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6.4 - Directional Derivative & Gradient In #1-4, for the given function & point p & vector ?v (a) Find the gradient of the function

6.4 - Directional Derivative & Gradient

In #1-4, for the given function & point p & vector ?v

(a) Find the gradient of the function (b) find

the gradient at the point p (c) find the maximum rate of change at the point p (d) find the unit

vector in the direction of the greatest rate of change at p (e) Find the directional derivative of the

function at the given point p in the direction of the vector ?v

1. f(x,y) = y/x ... p=(3,2) ... ?v

= (-1,-2)

2. g(x,y) = x4y2 - 3x2y3 ... p=(3,-2) ... ?v

= (4,3)

3. f(x,y) = x

x2+y2 ... p=(2,1) ... ?v

=(3,5)

4. h(r,s) = ln(r2+2s) ... p=(4,-1) ... ?v

= (0.6, 0.8)

Also,

5. For 32 = x2+y4 at the point (-4,2), find the normal vector & the tangent vector.

6. For z = 2xy2 - ln(y), at the point (3,1,6), (a) find the normal vector (b) find the equation of

the normal line ... bonus: (c) find the equation of the tangent plane.

7. Find the derivative of f(x,y) = x ln(x+y) in the direction of the given angle, assuming the

gradient is angle 0 [in other words, this is the measure of the angle between the gradient

and the direction vector] (a) 0 (b) ? (c) ?/4 (d) ?/3

image text in transcribed
HW 6.4 - Directional Derivative & Gradient In #1-4, for the given function & point p & vector 17' {a} Find the gradient of the function [b] nd the gradient at the point p [(3] nd the maximum rate of change at the point p [(1] nd the unit vector in the direction of the greatest rate of change at p [e] Find the directional derivative of the function at the given point p in the direction of the vector v 1. 2. 3. 4-. Also, 5. 6. 7. _. f(x,y] = 31/): p=[3,2] 1T: = [-1,-2] g[x,y] 2 y2 3x23:3 p=[3,-2] f! = (4,3) f[x,y] = 2x 2 p={2,1) v =[3,5] x +y h[r,s] = ln[r2+25] p=[4-,1] a; = [0.6, 0.8) For 32 2 113+}? at the point [4,2], nd the normal vector & the tangent vector. For 2 = 21;}?2 ln[y], at the point (3,1,6), [a] nd the normal vector [b] nd the equation of the normal line bonus: [c] nd the equation of the tangent plane. Find the derivative of ffxy) = X ln(x+y) in the direction of the given angle, assuming the gradient is angle 0 [in other words, this is the measure of the angle between the gradient and the direction vector] [a] 0 (b) TI: [(3] Tt/4 [d] 11/3

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