Question

Use the function $$f(x, y)=3-\frac{x}{3}-\frac{y}{2}$$ Find \(D_{\mathbf{u}} f(3,2),\) where \(\mathbf{u}=\frac{\mathbf{v}}{\|\mathbf{v}\|}\) (a) \(\mathbf{v}\) is the vector from (1,2) to (-2,6) . (b) \(\mathbf{v}\) is the vector from (3,2) to (4,5) .

Short answer

(a) The directional derivative of the function at the point (3,2) in the direction of the unit vector (-3/5, 4/5) is 1/5. \n(b) The directional derivative of the function at the point (3,2) in the direction of the unit vector (1/\sqrt{10}, 3/\sqrt{10}) is -\sqrt{10}/5.

Step-by-step solution

Find the Partial Derivatives

Let's calculate the partial derivatives of f with respect to x and y.The partial derivative of f with respect to x, \(f_x(x,y)\), is obtained by differentiating f with respect to x, treating y as a constant. Doing so gives \(f_x(x, y) = -\frac{1}{3}\).Similarly, the partial derivative of f with respect to y, \(f_y(x,y)\), is obtained by differentiating f with respect to y, treating x as a constant. Doing this gives \(f_y(x, y) = -\frac{1}{2}\).

Find the Vector \(\mathbf{v}\)

The vector \(\mathbf{v}\) is obtained by subtracting the coordinates of the first point from the coordinates of the second point.(a) For \(\mathbf{v}\) from (1,2) to (-2,6), the vector \(\mathbf{v}\) is (-2-1, 6-2) = (-3,4).(b) For \(\mathbf{v}\) from (3,2) to (4,5), the vector \(\mathbf{v}\) is (4-3, 5-2) = (1,3).

Normalize the Vector \(\mathbf{v}\)

We must normalize the vector \(\mathbf{v}\) to obtain the unit vector \(\mathbf{u}\), which is the vector in the direction of \(\mathbf{v}\) with magnitude 1. This is done by dividing each component of the vector \(\mathbf{v}\) by its magnitude. (a) The magnitude of (-3, 4) is \(\sqrt{(-3)^2+4^2} = 5\). Dividing each component by the magnitude gives the unit vector \(\mathbf{u} = (-3/5, 4/5)\).(b) The magnitude of (1, 3) is \(\sqrt{1^2+3^2} = \sqrt{10}\). Dividing each component by the magnitude gives the unit vector \(\mathbf{u} = (1/\sqrt{10}, 3/\sqrt{10})\).

Find the Directional Derivative

Now, we can calculate the directional derivative using the formula mentioned in the analysis. We substitute the point (3,2), the unit vector \(\mathbf{u}\), and the calculated partial derivatives \(f_x(3,2)= -\frac{1}{3}\) and \(f_y(3,2)=-\frac{1}{2}\) into the expression for \(D_{\mathbf{u}} f(a, b)\) to calculate the directional derivative.(a) \(D_{\mathbf{u}} f(3,2) = [-\frac{1}{3}, -\frac{1}{2}] \cdot [-3/5, 4/5] = \frac{1}{5}\)(b) \(D_{\mathbf{u}} f(3,2) = [-\frac{1}{3}, -\frac{1}{2}] \cdot [1/\sqrt{10}, 3/\sqrt{10}] = -\frac{1}{\sqrt{10}}-\frac{3}{2\sqrt{10}}=-\frac{2}{\sqrt{10}}=-\frac{\sqrt{10}}{5}\)