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UT M 408D - Homework 14.6-solutions

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ha lvh262 Homework 14 6 karakurt 56295 This print out should have 10 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Find the directional derivative of f x y xe 8y at the point 3 0 in the direction indi cated by the angle 2 1 24 1 1 f 6x2 y y 2 2xy 2x3 2 f y 2 6x2 y 2x3 2xy 3 f 2xy 2x3 y 2 6x2 y 4 f y 2 6x2 y 2xy 2x3 correct 5 f 2xy 2x3 6x2 y y 2 6 f 2x3 2xy y 2 6x2 y Explanation Since 2 8 f x y 3 0 f f x y we see that 4 24 correct f 5 8 Explanation The directional derivative of f x y at x0 y0 in the direction of a unit vector u where u makes an angle with the positive x axis is Du f x0 y0 fx x0 y0 cos fy x0 y0 sin When f x y xe 8y it follows that fx x y e 8y and fy x y 8xe 8y Consequently Du f 3 0 e 8 0 cos 8 3 e 8 0 sin 2 2 24 keywords 002 f x y xy 2 2x3 y keywords 003 10 0 points Find the directional derivative fv of the function f x y 6 2x y at the point P 3 4 in the direction of the vector v h 3 4 i 1 fv 1 2 fv 4 5 3 fv 9 10 4 fv 11 10 5 fv 6 correct 5 10 0 points Find the gradient of y 2 6x2 y 2xy 2x3 Explanation ha lvh262 Homework 14 6 karakurt 56295 Now for an arbitrary vector v v fv f v 6 Dv f P where we have normalized so that the direction vector has unit length But when f x y 6 2x y then f 2 y i x y j P Explanation For an arbitrary vector v v Dv f f v where we have normalized so that the direction vector has unit length But when y f x y z x tan 1 z 4i f f f i j k x y z y x i tan 1 j z z 1 y z 2 3 j 2 f Consequently when v h 3 4i fv 3 4 3 4 2 1 4 then At P 3 4 therefore f 2 v 6 v 5 xy z 2 1 y z 2 k Thus keywords 1 f tan 004 10 0 points Find the directional derivative Dv f of y f x y z x tan 1 z 1 Dv f P 1 4 1 2 Dv f P correct 12 1 3 Dv f P 12 4 Dv f P 1 3 5 Dv f P 1 3 z i z2 xz xy j 2 k 2 y z y2 At P 1 1 1 therefore f P 1 1 i j k 4 2 2 Consequently when at the point P 1 1 1 in the direction of the vector v i 2j 2k y v i 2j 2k we see that v and Dv f P p 1 22 22 3 1 1 1 i j k i 2j 2k 3 4 2 2 Consequently Dv f P 1 1 1 1 3 4 12 keywords directional derivative gradient dot product unit vector ha lvh262 Homework 14 6 karakurt 56295 005 10 0 points Find the maximum slope on the graph of f x y 3 sin xy 006 3 10 0 points Suppose that over a certain region of space the electrical potential V is given by V x y z 4x2 5xy xyz at the point P 5 0 Find the rate of change of the potential at P 1 1 5 in the direction of the vector 1 max slope 15 v i j k 2 max slope 1 3 max slope 1 4 max slope 5 7 3 5 max slope 15 correct 2 7 6 max slope 3 7 3 correct 3 7 max slope 3 4 7 8 max slope 5 7 5 3 Explanation At P 5 0 0 the slope in the direction of v is given by v f v 5 0 But when Explanation The rate of change at P 1 1 5 is given by Du V V 1 1 5 v v Now when f x y 3 sin xy V x y z 4x2 5xy xyz the gradient of f is f x y 3y cos xy i 3x cos xy j it follows that V h8x 5y yz 5x xz xyi so at P 5 0 and f 5 0 15 j Consequently the slope at P will be maximized when v j in which case max slope 15 V 1 1 5 h8 0 1i Consequently Du V h8 0 1i h1 1 1i 7 3 3 keywords keywords slope gradient trig function maximum slope 007 10 0 points ha lvh262 Homework 14 6 karakurt 56295 Find the equation of the tangent plane to the surface at the point 2 0 0 1 x 2y z 2 4x2 5y 2 5z 2 66 2 x 2y z 2 at the point 2 1 3 3 x 2y z 2 1 8x 5y 15z 56 4 x 2y z 2 correct 2 8x 5y 15z 66 5 x 2y z 2 3 8x 5y 15z 56 Explanation Note that 4 4x 5y 5z 66 xey cos z z 2 Let 5 8x 5y 15z 66 correct F x xey cos z z Explanation Let The equation to the tangent plane to the surface at the point P 2 0 0 is given by F x 4x2 5y 2 5z 2 The equation to the tangent plane to the surface at the point P 2 1 3 is given by Fx P x 2 Fy P y 1 Fz P Fx Fx 8x Fx Fy 10y Fy P z 3 0 16 10 P and Fz 10z Fz P 30 it follows that the equation of the tangent plane is 8x 5y 15z 66 P x 2 Fy Find an equation for the tangent plane to the graph of z xey cos z 2 P z 0 Fx Fy xey cos z Fy P P 1 2 and Fz xey sin z 1 Fz P 1 it follows that the equation of the tangent plane is x 2y z 2 keywords If 10 0 points y 0 Fz Fx ey cos z 009 keywords P Since Since 008 4 10 0 points f x y x2 3y 2 use the gradient vector f 4 3 to find the tangent line to the level curve f x y 43 at the point 4 3 1 4x 9y 43 ha lvh262 Homework 14 6 karakurt 56295 5 The graph of 2 8x 18y 43 z f x y 3x2 y 2 2x 3y 3 4x 9y 43 correct is the set of all points …


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