ha (lvh262) – Homework 14.6 – karakurt – (56295) 1This print-out should have 10 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsFind the directional derivative of f (x, y) =xe−8yat the po int (3, 0) in the direction indi-cated by the angle θ =π2.1. 242. −83. 04. −24 correct5. 8Explanation:The directional derivative of f(x, y) at(x0, y0) in the direction of a unit vector uwhere u makes an angle θ with the positi vex-axis isDuf(x0, y0) = fx(x0, y0) cos θ+fy(x0, y0) sin θ .When f(x, y) = xe−8yit follows thatfx(x, y) = e−8yandfy(x, y) = −8xe−8y.Consequently,Duf(3, 0) = e−8·0cosπ2− 8(3)e−8·0sinπ2=−24 .keywords:002 10.0 pointsFind the gradient off(x, y) = xy2− 2x3y .1. ∇f =6x2y − y2, 2xy − 2x32. ∇f =y2− 6x2y, 2x3− 2xy3. ∇f =2xy − 2x3, y2− 6x2y4. ∇f =y2− 6x2y, 2xy − 2x3correct5. ∇f =2xy − 2x3, 6x2y − y26. ∇f =2x3− 2xy, y2− 6x2yExplanation:Since∇f(x, y) =∂f∂x,∂f∂y,we see that∇f =y2− 6x2y, 2xy − 2x3.keywords:003 10.0 pointsFind the directio nal derivati ve, fv, of thefunctionf(x, y) = 6 + 2x√yat the point P (3, 4) i n the direction of thevectorv = h3, −4 i.1. fv= 12. fv=453. fv=9104. fv=11105. fv=65correctExplanation:ha (lvh262) – Homework 14.6 – karakurt – (56295) 2Now for an arbitrary vector v,fv= ∇f ·v|v|,where we have normalized so that the direc-tion vector has unit length. But w henf(x, y) = 6 + 2x√y ,then∇f = (2√y) i +x√yj .At P (3, 4), therefore,∇fP= 4 i +32j .Consequently, when v = h3, −4i,fv(3, 4) =4,32·v|v|=65.keywords:004 10.0 pointsFind the directional derivat ive, Dvf, off(x, y, z) = x tan−1yzat the point P = (1, 1, 1) in the direction ofthe vectorv = i + 2j + 2k .1. Dvf|P=142. Dvf|P=112π correct3. Dvf|P=1124. Dvf|P=13π5. Dvf|P=136. Dvf|P=14πExplanation:For an arbitrary vector v,Dvf = ∇f ·v|v|,where we have normalized so that the direc-tion vector has unit length. But whenf(x, y, z) = x tan−1yz,then∇f =∂f∂xi +∂f∂yj +∂f∂zk= ta n−1yzi +xz(1 + (y/z)2)j−xyz2(1 + (y/z)2)k .Thus∇f = tan−1yzi +xzz2+ y2j −xyz2+ y2k .At P = (1, 1, 1), therefore,∇f|P=π4i +12j −12k .Consequently, whenv = i + 2j + 2k ,we see that|v| =p1 + 22+ 22= 3 ,andDvf|P=13π4i +12j −12k· (i + 2j + 2k) .Consequently,Dvf|P=13π4+ 1 − 1=112π.keywords: directional derivative, gradient,dot product, unit vector,ha (lvh262) – Homework 14.6 – karakurt – (56295) 3005 10.0 pointsFind the maximum slope on the graph off(x, y) = 3 sin(xy)at the point P (5, 0).1. max slope = 15π2. max slope = 13. max slope = π4. max slope = 55. max slope = 15 correct6. max slope = 37. max slope = 3π8. max slope = 5πExplanation:At P (5 , 0, 0) the slope in the di r ection of vis g iven by∇f(5,0)·v|v|.But whenf(x, y) = 3 sin(xy) ,the gradient of f is∇f(x, y) = 3y cos(xy) i + 3x cos(xy) j ,so at P (5, 0)∇f(5,0)= 15 j .Consequently, the sl ope at P will be maxi -mized when v = j in which casemax slope = 15 .keywords: sl o pe, gradient, trig function, max-imum slope006 10.0 pointsSuppose that over a certain region of spacethe electrical potential V is gi ven byV (x, y, z) = 4x2− 5xy + xyz .Find the rate of change o f t he potential atP (1, 1, 5) in the direction of the vectorv = i + j − k .1. −732. −73.7√3correct4. 75. −7√3Explanation:The rate of change at P (1, 1, 5) is given byDuV = ∇V (1, 1, 5) ·v|v|.Now, whenV (x, y, z) = 4x2− 5xy + xyzit follows that∇V = h8x − 5y + yz, −5x + xz, xyiand∇V (1, 1, 5) = h8, 0, 1i.Consequently,DuV = h8 , 0, 1i ·h1, 1, −1i√3=7√3.keywords:007 10.0 pointsha (lvh262) – Homework 14.6 – karakurt – (56295) 4Find the equation of the tangent plane tothe surface4x2+ 5y2+ 5z2= 66at the point ( 2, −1, 3) .1. 8x − 5y + 15z = 562. 8x + 5y + 15z = 663. 8x + 5y + 15z = 564. 4x − 5y + 5z = 665. 8x − 5y + 15z = 66 correctExplanation:LetF (x) = 4 x2+ 5y2+ 5z2.The equation to the tangent plane to the sur-face at the point P (2, −1, 3) is given byFxP(x −2) + FyP(y + 1) + FzP(z −3) = 0 .SinceFx= 8x , FxP= 16,Fy= 10y , FyP= −10,andFz= 10z , FzP= 30it follows that the equation of the t angentplane is8x − 5y + 15z = 66 .keywords:008 10.0 pointsFind an equation for the tangent plane tothe graph ofz = xeycos z − 2at the point (2, 0, 0).1. x + 2y + z = −22. x + 2y + z = 23. x + 2y − z = −24. x + 2y − z = 2 correct5. x − 2y − z = 2Explanation:Note thatxeycos z − z = 2LetF (x) = xeycos z − z .The equation to the tangent plane to the sur-face at the point P (2, 0, 0) is given byFxP(x − 2) + FyP(y − 0) + FzP(z − 0) .SinceFx= eycos z , FxP= 1,Fy= xeycos z , FyP= 2,andFz= −xeysin z − 1 , FzP= −1it follows that the equation of the tangentplane isx + 2 y − z = 2 .keywords:009 10.0 pointsIff(x, y) = x2+ 3y2,use the gradient vector ∇f(4, 3) to find thetangent line to t he level curve f(x, y) = 43 atthe point (4, 3).1. 4x − 9y = 43ha (lvh262) – Homework 14.6 – karakurt – (56295) 52. 8x + 18y = 433. 4x + 9y = 43 correct4. 8x − 18y = 435. 2x + 6y = 43Explanation:The equation of the tangent line is given by∇f(4, 3) · hx − 4, y − 3i = 0 .Whenf(x, y) = x2+ 3y2it follows that∇f = h2x, 6yi and ∇f(4, 3) = h8, 18i.Consequently, the equation of the tangent lineis4x + 9y = 43 .keywords:010 10.0 pointsIf r(x) is t he vector function whose graph istrace of t he surfacez = f(x, y) = 3x2− y2− 2x − 3yon the plane y+2x = 0, determine the tangentvector t o r(x) at x = 1.1. tangent vector = h1, 0, 2 i2. tangent vector = h2, 0, 4 i3. tangent vector = h1, −2, 4 i4. tangent vector = h2, 1, 4 i5. tangent vector = h1, −2, 2 i correct6. tangent vector = h2, 0, 2 iExplanation:The graph ofz = f(x, y) = 3x2− y2− 2x − 3yis the set of all points(x, y, f(x, y))as x, y vary in 3-space. So t he intersection ofthe surface with the plane y + 2x = 0 is theset of all points(x, −2x, f(x, −2x)) , −∞ < x < ∞.Butf(x, −2x) = −x2+ 4x .Thus the surface and the plane y = 2x inter-sect in the graph ofr(x) = hx, −2x, −x2+ 4x i.Now the tangent vector to the graph of r(x)is the derivativer′(x) = h1, −2, −2x + 4 i.Consequently, at x = 1 the graph of