# Math Help - climber

1. ## climber

The surface of mountain is modeled by the equation h(x,y)=5000 -0.001x^2-0.004y^2. A mountain climber is at the point (500, 300, 4390). In what direction should the climber move in order to ascend at the greatest rates ? Give direction as a vector only i & j components needed.

From that, give the direction that climber should move so that his altitude does not change (h doesn't change). The climber is still assumed to be at the point (500,300,4390).

2. Originally Posted by ggw
The surface of mountain is modeled by the equation h(x,y)=5000 -0.001x^2-0.004y^2. A mountain climber is at the point (500, 300, 4390). In what direction should the climber move in order to ascend at the greatest rates ? Give direction as a vector only i & j components needed.
What exactly is your problem?
1)Just find the gradient at the point.
2)Then evaluate the gradient at x=500 and y=300.

3. yes, solve for gradiantsl at giving the direction as a vector (only i & j components needed)

4. i am not sure if i get it right, but thanks anyway.

5. Originally Posted by ggw
The surface of mountain is modeled by the equation h(x,y)=5000 -0.001x^2-0.004y^2. A mountain climber is at the point (500, 300, 4390). In what direction should the climber move in order to ascend at the greatest rates ? Give direction as a vector only i & j components needed.

From that, give the direction that climber should move so that his altitude does not change (h doesn't change). The climber is still assumed to be at the point (500,300,4390).
grad(h) = (D_x h(x,y), D_y h(x,y)) = (-0.002 x, -0.008 y)

So the direction of steepest ascent is parallel to:

(-0.002*500, -0.008*300) = (1, 2.4)

reducing this to a unit vector gives:

u = (1, 2.4)/sqrt(1+2.4^2) ~= (0.3846, 0.9231) = 0.3846 i + 0.9231 j

The rate of ascent in direction v is v.grad(h), so to maintain a constant heigh the
climber must move off in a direction such that v.grad(h) = 0, so if we put:

v = -0.9231 i + 0.3846 j

this will do the job.

RonL