ha lvh262 Homework 14 4 karakurt 56295 This print out should have 8 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Find the linearization of z f x y at P 3 1 when f 3 1 1 and fx 3 1 2 fy 3 1 1 1 L x y 1 2x y 2 L x y 1 2x y 3 L x y 8z 2x y 4 L x y 8z 2x y 5 L x y 1 2x y keywords linearization partial derivative radical function square root function 002 fx 2 3 3 and so at P 3 1 L x y f 3 1 fx 3 1 x 3 fy 3 1 y 1 Consequently the linearization of f at P is 2 x 3y z 8 0 3 3x y z 8 0 4 3x y z 1 0 correct 5 x 3y z 8 0 6 3x y z 1 0 Explanation The equation of the tangent plane to the graph of z f x y at the point P a b f a b is given by z f a b fx a b x a fy a b y b But when P a b f a b 2 3 8 we see that a b 2 3 which after rearrangement becomes L x y 8 2x y f a b 8 while fx a b 3 L x y 1 2 x 3 y 1 fy 2 3 1 1 x 3y z 1 0 L x y f a b fx a b x a fy a b y b 10 0 points Find an equation for the tangent plane at the point P 2 3 8 on the graph of z f x y when 6 L x y 8 2x y correct Explanation The linearization of z f x y at P a b is given by 1 fy a b 1 So at P the tangent plane has equation z 8 3 x 2 y 3 which after rearrangement becomes 3x y z 1 0 ha lvh262 Homework 14 4 karakurt 56295 keywords tangent plane partial derivative 003 2 keywords linearization partial derivative Linear Approximation 10 0 points Use Linear Approximation to estimate the value of f 0 9 1 9 when f 1 2 3 004 Use Linear Approximation to estimate the change f f a x b y f a b and fx 1 2 1 fy 1 2 4 1 f 0 9 1 9 3 5 correct in f when x 0 3 4 f 0 9 1 9 3 6 5 f 0 9 1 9 3 9 Explanation Linear Approximation uses the Linearization L x y f a b fx a b x a fy a b y b of z f x y at P a b to write f x y L x y The value of L x y is usually much easier to calculate and for x y near a b it gives a good estimate of f x y fx a b 2 2 f 0 4 3 f 0 3 4 f 0 5 correct 5 f 0 2 Explanation By Linear Approximation of f at a b f a x b y L a x b y f a b fx a b x fy a b y Thus f f a x b y f a b fx a b x fy a b y Consequently when x 0 3 fx 1 2 x 1 fy 1 2 y 2 Consequently the linearization of f at P is L x y 3 x 1 4 y 2 But then f 0 9 1 9 L 0 9 1 9 3 5 fy a b 1 1 f 0 6 Now at P 1 2 L x y f 1 2 y 0 1 and 2 f 0 9 1 9 3 8 3 f 0 9 1 9 3 7 10 0 points y 0 1 and fx a b 2 fy a b 1 Linear Approximation gives the estimate f 2 0 3 1 0 1 0 5 ha lvh262 Homework 14 4 karakurt 56295 3 while keywords linearization partial derivative Linear Approximation 005 10 0 points f x 2 1 at the point P 2 1 2 L x y 7 2x 2y correct 3 L x y 7 2x 2y L x y 1 2 x 2 2 y 1 which after rearrangement becomes keywords linearization partial derivative radical function square root function 006 4 L x y 5 x 2y 6 L x y 7 x 2y at the point 9 1 Explanation The linearization of z f x y at the point P a b is given by 1 L x y 2 L x y L x y f a b a b f y 3 L x y a b y b 4 L x y Now when f x y p 10 0 points Find the linearization L x y of f x y y x 5 L x y 5 2x 2y x a 2 L x y 7 2x 2y 1 L x y 5 2x 2y f x 2 1 So at P the linearization of f is Find the linearization L x y of p f x y 7 x2 2y 2 f y 2 7 x2 2y 2 5 L x y 6 L x y we see that 3 1 x 3y correct 2 6 3 1 3x y 2 6 3 1 3 x y 2 6 1 3 3 x y 6 2 3 1 3x y 2 6 3 1 x 3y 2 6 f x p x 7 x2 2y 2 Explanation The linearization of f f x y at a point a b is given by f 2y p y 7 x2 2y 2 L x y f a b x a while a b y b But when f x y y x Thus at P f 2 1 1 f x y f x 2 x f x y f y a b ha lvh262 Homework 14 4 karakurt 56295 and thus when a b 9 1 f x 1 a b 6 f y f y 3 a b 1 0 4 4 Consequently while f a b 3 Consequently L x y 3 1 L x y x 3y 2 6 1 1 x 4y 2 4 2 keywords keywords 007 008 10 0 points Find the linearization L x y of the function f x y tan 1 x 8y 10 0 points Find an equation for the plane passing through the origin that is parallel to the tangent plane to the graph of z f x y 3x2 2y 2 3x 3y at the point 1 0 1 L x y x 4y 1 4 2 at the point 1 1 f 1 1 1 z 3x y 3 0 1 1 2 L x y x 4y 2 2 2 z 3x y 5 …
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