Question

Use the gradient to find a unit normal vector to the graph of the equation at the given point. Sketch your results $$ x e^{y}-y=5,(5,0) $$

Short answer

The unit normal vector to the graph of the equation at point (5,0) is \( \frac{(1,4)}{\sqrt{17}} \). You can sketch this vector starting at the point (5,0) and ending at \(\left(5+\frac{1}{\sqrt{17}}, 0+\frac{4}{\sqrt{17}}\right)\) on a 2-dimensional Cartesian plane.

Step-by-step solution

Find the gradient

The first part involves determining the gradient of the given function. This is done by calculating the partial derivatives of the function with respect to x and y. So, \( \nabla F = (F_x, F_y) \) where \( F(x, y) = x e^{y} - y - 5 \). Taking the partial derivatives, we have \( F_x = e^{y} \) and \( F_y = x e^{y} - 1 \). Thus, the gradient at any point (x, y) on the surface F=0 is \( \nabla F = (e^{y}, x e^{y} - 1) \).

Evaluate the gradient at the given point

Next, we plug in the given point into the gradient that we found, therefore, evaluating the gradient at point (5,0). Given \( \nabla F=(e^{y}, x e^{y} - 1) \), we substitute the values of x=5 and y=0 to obtain the vector \( \nabla F(5,0)=(1,4) \).

Normalize the vector

A unit vector is obtained through the normalization of \( \nabla F \). This is done by dividing the vector by its magnitude, which in this case is \( \sqrt{(1^2 + 4^2)} = \sqrt{17} \). Therefore, the unit normal vector is \( \frac{\nabla F}{|\nabla F|} = \frac{(1,4)}{\sqrt{17}} \).

Sketching the results

Remember the 2-dimensional diagram; draw a Cartesian plane and depict the point (5,0). Drawing the unit normal vector, it starts from (5,0) and finishes at the point \(\left(5+\frac{1}{\sqrt{17}}, 0+\frac{4}{\sqrt{17}}\right)\). This vector is perpendicular to the tangent plane at point (5,0).