if (z=(x + y)e^{y}), (x = 3t), (y=4 - t^{2}), find (dz/dt) using the chain - rule. assume the variables are…

Explanation:

Step1: Recall the chain - rule formula

The chain - rule for $z = z(x,y)$ where $x = x(t)$ and $y = y(t)$ is $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$.

Step2: Calculate $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Given $z=(x + y)e^{y}$, then $\frac{\partial z}{\partial x}=e^{y}$, $\frac{\partial z}{\partial y}=e^{y}+(x + y)e^{y}=(x + y + 1)e^{y}$. Given $x = 3t$, so $\frac{dx}{dt}=3$. Given $y = 4 - t^{2}$, so $\frac{dy}{dt}=-2t$.

Step3: Substitute into the chain - rule formula

$\frac{dz}{dt}=e^{y}\times3+(x + y + 1)e^{y}\times(-2t)$. Substitute $x = 3t$ and $y = 4 - t^{2}$ into the above formula: [ \begin{align*} \frac{dz}{dt}&=3e^{4 - t^{2}}+(3t+4 - t^{2}+1)e^{4 - t^{2}}\times(-2t)\ &=3e^{4 - t^{2}}+(5 + 3t - t^{2})e^{4 - t^{2}}\times(-2t)\ &=e^{4 - t^{2}}(3-10t - 6t^{2}+2t^{3}) \end{align*} ]

Answer:

$e^{4 - t^{2}}(2t^{3}-6t^{2}-10t + 3)$