# How does one project the gradient at a point on a surface into a plane?

I am studying Multivariable Calculus and have come to the following excerpt in my book:

I can see clearly how they get from the given function to

$$y'(x)\ =\ \frac{3y}{x}$$

And understand that this slope passes through the given point. The following line leaves me totally lost, however:

You can verify that the solution to this differential equation is $$y\ =\ \frac{4x^3}{27}$$ and the projection of the path of steepest descent in the xy-plane is the curve $$y\ =\ \frac{4x^3}{27}$$

How did they get from any of the given information--the initial equation, its gradient, the slope of the gradient, etc.--to the above equation for y? Furthermore, how am I to find the projection of the gradient in the xy-plane?

• You have a Separable DEQ that leads to $\displaystyle \int \dfrac{1}{y}~dy = \int \dfrac{3}{x}~dx$ with initial condition $y(3) = 4$. – Moo Mar 3 '16 at 14:35
• @Moo - Fist bump for getting it right--and an awesome username – StudentsTea Mar 3 '16 at 17:48
• For the rest of your questions, I think this write-up and Examples would do it: aleph0.clarku.edu/~djoyce/ma131/directional.pdf – Moo Mar 3 '16 at 18:02