Question

Suppose that you are climbing a hill whose shape is given by z=484−0.1x²−0.09y², and that you are at the point (20,40,300). In which direction should you proceed initially in order to reach the top of the hill fastest? If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?

Answer 1

To reach the top of the hill fastest, you should proceed in the direction of the gradient vector of the hill at that point. The gradient vector is a vector that points in the direction of the greatest increase in the function (in this case, z) and has a magnitude equal to the rate of change in that direction.

The gradient vector is given by:

∇z = <∂z/∂x, ∂z/∂y>

So for this problem, we have:

z = 484 - 0.1x² - 0.09y²

∂z/∂x = -0.2x
∂z/∂y = -0.18y

At the point (20, 40, 300), we have:

∇z = <-4, -7.2>

To climb the hill fastest, you should proceed in the direction of this vector.

The angle above the horizontal that you will be climbing initially can be found using the dot product of the gradient vector and a unit vector in the direction of motion. Let u be a unit vector in the direction of motion. Then:

cos θ = (∇z · u) / ||∇z|| ||u||

where ||v|| denotes the magnitude of vector v.

To maximize your initial rate of ascent, you should choose u to be a unit vector in the direction of ∇z. Then:

cos θ = (∇z · ∇z / ||∇z||²

= (-4)(-4) + (-7.2)(-7.2) / (16 + 51.84)

= 65.44 / 67.84

= 0.964

θ ≈ 0.27 radians or about 15.5 degrees above horizontal.

I hope this helps! Let me know if you have any other questions.