Question

Find a normal vector to the level curve \(f(x, y)=c\) at \(P.\) $$ \begin{array}{l} f(x, y)=x y \\ c=-3, \quad P(-1,3) \end{array} $$

Short answer

The normal vector to the level curve \(f(x, y) = -3\) at the point \(P(-1,3)\) is (3, -1).

Step-by-step solution

Calculate the gradient vector

First find the partial derivatives of the given function \(f(x, y) = xy\) with respect to \(x\) and \(y\). So: \(\frac{df}{dx} = y\) and \(\frac{df}{dy} = x\). The gradient vector \(\nabla f\) at any point \((x, y)\) is given by \(\nabla f = (\frac{df}{dx}, \frac{df}{dy})\). Therefore, substitute \(x = -1\) and \(y = 3\) into the equations to find the gradient vector at the point \((-1, 3)\). This yields \(\nabla f = (3, -1)\).

Using the direction of the gradient vector to find the normal vector

The gradient vector \(\nabla f\) gives the direction of the maximum rate of change of \(f(x, y)\) at a given point, and it is always perpendicular to level curves. Hence, the gradient vector \(\nabla f\) at the point \((-1, 3)\) is the normal vector to the level curve \(f(x, y) = -3\) at \(P = (-1, 3)\). Therefore, the normal vector at point \((-1, 3)\) is \(\nabla f = (3, -1)\).