gradient vectors and directional derivatives

[Guest]

Okay for homework in Calc 3 im trying to solve the following story problem:

Near a buoy, the depth of a lake at point with coordinates (x,y) is z=200+0.02x^2-0.001y^3, where x,y and z are measured in meters. A fisherman in a boat starts at (80,60) and moves towards the buoy which is at (0,0). Is the water under the boat getting deeper or shallower when he departs?

Now im thinking I need to find the gradient vector. So Im also thinking I need the partial derivatives with respect to x,y, and z. Then I dot this with the unit vector. Im thinking the unit vector in this case is like <80-0,60-0> but since he moves towards (0,0) im thinking maybe this is negative.

Is my thinking correct? Do I need the gradient vector? Is -80,-60 the vector I need to normalize? Am I missing anything?
Thanks for any help cause Im not sure if im right or not.

 

[tridelta246]

Okay for homework in Calc 3 im trying to solve the following story problem:

Near a buoy, the depth of a lake at point with coordinates (x,y) is z=200+0.02x^2-0.001y^3, where x,y and z are measured in meters. A fisherman in a boat starts at (80,60) and moves towards the buoy which is at (0,0). Is the water under the boat getting deeper or shallower when he departs?

Now im thinking I need to find the gradient vector. So Im also thinking I need the partial derivatives with respect to x,y, and z. Then I dot this with the unit vector. Im thinking the unit vector in this case is like <80-0,60-0> but since he moves towards (0,0) im thinking maybe this is negative.

Is my thinking correct? Do I need the gradient vector? Is -80,-60 the vector I need to normalize? Am I missing anything?

Thanks for any help cause Im not sure if im right or not.

GRADIENT(z) is the vector whose direction is that for which the Directional Derivative has maximum value at point (x,y), and whose magnitude |GRADIENT(z)| equals that maximum Directional Derivative at (x,y). In effect, GRADIENT(z) is normal to the contours of "z" on the (x,y) plane.

To obtain the (scalar) Directional Derivative in another direction, calculate the DOT product of GRADIENT(z) with the (normalized) UNIT vector in the direction of this (new) desired Directional Derivative. In this case, that unit vector will be:
<(0 - 80), (0 - 60)>/sqrt((80^2) + (60^2))

Calculate GRADIENT(z) using standard definition involving partials of "z" with respect to "x" and "y" and proceed with the DOT product to obtain the desired (scalar) Directional Derivative.

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