Let f(x, y. z) = x1 + xy + yz. Grading Policy: Points 7+7+2+2+2. Calculate Delta

Question

Let f(x, y. z) = x1 + xy + yz. Grading Policy: Points 7+7+2+2+2. Calculate Delta (f)(l. 2,5). Grading Policy: gradient 4 and evaluation 3. Answer: Gradient = evaluates to . (b) Calculate the directional derivative of / in the direction of at the point (1.2.5) (In notation: Dv(f)( 1,2.5)). Grading Policy: Gradient 4. evaluation of formula 3. One off for forgetting to length. Answer: The derivative is -2/ square root 14 Find a vector v such that Dv(f)(1,2,5) = 0. Answer: Perpendicular to the gradient: Many possible answers. For example

Explanation / Answer

c)

let (a,b,c) be the vector v

f. v/|v|=0

<4,6,2>.<a,b,c>/(a2+b2+c2) =0

==>4a+6b+2c=0

any (a,b,c) satisfying 4a+6b+2c=0 is answer

let a=6 ,b=-4 ==>4*6 +6*(-4) +2c=0==>c=0

<a,b,c>=<6,-4,0>

d)directional derivative is maximum when v=f

==>v=<4,6,2>

Du=f.f/|f|

Du=|f||f|cos0/|f|

Du=|f||f|*1/|f|

Du=|f|

Du=sqrt(42+62+22)

Du=2sqrt14

e)directional derivative is minimum when v=-f

==>v=-<4,6,2>

Du=f.(-f)/|-f|

Du=|f||f|cos180o/|f|

Du=-|f||f|*1/|f|

Du=-|f|

Du=-sqrt(42+62+22)

Du=-2sqrt14

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