Gradient Vector: Find the Projection of Steepest Ascent Path on xy-Plane

[plexus0208]

Homework Statement

A hiker climbs a mountain whose height is given by z = 1000 - 2x² - 3y².

When the hiker is at point (1,1,995), she moves on the path of steepest ascent. If she continues to move on this path, show that the projection of this path on the xy-plane is y = x^3/2

Homework Equations

The Attempt at a Solution

The path of steepest ascent is in the direction in which she would ascent as rapidly as possible, aka the gradient vector.

gradf = f_x i + f_y j = -4x i -6y j
gradf at (1,1,995) = -4 i - 6 j

What now?

 

[HallsofIvy]

Okay, since the gradient vector always points in the direction of fastest ascent, she should be moving on a curve whose tangent vector is -4xi- 6yj. That is, dx/dt is a multiple of -4x and dy/dt is the same multiple of -6y. That means that dy/dx= (-4x)/(-6y)= 2x/dy. Solve the differential equation dy/dx= (2x)/(3y) with intial value y(1)= 1.

 

[plexus0208]

You probably mean dy/dx = 3y/2x, not 2x/3y. But thanks a lot!

 

[HallsofIvy]

yes, of course. Sorry for that.