Question

In Exercises 27-32, use the function $$f(x, y)=3-\frac{x}{3}-\frac{y}{2}$$ Find \(D_{\mathrm{u}} f(3,2),\) where \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\) (a) \(\theta=\frac{\pi}{4}\) (b) \(\theta=\frac{2 \pi}{3}\)

Short answer

The directional derivative \(D_{\mathbf{u}} f(3,2)\) of the function at the point (3,2) in the direction of \(\mathbf{u}\) is: \n(a) \(-\frac{5\sqrt{2}}{12}\) for \(\theta = \frac{\pi}{4}\),\n(b) \(\frac{\sqrt{3}-3}{24}\) for \(\theta = \frac{2\pi}{3}\).

Step-by-step solution

Calculating the Gradient Vector

The gradient vector of the function \(f(x, y)\), denoted as \(\nabla f(x, y)\), is calculated by taking the partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\). Given the function \(f(x, y) = 3 - \frac{x}{3} - \frac{y}{2}\), the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = -\frac{1}{3}\), and the partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = -\frac{1}{2}\). Hence, the gradient vector at (3,2) is \(\nabla f(3,2) = (-1/3, -1/2)\).

Formulating the Vector \(\mathbf{u}\)

The vector \(\mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\). The exercise requires us to calculate \(D_{\mathrm{u}} f(3,2)\) for \(\theta=\frac{\pi}{4}\) and \(\theta=\frac{2 \pi}{3}\). When \(\theta=\frac{\pi}{4}\), \(\mathbf{u} = \cos(\frac{\pi}{4})\mathbf{i} + \sin(\frac{\pi}{4})\mathbf{j} = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\). Similarly, when \(\theta=\frac{2\pi}{3}\), \(\mathbf{u} = \cos(\frac{2\pi}{3})\mathbf{i} + \sin(\frac{2\pi}{3})\mathbf{j} = (-\frac{1}{2}, \frac{\sqrt{3}}{2})\).

Calculating the Directional Derivative

The directional derivative in the direction of \(\mathbf{u}\), at the point (3,2) is given by \(D_{\mathbf{u}} f(3,2) = \nabla f(3,2) \cdot \mathbf{u}\) (the dot product of the gradient of \(f\) at (3,2) and the vector \(\mathbf{u}\)). For \(\theta=\frac{\pi}{4}\), \(D_{\mathbf{u}} f(3,2) = (-1/3, -1/2) \cdot (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) = -\frac{5\sqrt{2}}{12}\). For \(\theta=\frac{2\pi}{3}\), \(D_{\mathbf{u}} f(3,2) = (-1/3, -1/2) \cdot (-\frac{1}{2}, \frac{\sqrt{3}}{2}) = -\frac{1}{12} + \frac{\sqrt{3}}{8} = \frac{\sqrt{3}-3}{24}\).