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UIUC MATH 241 - Worksheet_030614_sol

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Thursday, March 6 ∗ Solutions ∗ Integrating vector fields.1. Consider the vector field F =(y,0) on R2.(a) Draw a sketch of F on the region where −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2. Check your answer with theinstructor.SOLUTION:Below is the image for parts (a) and (b)(b) Consider the following two curves which start at A = (−2, 0) and end at B = (2,0), namely the linesegment C1and upper semicircle C2.Add these curves to your sketch, and compute bothRC1F·dr andRC2F·dr. Check you answers withthe instructor.SOLUTION:Parametrize C1by r1(t ) = (t,0),0 ≤ t ≤ 2 and parametrize C2by r2(t ) = (−2cos t,2sin t ), 0 ≤ t ≤ π.We haveZC1F ·dr =Z20F (r1(t )) ·r01(t ) d t =Z20(0,0) ·(1,0)d t =0ZC2F ·dr =Zπ0F (r2(t )) ·r02(t ) d t =Zπ0(2sin t , 0) ·(2sin t , 2 cos t)d t =4Zπ0sin2(t ) d t=4 ·12·t −12sin(2t )¸π0=2π(c) Based on your answer in (b), could F be ∇f for some f : R2→R? Explain why or why not.SOLUTION:By the Fundamental Theorem of Line Integrals, if F = ∇f for some f : R2→R thenRCF ·dr is pathindependent for any curve C starting at A = (−2,0) and ending at B = (2,0). Since we obtaineddifferent answers for the paths C1and C2, F cannot be of this form.2. Consider the curve C and vector field F shown below.(a) Calculate F·T, where here T is the unit tangent vector along C. Without parameterizing C , evaluateRCF ·dr by using the fact that it is equal toRCF ·Tds.SOLUTION:From the picture we suppose that F(x, y) =(1,1). We have T =1p5(−2,−1), so F ·T =−3p5. SoZCF ·dr =ZCF ·Tds =−3p5ZCds =−3sinceRCds is simply the distance between (1,1) and (3,2).(b) Find a parameterization of C and a formula for F. Use them to check your answer in (a) by com-putingRCF ·dr explicitly.SOLUTION:Parametrize C by r(t) =(3 −2t,2 −t),0 ≤ t ≤1. We already have F =(1, 1). SoZCF ·dr =Z10(1,1) ·(−2,−1)d t =−33. Consider the points A = (0, 0) and B = (π,−2). Suppose an object of mass m moves from A to B andexperiences the constant force F =−mgj, where g is the gravitational constant.(a) If the object follows the straight line from A to B, calculate the work W done by gravity using theformula from the first week of class.SOLUTION:Recall that the work done on an object moving along a straight line subject to a constant force Fis W = F ·D, where D is the displacement vector. In this case D = (π,−2) and F = (0,−mg). SoW =(π, −2) ·(0, −mg) =2mg .(b) Now suppose the object follows half of an inverted cycloid C as shown below. Explicitly parameter-ize C and use that to calculate the work done via a line integral.SOLUTION:A parametrization for the inverted cycloid C is r(t) =(t −sin t,cos t −1),0 ≤t ≤π. SoW =ZCF ·dr =Zπ0(0,−mg ) ·(1 −cos t,−sin t ) d t =Zπ0mg sin t dt =mg[−cos t]π0=2mg(c) Find a function f : R2→R so that ∇f =F. Use the Fundamental Theorem of Line Integrals to checkyour answers for (a) and (b). Have you seen the quantity −f anywhere before? If so, what was itsname?SOLUTION:If such an f exists, we must have fx=0 and fy=−mg . Integrating −mg with respect to y we obtainf =−mg y +C (x), where C(x) is some function of x. Differentiating this with respect to x we obtainfx=C0(x) =0, so f =−mg y +K , where K is a constant, is a potential function for F.By the Fundamental Theorem of Line integrals, both (a) and (b) must have the same answer, namelyZLF ·dr =ZL∇(f ) ·d r = f (B) − f (A) = f (π,−2) − f (0, 0) =(−mg (−2) +K ) −K =2mgwhere L is the line segment from A to B andZCF ·dr =ZC∇(f ) ·dr = f (B)− f (A) =2mgThe quantity −f is called the potential energy.4. If you get this far, work #52 from Section 16.2:SOLUTION:We are assuming that B has magnitude which only depends on the distance from the wire. So B = |B|is constant along any circle centered around the wire in a plane perpendicular to the wire. Let C =r(t ) be such a circle with radius r parametrized in the counterclockwise direction and let B denote themagnitude of B along C. Note that B(r(t)) is a positive multiple of r0(t ) by definition. So it follows thatT(t ), the unit tangent vector to C, is given by T(t) =B(r(t ))B. We haveZCB ·dr =ZCB ·Tds =ZCB ·BBds =BZCds =2πr BBy Ampere’s Law,RCB ·dr =µ0I , so we have 2πr B =µ0I , or B


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UIUC MATH 241 - Worksheet_030614_sol

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