psCalculus2Ans

psCalculus2Ans - Fall 2007 Problem Set #10- Answer key...

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Fall 2007 ARE211 Problem Set #10- Answer key Second Calculus Problem Set (1) Consider the function f ( x, y, z ) = xyz , with y = x 2 and z = x 1 / 3 . (a) Rewrite f as a function g : R R alone and compute g p ( · ). Using g p , approximate the change in f when x increases by 0.1 units, starting from (8 , 64 , 2). Ans: g ( x ) = f ( x, x 2 , x 1 / 3 ) = x × x 2 × x 1 / 3 = x 10 / 3 so that g p ( x ) = 10 / 3 x 7 / 3 g p (8) = 10 / 3 × 8 7 / 3 = 10 / 3 * 2 7 = 426 . 6667 dg = df = 0 . 1 × 10 / 3 × 128 = 42 . 6 . (b) Compute the total derivative of f with respect to x . Using the total derivative, ap- proximate the change in f when x increases by 0.1 units, starting from (8 , 64 , 2). Ans: d f d x = f x + f y d y d x + f z d z d x = yz + xz × 2 x + yx × 1 / 3 x - 2 / 3 = x 7 / 3 + x 4 / 3 × 2 x + x 3 × 1 / 3 x - 2 / 3 = x 7 / 3 (1 + 2 + 1 / 3) = 10 / 3 x 7 / 3 so that df = d f (8) d x = 0 . 1 × 10 / 3 × 8 7 / 3 = 42 . 6 (c) Write down the diFerential of f at (8 , 64 , 2). Using the diFerential, approximate the change in f when x increases by 0.1 units, starting from (8 , 64 , 2).
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2 Ans: f ( x, y, z ) = b yz xz yx B so that f (8 , 64 , 2) = b 128 16 512 B df = f ( x, y, z ) dx y p ( x ) dx z p ( x ) dx p = b 128 16 512 B 0 . 1 16 × 0 . 1 0 . 0833 × 0 . 1 = 12 . 8 + 16 × 0 . 16 + 512 × 0 . 025 = 42 . 7 (d) Identify the direction h * that ( x, y, z ) moves in, starting from (8 , 64 , 2), when x in- creases. Write down the directional derivative of f in the direction h * , i.e., f h * ( · , · , · ), and evaluate this derivative at (8 , 64 , 2). Using f h * (8 , 64 , 2), approximate the change in f when x increases by 0.1 units, starting from (8 , 64 , 2). Ans: When x increases by one, the vector ( x, y, z ) increases in the direction ( dx, y p ( x ) dx, z p ( x ) dx ) . When x = 8 and dx = 1 , therefore, ( x, y, z ) increases in the direction h * = (1 , 16 , 0 . 0833) . The unit length vector pointing in this direction is h * / || h * || = (0 . 0624 , 0 . 9980 , 0 . 0052) . Therefore, using the diFerential to compute the directional derivative, f h * (8 , 64 , 2) = b 128 16 512 B 0 . 0624 0 . 9980 0 . 0052 = 26 . 613 . Now a dx of 0.1 induces a shift in R 3 of 0 . 1 h * , which has length 0 . 1 || h * || . df = f h * (8 , 64 , 2) × 0 . 1 || h * || = 42 . 67 (e) Check to see that all four of these distinct methods give you the same answer! Ans: Amazingly, they do! (2) Recall that a function f : R n R m is nothing more than m functions, f 1 ...f m , each mapping R n R , and stacked on top of each other. (a) Using this fact, write down a formal deFnition of the directional derivative of f at x 0 in the direction h R n , for a function f : R n R m . Your deFnition should be of the
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3 form blah, blah = lim k →∞ blah blah blah, blah Ans: Defnition: Given f : R n R m and h R n , the directional derivative oF f at x 0 in the direction h is given by, For i = 1 , ...m , f i h = lim | k |→∞ (
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psCalculus2Ans - Fall 2007 Problem Set #10- Answer key...

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