Suppose $f(x, y)=\frac{x}{y}, P=(0,2)$ and $\mathbf{v}=1 \mathbf{i}+1 \mathbf{j}$. A. Find the gradient of $f$. $(\nabla f)(x, y)=$ i+ $\mathrm{j}$ Note: Your answers should be expressions of $x$ and $y$; e.g. " $3 x-4 y "$ B. Find the gradient of $f$ at the point $P$. $(\nabla f)(P)=$ i+ Note: Your answers should be numbers C. Find the directional derivative of $f$ at $P$ in the direction of $\mathbf{v}$. $\left(D_{\mathbf{u}} f\right)(P)=$ Note: Your answer should be a number D. Find the maximum rate of change of $f$ at $P$. Note: Your answer should be a number E. Find the (unit) direction vector $w$ in which the maximum rate of change occurs at $P$. $\mathbf{w}=\square \mathbf{i}+\square \mathbf{j}$ Note: Your answers should be numbers

Step 1 ：Find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\): \(\frac{\partial f}{\partial x} = \frac{1}{y}\) and \(\frac{\partial f}{\partial y} = -\frac{x}{y^2}\)

Step 2 ：The gradient of \(f\) is given by: \((\nabla f)(x, y) = \frac{1}{y} \mathbf{i} - \frac{x}{y^2} \mathbf{j}\)

Step 3 ：Substitute \(x=0\) and \(y=2\) into the gradient to find the gradient of \(f\) at the point \(P=(0,2)\): \((\nabla f)(P) = \frac{1}{2} \mathbf{i}\)

Step 4 ：Find the unit vector in the direction of \(\mathbf{v}=1 \mathbf{i}+1 \mathbf{j}\): \(\mathbf{u} = \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j}\)

Step 5 ：The directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{v}\) is given by the dot product of the gradient of \(f\) at \(P\) and the unit vector in the direction of \(\mathbf{v}\): \(\left(D_{\mathbf{u}} f\right)(P) = \frac{1}{2\sqrt{2}}\)

Step 6 ：The maximum rate of change of \(f\) at \(P\) is given by the magnitude of the gradient at \(P\): \(||(\nabla f)(P)|| = \frac{1}{2}\)

Step 7 ：The direction in which the maximum rate of change occurs at \(P\) is given by the direction of the gradient at \(P\). Since the gradient at \(P\) is \(\frac{1}{2} \mathbf{i}\), the direction vector is \(\mathbf{w} = 1 \mathbf{i} + 0 \mathbf{j}\)