Question

if (z=(x + y)e^{y}), (x = 3t), (y=4 - t^{2}), find (dz/dt) using the chain - rule. assume the variables are restricted to domains on which the functions are defined. (\frac{dz}{dt}=)

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 $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

Given $z=(x + y)e^{y}$, then $\frac{\partial z}{\partial x}=e^{y}$ and $\frac{\partial z}{\partial y}=e^{y}+(x + y)e^{y}=(x + y + 1)e^{y}$.

Step3: Calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Since $x = 3t$, then $\frac{dx}{dt}=3$. Since $y = 4 - t^{2}$, then $\frac{dy}{dt}=-2t$.

Step4: 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:
$\frac{dz}{dt}=3e^{4 - t^{2}}+(3t+4 - t^{2}+1)e^{4 - t^{2}}\times(-2t)$
$=3e^{4 - t^{2}}+(5 + 3t - t^{2})e^{4 - t^{2}}\times(-2t)$
$=3e^{4 - t^{2}}-2t(5 + 3t - t^{2})e^{4 - t^{2}}$
$=e^{4 - t^{2}}(3-10t - 6t^{2}+2t^{3})$.

Answer:

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