Following Theorem 12.5.1, we find \begin{equation*} f_x(x,y) = 2xy+1,\qquad f_y(x,y) = x^2,\qquad \frac{dx}{dt} = \cos(t) ,\qquad \frac{dy}{dt}= 5e^{5t}. \end{equation*}

Applying the theorem, we have \begin{equation*} \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t}. \end{equation*}

This may look odd, as it seems that \(\frac{dz}{dt}\) is a function of \(x\text{,}\) \(y\) and \(t\text{.}\) Since \(x\) and \(y\) are functions of \(t\text{,}\) \(\frac{dz}{dt}\) is really just a function of \(t\text{,}\) and we can replace \(x\) with \(\sin(t)\) and \(y\) with \(e^{5t}\text{:}\) \begin{equation*} \frac{dz}{dt} = (2xy+1)\cos(t) + 5x^2e^{5t} = (2\sin(t)e^{5t}+1)\cos(t) +5e^{5t}\sin^2(t) . \end{equation*}

The Multivariable Chain Rule states that \begin{align*} \frac{dz}{dt} \amp = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\\ \amp = 5(3)+(-2)(7)\\ \amp =1. \end{align*}

By knowing certain rates–of–change information about the surface and about the path of the particle in the \(x\)-\(y\) plane, we can determine how quickly the object is rising/falling.

It is straightforward to compute \begin{equation*} f_x(x,y) = 2x-y,\qquad f_y(x,y) = 2y-x,\qquad \frac{dx}{dt} = -\sin(t) ,\qquad \frac{dy}{dt} = \cos(t) . \end{equation*}

Combining these according to the Chain Rule gives: \begin{equation*} \frac{dz}{dt} = -(2x-y)\sin(t) + (2y-x)\cos(t) . \end{equation*}

When \(t=0\text{,}\) \(x=1\) and \(y=0\text{.}\) Thus \(\ds\frac{dz}{dt} = -(2)(0)+ (-1)(1) = -1\text{.}\) When \(t=0\text{,}\) the particle is moving down, as shown in Figure 12.5.6.

To find where \(z\)-value is maximized/minimized on the particle's path, we set \(\frac{dz}{dt}=0\) and solve for \(t\text{:}\) \begin{align*} \frac{dz}{dt} =0 \amp = -(2x-y)\sin(t) + (2y-x)\cos(t) \\ 0\amp = -(2\cos(t) -\sin(t) )\sin(t) +(2\sin(t) -\cos(t) )\cos(t) \\ 0\amp = \sin^2(t) -\cos^2(t) \\ \cos^2(t) \amp =\sin^2(t) \\ t\amp = n\frac{\pi}4 \text{ (for odd \(n\)) } \end{align*}

We can use the First Derivative Test to find that on \([0,2\pi]\text{,}\) \(z\) has reaches its absolute minimum at \(t=\pi/4\) and \(5\pi/4\text{;}\) it reaches its absolute maximum at \(t=3\pi/4\) and \(7\pi/4\text{,}\) as shown in Figure 12.5.6.

Following Theorem 12.5.7, we compute the following partial derivatives: \begin{equation*} \frac{\partial f}{\partial x} = 2xy+1\qquad\qquad \frac{\partial f}{\partial y} = x^2, \end{equation*} \begin{equation*} \frac{\partial x}{\partial s} = 2s \qquad\qquad \frac{\partial x}{\partial t} = 3\qquad\qquad \frac{\partial y}{\partial s} = 2 \qquad\qquad \frac{\partial y}{\partial t} = -1. \end{equation*}

Thus \begin{equation*} \ds \frac{\partial z}{\partial s} = (2xy+1)(2s) + (x^2)(2) = 4xys+2s + 2x^2, \text{ and } \end{equation*} \begin{equation*} \ds \frac{\partial z}{\partial t} = (2xy+1)(3) + (x^2)(-1) = 6xy-x^2+3. \end{equation*}

When \(s=1\) and \(t=2\text{,}\) \(x= 7\) and \(y= 0\text{,}\) so \begin{equation*} \frac{\partial z}{\partial s} = 100\qquad \text{ and } \qquad \frac{\partial z}{\partial t} = -46. \end{equation*}

Following Theorem 12.5.7, we compute the following partial derivatives: \begin{equation*} \frac{\partial f}{\partial x} = y\qquad\qquad \frac{\partial f}{\partial y} = x\qquad\qquad \frac{\partial f}{\partial z} = 2z, \end{equation*} \begin{equation*} \frac{\partial x}{\partial t} = 2te^s\qquad\qquad \frac{\partial y}{\partial t} = \cos(s) \qquad\qquad \frac{\partial z}{\partial t} = s\cos(t) . \end{equation*}

Thus \begin{equation*} \ds \frac{\partial w}{\partial t} = y(2te^s) + x(\cos(s) ) + 2z(s\cos(t) ). \end{equation*}

When \(s=0\) and \(t=\pi\text{,}\) we have \(x=\pi^2\text{,}\) \(y=\pi\) and \(z=0\text{.}\) Thus \begin{equation*} \frac{\partial w}{\partial t} = \pi(2\pi) + \pi^2 = 3\pi^2. \end{equation*}