Suppose f(x, y) = x/y, P = (1, 3) and v = -1i

Suppose f(x, y) = x/y, P = (1, 3) and v = -1i - 1j. A. Find the gradient of f. ?f = [ ]i + [ ]j B. Find the gradient of f at the point P. (?f)(1, 3) = [ ]i + [ ]j C. Find the directional derivative of f at P in the direction of v, where the direction u of a vector v is the unit vector obtained by normalizing that vector, i.e., u = v/||v||. (D_u f)(1, 3) = [ ] D. Find the maximum rate of change of f at P. [ ] E. Find the (unit) direction vector in which the maximum rate of change occurs at P. [ ]i + [ ]j

Transcript

00:01 In this problem, consider a 2d dimensional function f of xy equal to x over y, and consider the point 1 3 and the vector minus i minus j.

00:13 In question a, we want to find the gradient of our function f.

00:26 So the gradient of f will give us a vector containing the partial derivatives of f with vector x in the i direction, plus the partial derivative of f with step to y in the j direction.

00:45 So let's calculate our partial derivatives to insert it into our gradient expression.

00:52 So del f over del x will give us 1 over y, and del f over del y will give us minus x over y squared, which means that our gradient at any point x and y will be equal to 1 over y in the i direction minus x over y.

01:16 Square in the j direction in question b we want to find the gradient of f at point p one three so all we to do is replace x with one and y with three in our formula above for the gradient of f so we will obtain 1 over 3 in the i direction minus 1 over 9 in the j direction in question c, we want to find the directional derivative of f in direction u, where u is a unit vector in the direction of our given vector v.

02:13 So our directional derivative corresponds to the dot product of our gradient with the unit vector u.

02:26 So let's first start by finding our unit vector.

02:30 To find you, we will take v and divide you.

02:37 By its magnitude.

02:42 So you will be equal to the vector minus i minus j divided by the square root of 1 plus 1, so square root of 2.

02:55 So this is our direction vector.

02:58 And the reason i want unit vector is because we don't want the vector to affect the magnitude of our gradient.

03:09 So the directional derivative will give us the dot product between 1 over y i minus x x over y square j dot minus 1 over the square root of 2 i minus 1 over the square root of 2 j which will be equal to minus 1 over square root of 2 y plus x over square root of 2 y squared now in this problem specific we want to find the directional derivative at our point p so at the point p so at the 0 .1, 3.

04:02 So let's take this expression and evaluate at x is equal to 1 and y is equal to 3.

04:08 We'll find minus 1 over 3 times square root of 2, plus 1 over 9 times square root of 2...

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