lewis (scl876) – HW 14.6 – radin – (55315) 1This print-out should have 10 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsFind the gradient o ff(x, y) = xy2+ 3x3y .1. ∇f =9x2y − y2, 2xy + 3x32. ∇f =3x3− 2xy, y2+ 9x2y3. ∇f =y2+ 9x2y, 3x3− 2xy4. ∇f =2xy + 3x3, y2+ 9x2y5. ∇f =2xy + 3x3, 9x2y − y26. ∇f =y2+ 9x2y, 2xy + 3x3correctExplanation:Since∇f(x, y) =∂f∂x,∂f∂y,we see that∇f =y2+ 9x2y, 2xy + 3x3.keywords:002 10.0 pointsThe contour map given below for a functionf shows al so a path r(t) traversed counter-clockwise as indicat ed.0123-3-2-10QPRWhich of the following properties does thederivativeddtf(r(t))have?I positive at Q,II positive at P ,III zero at R.1. I only2. II only3. all of them4. I and II only5. II and III only6. none of them7. III only8. I and III o nl y correctExplanation:By the multi-va riable Chain Rule,ddtf(r(t)) = (∇f )(r(t))·r′(t) .Thus the sign ofddtf(r(t))lewis (scl876) – HW 14.6 – radin – (55315) 2will be the si gn of the slope of the surface inthe direction of the tangent to the curve r(t),and we have to know which way the curveis being traversed to know the direction thetangent points. In other words, if we think ofthe curve r(t) as defining a path on the graphof f, then we need to know the slope of thepath as we travel around that path - are wegoing uphill , downhill, or on the level. Thatwill depend on whi ch way we ar e walking!From the contour map we see thatI TRUE: at Q we are ascending - the con-tours are increasing in the counter-clockwisedirection.II FALSE: at P we are descending - the con-tours are decreasing i n the counter-clockwisedirection.III TRUE: at R we are on the level - we arefollowing the contour.keywords: contour map, contours, slope,curve on surface, tangent, Chain Rule, multi-variable Chain Rule,003 10.0 pointsFind the directional derivative, fv, off(x, y) = 2yx1/2at P = (1, 1) in the di rection of the vector−−→P Q w hen Q = (4, 5).1. fv=15correct2. fv= −1103. fv= −154. fv= 05. fv=110Explanation:The directional derivative of f(x, y) at Pin the direction of v =−−→P Q is given by the dotproductfv|P= ∇f |P·v|v|.But whenf(x, y) = 2yx1/2,we see that∂f∂x= −yx31/2,∂f∂y=1xy1/2,so that at P (1, 1),(∇f )(1, 1) = −i + j =D−1, 1E.On the other hand, v = h3, 4i which as avector of unit length becomesv|v|=D35,45E.Consequently,∇f|P=D−1, 1E.D35,45E=15.004 10.0 pointsFind the ma ximum slope on the graph off(x, y) = 2 sin(xy)at the point P (5, 0).1. max slope = 12. max slope = 5π3. max slope = 24. max slope = π5. max slope = 10π6. max slope = 10 correct7. max slope = 58. max slope = 2πlewis (scl876) – HW 14.6 – radin – (55315) 3Explanation:At P (5, 0, 0) the slope in the di rection of vis g iven by∇f(5,0)·v|v|.But whenf(x, y) = 2 sin(xy) ,the gradient of f is∇f(x, y) = 2y cos(xy) i + 2x cos(xy) j ,so at P (5, 0)∇f(5,0)= 10 j .Consequently, the slope at P will be maxi-mized when v = j in which casemax slope = 10 .keywords: slope, gradient, trig function, max-imum slope005 10.0 pointsSuppose that over a certain region of spacethe electrical potential V is given byV (x, y, z) = 2x2− 4xy + xyz .Find the rate of change of the potential atP (4, 3, 7) in the direction of the vectorv = i + j − k .1. −252.25√3correct3. −25√34. −2535. 25Explanation:The rate of change at P (4, 3, 7) is given byDuV = ∇V (4, 3, 7) ·v|v|.Now, whenV (x, y, z) = 2x2− 4xy + xyzit follows that∇V = h4x − 4y + yz, −4x + xz, xyiand∇V ( 4 , 3, 7) = h25, 12, 12i.Consequently,DuV = h25, 12, 12i ·h1, 1, −1i√3=25√3.keywords:006 10.0 pointsIf 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 inthe direction of t he vectorv = −3 i + 4 j .1. sl ope = −252. sl ope = −453. sl ope = −354. sl ope = −65correct5. sl ope = −1Explanation:lewis (scl876) – HW 14.6 – radin – (55315) 4The slope of the graph of z = f (x, y) i n thex-direction at P = (1, 2) is the valuefx(1, 2) =∂f∂x(1, 2)= −2of the partial derivative fxat P , while theslope of the graph of z = f (x, y) in the y-direction at P is the valuefy(1, 2) =∂f∂y(1, 2)= −3of the partial derivative fyat P = (1, 2).On the other hand, the slope at P in thedirection of any vectorv = a i + b jis g iven by∇f(1,2)·v|v|=1|v|fx(1, 2) i + fy(1, 2) j·(a i + b j) .Whenv = −3 i + 4 j ,therefore, the graph of f hasslope =15(−2 i − 3 j)·(−3 i + 4 j) = −65at P in the direction of v.keywords:007 10.0 pointsFind an equation for the tangent plane tothe graph ofz = xeycos z − 7at the point (7, 0, 0).1. x + 7y + z = 72. x + 7y − z = −73. x + 7y − z = 7 correct4. x + 7y + z = −75. x − 7y − z = 7Explanation:Note thatxeycos z − z = 7LetF (x) = xeycos z − z .The equation to the tangent plane to the sur-face at the po int P (7, 0, 0) is given byFxP(x − 7) + FyP(y − 0) + FzP(z − 0) .SinceFx= eycos z , FxP= 1,Fy= xeycos z , FyP= 7,andFz= −xeysin z − 1 , FzP= −1it follows that the equation of the tangentplane isx + 7 y − z = 7 .keywords:008 10.0 pointsIff(x, y) = x2+ 2y2,use the gradient vector ∇f (3, 1) to find thetangent line to the level curve f (x, y) = 11 atthe point (3, 1).1. 6x − 4y = 112. 3x − 2y = 113. 3x + 2y = 11 correct4. 6x + 4y = 11lewis (scl876) – HW 14.6 – radin – (55315) 55. 2x + 4y = 11Explanation:The equation of the tangent line is given by∇f(3, 1) · hx − 3, y − 1i = 0 .Whenf(x, y) = x2+ 2y2it follows that∇f = h2x, 4yi and ∇f(3, 1) = h6 , 4i.Consequently, the equation of the tangent lineis3x + 2y = 11 .keywords:009 10.0 pointsIf r(x) is t he vector function whose graph istrace o f the surfacez = f (x, y) = 4 x2− 2y2+ 2x − 3yon the plane y+2x = 0, determine the tangentvector to r(x) at x = 1.1. tangent vector = h1, −2, 0 i correct2. tangent vector = h2, 1 , 8 i3. tangent vector = h2, 0 , 0 i4. tangent vector = h1, 0 , 0 i5. tangent vector = h2, 0 , 8 i6. tangent vector = h1, −2, 8 iExplanation:The graph ofz = f (x, y) = 4 x2− 2y2+ 2x − 3yis the set of all po ints(x, y, f (x, y))as x, y vary in 3-space. So the intersection ofthe surface …