Question

Find a normal vector to the level curve \(f(x, y)=c\) at \(P.\) $$ \begin{array}{l} f(x, y)=6-2 x-3 y \\ c=6, \quad P(0,0) \end{array} $$

Short answer

The normal vector to the level curve at point \(P(0,0)\) is <-2, -3>.

Step-by-step solution

Find the partial derivatives of f

First, calculate the partial derivative of \(f\) with respect to \(x\) and \(y\). \[f_x = \frac{\partial f}{\partial x} = -2\] \[f_y = \frac{\partial f}{\partial y} = -3\] These derivatives represent the slope of the function \(f\) in the x and y directions, respectively.

Evaluate the gradient vector at P(0,0)

Next, we evaluate the gradient of \(f\) at the point \(P\), the result represents the normal vector. The gradient of \(f\), \(\nabla f\), is given by \[\nabla f = \] Substituting the values of \(f_x\) and \(f_y\) in the gradient we get \[\nabla f(0,0) = <-2, -3>\]

Conclusion: Normal Vector

Finally, we find that \(\nabla f(0,0)\) is the normal vector to the level curve at point \(P(0,0)\).