The problem asks us to find the partial derivative of $z$ with respect to $t$, denoted as $\frac{\partial z}{\partial t}$, given $z = x^2y$, $x = 2t + s$, and $y = 1 - st^2$, evaluated at $s = 1$ and $t = -2$.

1. Problem Description

The problem asks us to find the partial derivative of $z$ with respect to $t$, denoted as $\frac{\partial z}{\partial t}$, given $z = x^2y$, $x = 2t + s$, and $y = 1 - st^2$, evaluated at $s = 1$ and $t = -2$.

2. Solution Steps

We will use the chain rule to find $\frac{\partial z}{\partial t}$. First, we find the partial derivatives of $z$ with respect to $x$ and $y$.

$\frac{\partial z}{\partial x} = 2xy$

$\frac{\partial z}{\partial y} = x^2$

Next, we find the partial derivatives of $x$ and $y$ with respect to $t$.

$\frac{\partial x}{\partial t} = 2$

$\frac{\partial y}{\partial t} = -s(2t) = -2st$

Now, we apply the chain rule:

$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$

$\frac{\partial z}{\partial t} = (2xy)(2) + (x^2)(-2st) = 4xy - 2stx^2$

Substitute $x = 2t + s$ and $y = 1 - st^2$ into the equation:

$\frac{\partial z}{\partial t} = 4(2t+s)(1-st^2) - 2st(2t+s)^2$

Now we need to evaluate this at $s = 1$ and $t = -2$.

$x = 2(-2) + 1 = -4 + 1 = -3$

$y = 1 - (1)(-2)^2 = 1 - 4 = -3$

$\frac{\partial z}{\partial t}|_{s=1, t=-2} = 4(-3)(-3) - 2(1)(-2)(-3)^2$

$= 4(9) - 2(-2)(9)$

$= 36 + 36$

$= 72$

3. Final Answer

$\frac{\partial z}{\partial t}|_{s=1, t=-2} = 72$