Chapter Exercise 7. Exercise 8. VectorPlot[{-x, -y}, {x, -2, 2}, {y, -2, 2}]

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1 Chapter 5. Exercise 7 VectorPlot[{x, }, {x, -, }, {y, -, }] Exercise 8 VectorPlot[{-x, -y}, {x, -, }, {y, -, }]

2 ch5-hw-solutions.nb Exercise VectorPlot[{y, -x}, {x, -, }, {y, -, }] Exercise 5 ϕ = x + y ; F = { x, y}; vectorfield = VectorPlot[F, {x, -4, 4}, {y, -4, 4}]; levelcurves = ContourPlot[ϕ, {x, -4, 4}, {y, -4, 4}, ContourShading None, Contours ]; Show[vectorField, levelcurves]

3 ch5-hw-solutions.nb 3 Exercise 8 ϕ = x y; F = {D[ϕ, x], D[ϕ, y]}; vectorfield = VectorPlot[F, {x, -, }, {y, -, }]; levelcurves = ContourPlot[ϕ, {x, -, }, {y, -, }, ContourShading None, Contours ]; Show[vectorField, levelcurves] Exercise 37 a. If ϕ(x, y) = x + 3 y, then ϕ = F =, 3. b. ϕ(, ) = + 3 = 5. The level curve ϕ(x, y) = 5 is the line x + 3 y = 5. Two points on this line are P =, 5 and Q = 5,, so a vector parallel to it is PQ = 5, - 5. On the other hand, the vector 3 3 F(, ) =, 3. We calculate F(, ) PQ = = 5-5 =, so they are orthogonal. 3 c. At any point, the level curve x + 3 y = c has direction vector c, - c, and the force vector is constant 3, 3. Their dot product is c - c =, so they are orthogonal. d. Any two lines with direction vector, 3 will do. A hand sketch is fine here, but you could make composite sketches using Mathematica and the technique we used in the previous problems. Chapter 5. Exercise 33 F = x, y Curve C is parameterized by r(t) = 4 t, t, where t. F dr = 4 t, t 4, t dt = 6 t + t 3 dt The integral is 8 t + t4, and when evaluated at the bounds we get 8 + = 7.

4 4 ch5-hw-solutions.nb Exercise 35 F = y, x on the line segment from (, ) to (5, ), so r(t) =, + t 4, 9 = 4 t +, 9 t +. F dr = 9 t +, 4 t +. 4, 9 dt = 7 t + 3 dt = = 49. Exercise 38 F = x, x +y y x +y Curve C is given by r(t) = t, 4 t, where t. Note that on curve C we have x = t and y = 4 t, so that x + y = 7 t. This gives us t 4 t F(r(t)) =, =, 4. 7 t 7 t 7 t t F dr = 7, 4, 4 dt = t t 7 t 7 dt = ln() - ln() = ln(). t Or use Mathematica: 7 7 t dt Log[] One can even get Mathematica to carry out the computations like so: x x + y, y 7 t, 4 7 t /. {x t, y 4 t} x + y Integrate 7 t, 4.{, 4}, {t,, } 7 t Log[] Exercise 4 C is the path consisting of the line segment from (-,) to (,8) followed by the line segment from (,8) to (,8). F = x, y on C : the line segment from (-, ) to (, 8). So r (t) = -, + t, 8 = - + t, 8 t. F dr = t -, 8 t, 8 dt = ( t) dt = 63 F = x, y on C : the line segment from (, 8) to (, 8). So r (t) =, 8 + t, = t, 8. F dr = t, 8., dt = Therefore, the work required is: F dr = F dr + F dr = 63 + = 67.

5 ch5-hw-solutions.nb 5 Chapter 5.3 Exercise 9 With F = f, g =,, we have f y = g x (both are zero), so F is conservative. A potential function is ϕ = x + y. Exercise With F = f, g = x, y, we have f y = g x (both are zero), so F is conservative. A potential function is ϕ = x + y. Exercise 4 F = f(x, y), g(x, y) = x 3 + xy, y 3 + x y f y = xy g x = xy We see that f y = g x. Therefore F is conservative on R. Exercise 7 a. With F = f, g = ϕ = y, x, and r(t) = cos t, sin t for t π, we have the line integral: π sin t, cos t -sin t, cos t dt = π cos t - sin t dt = π cos(t) dt = (sin( π) - sin()) =. b. With F = ϕ, we know that F is conservative, so the line integral is ϕ(r(π)) - ϕ(r()) = ϕ(-, ) - ϕ(, ) = - =. This agrees with our computation in part a. Hooray! Exercise 9 a. With F = f, g = ϕ =, 3, and r(t) = - t, t for t, we have the simple line integral:, 3 -, dt = dt = 4 - = 4. b. With F = ϕ, we know that F is conservative, so the line integral is ϕ(r()) - ϕ(r()) = ϕ(, ) - ϕ(, ) = 6 - = 4. This agrees with our computation in part a. Hooray! Exercise 33 r(t) = 4 cos(t), 4 sin(t) for t π. So the integrand is F(r(t)) r' (t) = -6 sin(t) cos(t) + 6 sin(t) cos(t) =. Since the integrand is zero, the line integral over all of C, or even along any portion of C, is zero. We observe that F is conservative (with potential function ϕ = x + y ), so the line integral of F around any closed curve will be zero. Exercise 34 r(t) = 8 cos(t), 8 sin(t) for t π. So F(r(t)) r' (t) = -64 sin (t) + 64 cos (t). Note that by the double

6 6 ch5-hw-solutions.nb r(t) = 8 cos(t), 8 sin(t) for t π. So F(r(t)) r' (t) = -64 sin (t) + 64 cos (t). Note that by the double angle formula for cosines, we can simplify the integrand to 64 cos (t) - 64 sin (t) = 64 cos( t), making the resulting integral simple to calculate by had. But we could also just have Mathematica do it: 64 π Cos[t] - Sin[t] dt We observe that F is conservative (with potential function ϕ = xy), so the line integral of F around any closed curve will be zero. Exercise 35 We have to parameterize each side of the triangle, and then sum the three resulting line integrals. Since this is the same field as in #33, F is conservative, so the answer should be zero. The three line segments are: C : (, -) to (, ): r (t) = t, t - for t. C : (, ) to (, ): r (t) = - t, t for t. C 3 : (, ) to (, -): r 3 (t) =, - t for t. The three integrands are: F(r (t)) r ' (t) = t, t -, = t -. F(r (t)) r ' (t) = - t, t -, = t -. F(r 3 (t)) r 3 ' (t) =, - t, - = 4 t - = ( t - ). So we get: F dr = F dr + F dr + 3 F dr 3 = t - dt + t - dt + t - dt = 4 t - dt The integral evaluates to 4 t - t, and from to this is zero, as it should be. Chapter 5.4 Exercise F = f, g = x, y. a. The curl is g x - f y = - =. b. The line integral is since F is conservative (with potential function ϕ(x, y) = x + y ), and the curve is closed. The double integral is zero since the integrand is the curl, which is zero. c. F is conservative since f = g y x Exercise 4 (Theorem 5.3). F = f, g = -3 y, 3 x. R is the triangle with vertices (, ), (, ), (, ).

7 ch5-hw-solutions.nb 7 a. The curl is g - f = 3 - (-3) = 6. x y b. The line integral must be evaluated in three parts (much like #35 of 5.3). The three line segments are: C : (, ) to (, ): r (t) = t, for t. C : (, ) to (, ): r (t) = - t, t for t. C 3 : (, ) to (, ): r 3 (t) =, - t for t. The three integrands are: F(r (t)) r ' (t) =, 3 t, =. F(r (t)) r ' (t) = -6 t, 3-3 t -, = 6. F(r 3 (t)) r 3 ' (t) = 6 t - 6,, - =. So we get: F dr = F dr + F dr + 3 F dr 3 = dt + 6 dt + dt = 6 dt = 6. According to Green s therem (circulation form), this should be equal to the double integral of the curl. The region of integration is the given triangle: x, and for each x in this inteval, y - x x 6 dy dx = ( - x) dx = - 6 = 6. This agrees with the line integral. c. F is not conservative since f g y x (Theorem 5.3). Said another way, the curl is nonzero. Exercise The parabolas intersect when: t = - t 3 t = t = 4 t = ± A picture confirms that our interval is - t :

8 8 ch5-hw-solutions.nb ParametricPlott, t, t, - t, {t, -, } To do a line integral, we will traverse the entire closed curve in counterclockwise fashion. We have to break up the line integral as a sum of line integrals, one for each curve. So for the bottom curve C we have r (t) = t, t with - t. The top curve C then runs from right to left, so the parameterized curve that we are given (which runs from left to right) is -C. So r (t) = t, - t, - t, is -C. Finally, we need to choose an appropriate vector field F. Let s try, x, since its curl is (check this), and hence the double integral of its curl will be the area of region R. Note: We could also have chosen -y,, or - y, x for the vector field, as discussed in the text. The two integrands are: F(r (t)) r ' (t) =, t, 4 t = 4 t. F(r (t)) r ' (t) =, t, - t = - t. F dr = F dr + F dr = F dr - -C F dr = - 4 t dt t dt = 6 - t dt The integral is 6 3 t3 = t 3, and plugging in the bounds we get (8 - (-8)) = 3. To check our answer, we can evaluate the equivalent double integral and see that it agrees: - -x dy dx x 3

9 ch5-hw-solutions.nb 9 Exercise 9 By the flux form of Green s theorem, C f dy - g dx = f over the given square. C x + e y dy - 4 y + e x dx + g x y da. We evaluate the double integral = f x y ( f = x + ey, and g = 4 y + e x ) = ( + 8 y) da (integrand nonzero, so the vector field f, g is not source-free) = ( + 8 y) dx dy = ( x + 8 xy) = ( + 8 y) dy = (y + 4 y Exercise 3 dy = 6 By the flux form of Green s theorem, C f dy - g dx = f + g x y the right over triangular region with vertices (, ), (, ), (, 4). 4 da. We evaluate the double integral on With f, g =, xy, the integrand is ( + x) = x, which (since it is nonzero) tells us that the vector field is not source-free. So we have: 4- x x dy dx = x(4 - x) dx The integral evaluates to x - 3 x3, and plugging in the bounds we get 8. Mathematica agrees: x x dy dx

10 ch5-hw-solutions.nb Exercise 3 By the flux form of Green s theorem, C f dy - g dx = f + g x y the right over semicircular region shown below: In[3]:= RegionPlotImplicitRegion- x && y - x, {x, y}, AspectRatio Automatic da. We evaluate the double integral on Out[3]= We have the added twist here that the boundary curve is traversed clockwise. This changes the outcome by a sign, so we need to take the negative of the double integral in Green s theorem. With f, g = x, y, the integrand is x + 4 y, which (since it is nonzero) tells us that the vector field is not source-free. Because the region is a polar rectangle, we will evaluate the integral in polar coordinates: π ( r cos(θ) + 4 r sin(θ)) r dr dθ The inner integral is 3 r3 cos(θ) r3 sin(θ), and plugging in bounds it becomes 3 cos(θ) sin(θ). Performing the second integration we get sin(θ) - 4 cos(θ), and plugging in bounds we see that the 3 3 sine terms go to zero, so we get -4/3 (-) - (-4/3) () = 4/3 + 4/3 = 8. Since we are traversing the 3 boundary clockwise, the answer is Mathematica agrees: - π r Cos[θ] + 4 r Sin[θ] r dr dθ - 8 3

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