2\sin t+t^2$.
5. Differentiate directly: $\dfrac{dw}{dt}=-\sin t+2\cos t+2t$.
6. Both methods give the same result, $\dfrac{dw}{dt}=2t+2\cos t-\sin t$ — the chain-rule computation is verified.

(b) Newton's law of cooling

1. Newton's law: $\dfrac{dT}{dt}=-k(T-T_s)$, with solution $T(t)=T_s+(T_0-T_s)e^{-kt}$, where $T_0=100^\circ$C (initial temperature) and $T_s=15^\circ$C (room temperature).
2. So $T(t)-15=85\,e^{-kt}$.
3. Given $T=60$ at $t=5$: $60-15=85\,e^{-5k}\Rightarrow e^{-5k}=\dfrac{45}{85}=\dfrac{9}{17}$. …