Question

Use the function $$f(x, y)=3-\frac{x}{3}-\frac{y}{2}$$ Find a unit vector \(\mathbf{u}\) orthogonal to \(\nabla f(3,2)\) and calculate \(D_{\mathbf{u}} f(3,2) .\) Discuss the geometric meaning of the result.

Short answer

The computed unit vectors orthogonal to the gradient \(\nabla f(3,2)\) at the point (3,2) are \(\frac{1}{\sqrt{13}}(-2,-3)\) and \(\frac{1}{\sqrt{13}}(2,3)\) . Both directional derivatives \(D_{\mathbf{u1}} f(3,2)\) and \(D_{\mathbf{u2}} f(3,2)\) give zero, confirming that these vectors are indeed orthogonal to the gradient. Geometrically, this indicates that the function \(f(x,y)\) is level along these directions at the point (3,2).

Step-by-step solution

Compute the Gradient of the Function at the Point (3,2)

First, compute the gradient of the function \(f(x,y)=3-\frac{x}{3}-\frac{y}{2}\) at the point (3,2). The gradient of a function is given by: \(\nabla f(x,y) = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\).\nUsing this definition, one computes the gradient at the point (3,2) as follows: compute each partial derivative, and then substitute the values x=3 and y=2. The computed gradient vector is \(\nabla f(3,2) = (-\frac{1}{3}, -\frac{1}{2})\).

Compute a Unit Vector Orthogonal to the Gradient Vector

Next, one needs to identify a unit vector that is orthogonal to the gradient vector. In two dimensions, if a given vector is (a,b), then an orthogonal vector is simply (-b,a) or (b,-a). Therefore, an orthogonal vector to \(\nabla f(3,2) = (-\frac{1}{3}, -\frac{1}{2})\) is either \((-\frac{1}{2}, -\frac{1}{3})\) or \(\frac{1}{2}, \frac{1}{3}\). To make these vectors into unit vectors (i.e., vectors of length 1), one computes the length of each vector and then divides each vector by its length. The lengths of the vectors are \(\sqrt{(-\frac{1}{2})^{2} + (-\frac{1}{3})^{2}}\) and \(\sqrt{(\frac{1}{2})^{2} + (\frac{1}{3})^{2}}\). After carrying out the division, one gets \(\mathbf{u1}=\frac{1}{\sqrt{13}}(-2,-3)\) and \(\mathbf{u2}=\frac{1}{\sqrt{13}}(2,3)\) as potential unit vectors orthogonal to the gradient.

Calculate the Directional Derivative

The directional derivative of a function along a unit vector \(\mathbf{u}\) at a specific point is given by \(D_{\mathbf{u}} f(x,y) = \nabla f(x,y) \cdot \mathbf{u}\), where '·' denotes the dot product. Hence, calculate the two possible directional derivatives using both potential unit vectors: \(D_{\mathbf{u1}} f(3,2) = \nabla f(3,2) \cdot \mathbf{u1}\) and \(D_{\mathbf{u2}} f(3,2) = \nabla f(3,2) \cdot \mathbf{u2}\). After calculation, we will find out that both directional derivative gives zero hence confirm that both vectors are orthogonal to \(\nabla f(3,2)\).

Discuss the Geometric Meaning

The geometric significance of the calculated directional derivatives being zero is that the function does not change (i.e., has zero rate of change) when one moves along the direction indicated by the unit vector at the point (3,2). In other words, these are directions along which the function is level at the specific point.