ha (lvh262) – Homework 13.2 – karakurt – (56295) 1This print-out should have 11 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsWhich of the following equations hold for alldifferentiable vector functions u(t), v(t) andall differentiable scalar functions f (t)?A:(u(t) × v(t))′= u′(t) × v(t) + u(t) × v′(t) ,B:(f (t)u(t))′= f′(t)u′(t) + f(t)u(t) ,C:(u(t) · v(t))′= u′(t) · v′(t) + u(t) · v(t) .1. B and C only2. none of them3. B only4. A and B only5. A and C only6. all of them7. A only correct8. C onlyExplanation:A. The derivative of the cross productof two vector functions follows the productrule for derivati ves. So the stated equatio n isTRUE.B. The derivative o f the product of ascalar function and a vector function followsthe product rule for derivatives. Thus(f (t)u(t))′= f′(t)u(t) + f(t)u′(t) ,so the stated equation is FALSE.C. The derivative of the dot product oftwo vector functions follows the product rulefor derivatives. Thus,(u(t) · v(t))′= u′(t) · v(t ) + u(t) · v′(t)so the stated equation is FALSE.keywords: derivative, scalar function, vectorfunction, dot product, cross product,002 10.0 pointsFind the derivat ive ofr(t) = a + tb + t5c .1. r′(t) = a + b − 5t4c2. r′(t) = b − t4c3. r′(t) = a + b + 5t4c4. r′(t) = a + b − t4c5. r′(t) = b + 5t4c correctExplanation:The derivative of r(t) = a + tb + t5c withrespect to t is the sum of the derivatives ofeach of the terms. Nowddt(a) = 0 ,ddt(tb) = b ,whileddt(t5c) = 5t5−1c = 5t4c .Consequently,r′(t) = b + 5t4c .keywords: vector function, deriva tive,003 10.0 pointsha (lvh262) – Homework 13.2 – karakurt – (56295) 2Determine the vectorI =Z10r(t) dtwhenr(t) =D81 + t2,2t1 + t2,6(1 + t)2E.1. I = h8 ln 2, 2, 6π i2. I = h8, ln 2, 6π i3. I = h2π, 2, 3 ln 2 i4. I = h8, 2π, 6 ln 2 i5. I = h2π, ln 2, 3 i correct6. I = h2 ln 2, π, 3 iExplanation:For a vector functionr(t) = hf (t), g(t), h(t)i,the components of the vectorI =Z10r(t) dtare given byZ10f(t) dt,Z10g(t) dt,Z10h(t) dt ,respectively. But whenr(t) =D81 + t2,2t1 + t2,6(1 + t)2E,we see thatZ10f(t) dt =Z1081 + t2dt=h8 tan−1ti10= 2π ,whileZ10g(t) dt =Z102t1 + t2dt=hln(1 + t2)i10= ln 2 ,andZ10h(t) dt =Z106(1 + t)2dt=h−61 + ti10= 3 .Consequently,I = h2π, ln 2, 3 i .004 10.0 pointsIf r(t) =t, t4, t5, find r′′(t).1. r′′(t) =0, 4t2, 5t32. r′′(t) =1, 4t2, 5t33. r′′(t) =1, 4t3, 5t44. r′′(t) =0, 12t2, 20t3correct5. r′′(t) =1, 12t2, 20t36. r′′(t) =0, 12t3, 20t4Explanation:For a vector functionr(t) = hf(t), g(t), h(t)i,the components of the vector function r′′(t)are given byf′′(t), g′′(t), h′′(t),respectively. But, whenr(t) =t, t4, t5,we see thatf′(t) = 1, f′′(t) = 0 ,whileg′(t) = 4t3, g′′(t) = 12t2,ha (lvh262) – Homework 13.2 – karakurt – (56295) 3andh′(t) = 5t4, h′′(t) = 20t3.Consequently,r′′(t) =0, 12t2, 2 0t3.keywords:005 10.0 pointsFind an equation in para met ric form for thetangent line to the graph ofr(s) = hs3, s6, s5iat the point (1, 1, 1).1. x(t) = 3t2, y(t) = 6t5, z(t) = 5t42. x(t) = 1+2t, y(t ) = 1+5t, z(t) = 1+4t3. x(t) = 3t−1, y(t ) = 6t−1, z(t) = 5t−14. x(t) = 1−3t, y(t ) = 1+6t, z(t) = 1−5t5. x(t) = 2t − 1, y(t) = 5t − 1, 4t − 16. x(t) = 1 + 3t, y(t) = 1 + 6t, z(t) = 1 + 5tcorrectExplanation:The g r aph of r(s) has tangent vectorr′(s) = h3s2, 6s5, 5s4i.On the other hand, (1, 1, 1) = r(1). Thus,at (1, 1, 1) the equation of the tangent line invector form isL(t) = r(1) + tr′(1)= h1 + 3t, 1 + 6t, 1 + 5t i,which in parametric form becomesx(t) = 1 + 3t, y(t) = 1 + 6t, z(t) = 1 + 5t .keywords: tangent line, tangent vector, vectorfunction, derivative, vector form, parametricform,006 10.0 pointsFind an equation in vector form for thetangent line at the point (1, 3, 3) to the spacecurve given parametrically byr(s) = hcos s, 3e2s, si n s + 3e−2si.1. tangent line = ht, 3 − 3t, 3 + 6ti2. tangent line = h1, 3+6t, 3−5ti correct3. tangent line = h1, 3 − 3t, 3 + 5ti4. tangent line = ht, 3 − 2t, 3 − 5ti5. tangent line = ht, 3 + 2t, 3 − 6ti6. tangent line = h1, 3 + 6t, 3 − 6tiExplanation:Since (1, 3, 3) = r(0), the tangent line hasvector form r(0) + t r′(0) . Butr′(s) = h−sin s, 6e2s, cos s − 6e−2si,so at r(0) the t angent vector isr′(0) = h0, 6, 1 − 6 i= h0, 6, −5 i.Consequently, at (1, 3, 3) thetangent line = h1, 2 + 6t, 2 − 5t i .007 10.0 pointsFind an equation in para met ric form for thetangent line to the graph ofr(s) = hcos s, 3e2s, 3e−2siat the point (1, 3, 3).1. x = t, y = 3 + 6t, z = 3 − 6t2. x = t, y = 3 − 3t, z = 3 + 6tha (lvh262) – Homework 13.2 – karakurt – (56295) 43. x = 0, y = 3 − 2t, z = 3 − 2t4. x = 0, y = 3 + 2t, z = 3 − 2t5. x = 1, y = 3 + 6t, z = 3 −6t correct6. x = 1, y = 3 − 3t, z = 3 + 2tExplanation:Note first that (1, 3, 3) = r(0). On theother hand, t he tangent vector at an a r bi trarypoint r( s) is given byr′(s) = h−sin s, 6e2s, −6e−2si,so at r(0) the t angent vector isr′(0) = h0, 6 , −6 i.Thus in vector form the tangent line at r(0) isL(t) = r(0) + t r′(0) = h1, 2 + 6t, 2 − 6t i,which in parametric form becomesx(t) = 1, y(t) = 3 + 6t, z(t) = 3 − 6t .keywords:008 10.0 pointsThe curves g iven parametrically byx(t) = t, y(t) = t6, z(t ) = t3,andx(t) = sin t, y(t) = sin 4t, z(t) = tintersect at the orig in.Find the cosine of their angle, θ, of inter-section.1. cos θ =13√2correct2. cos θ =1√173. cos θ =2√194. cos θ =1√195. cos θ =2√176. cos θ =23√2Explanation:The angle of intersection of the two curvesis the angle between the tangent vectors atthe point of intersection. Now the tangentvector at the origin to the first curve isr1(0) = h1, 6t5, 3t2it=0= h1, 0, 0 i,while the tangent vector at the origin to thesecond curve isr2(0) = hcos t, 4 cos 4t, 1 it=0= h1, 4, 1 i.But thencos θ =r′1(0) · r′2(0)|r′1(0)||r′2(0)|=1(2 + 16)1/2.Consequently,cos θ =13√2.009 10.0 pointsEvaluate t he integralI =Z10a(t) · b(t) dtwhena(t) = 2et2i + 2 j − ln( 1 + t) kandb(t) = 4ti + 3 j − 3 k .1. I = 4e − 1 + 6 ln 2 correct2. I = 8e − 1 − 3 ln 23. I = 8e + 1 + 3 ln 2ha (lvh262) – Homework 13.2 – karakurt – (56295) 54. I = 4e − 1 − 3 ln 25. I = 8e + …