# UT M 408C - HW 14.6-solutions-1 (6 pages)

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## HW 14.6-solutions-1

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- Pages:
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- School:
- University of Texas at Austin
- Course:
- M 408c - Differential and Integral Calculus

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lewis scl876 HW 14 6 radin 55315 This print out should have 10 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 1 0 1 2 3 10 0 points Q 3 2 Find the gradient of P 2 3 f x y xy 3x y 9x2 y y 2 2xy 3x3 2 f 3x3 2xy y 2 9x2 y 3 f y 2 9x2 y 3x3 2xy 4 f 2xy 3x3 y 2 9x2 y I positive at Q 5 f 2xy 3x3 9x2 y y 2 II positive at P 6 f 2 y 9x y 2xy 3x 0 R 1 f 2 1 Which of the following properties does the derivative d f r t dt have III zero at R 3 correct 1 I only Explanation Since 2 II only f x y f f x y 3 all of them 4 I and II only we see that 5 II and III only f y 2 9x2 y 2xy 3x3 6 none of them 7 III only 8 I and III only correct Explanation By the multi variable Chain Rule keywords 002 10 0 points The contour map given below for a function f shows also a path r t traversed counterclockwise as indicated d f r t f r t r t dt Thus the sign of d f r t dt lewis scl876 HW 14 6 radin 55315 will be the sign of the slope of the surface in the direction of the tangent to the curve r t and we have to know which way the curve is being traversed to know the direction the tangent points In other words if we think of the curve r t as defining a path on the graph of f then we need to know the slope of the path as we travel around that path are we going uphill downhill or on the level That will depend on which way we are walking From the contour map we see that I TRUE at Q we are ascending the contours are increasing in the counter clockwise direction II FALSE at P we are descending the contours are decreasing in the counter clockwise direction III TRUE at R we are on the level we are following the contour keywords contour map contours slope curve on surface tangent Chain Rule multivariable Chain Rule 003 10 0 points Find the directional derivative fv of f x y 2 y 1 2 x at P 1 1 in the direction of the vector P Q when Q 4 5 1 fv 1 correct 5 2 fv 1 10 3 fv 1 5 2 The directional derivative of f x y at P in the direction of v P Q is given by the dot product v fv P f P v But when f x y 2 y 1 2 x we see that y 1 2 f 3 x x 1 1 2 f y xy so that at P 1 1 f 1 1 i j D f P D ED3 4E 1 1 1 5 5 5 004 10 0 points Find the maximum slope on the graph of f x y 2 sin xy at the point P 5 0 1 max slope 1 2 max slope 5 3 max slope 2 4 max slope 6 max slope 10 correct Explanation 1 1 Consequently 5 max slope 10 1 10 E On the other hand v h3 4i which as a vector of unit length becomes D3 4E v v 5 5 4 fv 0 5 fv 7 max slope 5 8 max slope 2 lewis scl876 HW 14 6 radin 55315 Explanation At P 5 0 0 the slope in the direction of v is given by v f 5 0 v 5 25 Explanation The rate of change at P 4 3 7 is given by Du V V 4 3 7 But when f x y 2 sin xy 3 v v Now when V x y z 2x2 4xy xyz the gradient of f is f x y 2y cos xy i 2x cos xy j it follows that V h4x 4y yz 4x xz xyi so at P 5 0 f 5 0 10 j Consequently the slope at P will be maximized when v j in which case max slope 10 keywords slope gradient trig function maximum slope and V 4 3 7 h25 12 12i Consequently Du V h25 12 12i 25 h1 1 1i 3 3 keywords 006 10 0 points Suppose that over a certain region of space the electrical potential V is given by If the graph of z f x y at P 1 2 has slope 2 in the x direction and slope 3 in the y direction find the slope at P in the direction of the vector V x y z 2x2 4xy xyz v 3 i 4 j 005 10 0 points Find the rate of change of the potential at P 4 3 7 in the direction of the vector v i j k 1 25 25 2 correct 3 25 3 3 4 25 3 1 slope 2 5 2 slope 4 5 3 slope 3 5 6 4 slope correct 5 5 slope 1 Explanation lewis scl876 HW 14 6 radin 55315 The slope of the graph of z f x y in the x direction at P 1 2 is the value fx 1 2 f x 1 2 2 of the partial derivative fx at P while the slope of the graph of z f x y in the ydirection at P is the value fy 1 2 f y 1 2 3 of the partial derivative fy at P 1 2 On the other hand the slope at P in the direction of any vector v ai bj is given by v f 1 2 v 1 fx 1 2 i fy 1 2 j a i b j v When v 3 i 4 j therefore the graph of f has slope 1 6 2 i 3 j 3 i 4 j 5 5 4 4 x 7y z 7 5 x 7y z 7 Explanation Note that xey cos z z 7 Let F x xey cos z z The equation to the tangent plane to the surface at the point P 7 0 0 is given by Fx P x 7 Fy P y 0 Fz P z 0 Since Fx ey cos z Fx Fy xey cos z Fy P P 1 7 and Fz xey sin z 1 Fz P 1 it follows that the equation of the tangent plane is x 7y z 7 keywords at P in the direction of v 008 keywords If 007 10 0 points Find an equation for the tangent plane to the graph of z xey cos z 7 at the point 7 0 0 1 x 7y z 7 10 0 points f x y x2 2y 2 use the gradient vector f 3 1 to find the tangent line to the …

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