Suppose that you are on a landscape whose elevation can be...

Suppose that you are on a landscape whose elevation can be modeled by the function f(x, y) = e xy − xy2 , and you are standing at the point (1, 2). (a) Find the rate of change of your elevation if you were to walk north (positive y-direction), or east (positive x-direction). (b) Find the rate of change of your elevation if you were to walk south, or west

Transcript

00:01 Okay, so before we go on with the solution of this exercise, let me make a remark.

00:06 Well, if we are walking north, then it means that we are moving following this direction, the direction v given by 0, 1.

00:21 If we are walking east, then the direction is given by the vector v equal to 1, 0.

00:32 Perfect.

00:33 Similarly, if we are walking south, the direction is going to be the vector, well, this one is going to be 0, negative 1.

00:49 If we are walking west, then the direction is going to be the vector 0, negative 1.

00:59 Perfect.

01:00 We are going to use this remark in the solution of our exercise because we just need to keep in mind that the rate of change of our function f is going to be the directional derivative.

01:13 Okay, so for part a of our exercise, the rate of change in the north direction, well, this one is going to be the directional derivative.

01:30 The directional derivative of f in the direction of v, this one, evaluated at p.

01:40 So we have the gradient of f evaluated at p dot product with v, which is 0, 1.

01:51 Okay, now before we go on, let's compute the gradient of f.

01:56 We are going to use it.

01:58 Well, the gradient of f is equal to the partial derivative of f with respect to x, which is y, e to the xy, minus y squared, comma, the partial derivative of f with respect to y, which is x, e to the xy, minus 2xy.

02:24 Perfect.

02:24 So the evaluation of our gradient at p is going to be, well, p is the point 1, 2.

02:32 So we are going to have 2 multiplied by e, minus 4.

02:42 This one is going to be, okay, 1 multiplied by e squared...

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