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

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.

• $P$ is not a point of the function. What do you mean by "slope at the point" $P$? – Raffaele Aug 19 '17 at 16:03
• The assignment is essentially that a mountain is described as the function $f(x,y)$ and a person is standing at the point $P$. The person wants to go down the mountain by going the furthest down in each point. The questions is in which direction the person should start to go (meaning which direction at the point $P$ the slope is the lowest). – osk Aug 19 '17 at 16:17

## 1 Answer

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.