Direction of the lowest slope in $f(x,y)=100-(x^2+8y^{5/2})^2$ at $P$: $(1/e,1/100)$

Question

We have the following function:

$f(x,y)=100-(x^2+8y^{5/2})^2$ when $x \ge 0, y \ge 0$

And the following point $P$: $(1/e,1/100)$.

We would like to determine in which direction the slope at the point $P$ is the lowest.

I can calculate the gradient, $\operatorname{grad}( f(x,y)) = (4x(x^2+8y^{5/2}),40y^{3/2}(x^2+8y^{5/2}))$. I'm not even sure how to insert the point in the gradient, it's kind of complicated calculations by paper. And even if I did, how would i get the lowest slope? I assume it would be in the opposite direction of the steepest slope.

Answer 1

The point gives you the numbers for (x,y). For example x=1/e. Use a calculator if 1/e doesn't look like a number.

Put those values into the gradient to find the direction of greatest increase. As you said, the opposite direction would be greatest decrease.

Direction might be given by a unit vector - it is not clear from the question.