#$&* course Mth 277 1/03 11 Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment `ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt. The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi). Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4. Thus our line integral has value 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Worked on paper, summarize,. initially started the integral by parts. took me a minute to realize to solve for for dx and substitute. ???? also I have a question, I'm not sure why you said that the Curve is the upper portion of the parabola. is it because its in the 1st quad, and that the points were positive? the integration should be -5.333 with the correction from your given solution, (-2/3 y^3)
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Given Solution: First verify that (1, 1) and (9, 3) are on the curve. This is easily verified for the first point, since 1^2 = 1, and for the second, since 3^2 = 9. This curve is part of the parabola x = y^2, which has vertex at the origin and opens to the right. The curve C is part of the 'upper half' of this parabola. If x = y^2, then dx = d(y^2) = 2 y dy. On this curve y changes from 1 to 3 (and x = y^2 changes from 1^2 = 1 to 3^2 = 9). Expressing our integrand -y dx + 3 y dy in terms of y we get -y ( 2 y dy) + 3 y dy = (-2 y^2 + 3 y) dy. Our integral is therefore integral((-2 y^2 + 3 y) dy, y from 1 to 3). Our antiderivative is -2/3 y^2 + 3/2 y^2. Between y = 1 and y = 3 this antiderivative changes from -2/3 + 3/2 to -6 + 27/2, a change of ... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): .antiderivative should be -2/3 y^3 not y^2.
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Given Solution: Our parameter here is theta, so our derivatives will be with respect to theta. Thus ' will indicate derivative with respect to theta. On this path we have dx = x ' dTheta = -2 sin(theta) dTheta and dy = y ' dTheta = 2 cos(theta) dTheta Thus (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta and x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta Thus our integral becomes integral( 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi) = integral( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta , theta from 0 to 2 pi). (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * (-2 sin(theta)) dTheta = -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta and x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta Thus our integral becomes integral( -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi) = integral( sin^3(theta) ) dTheta , theta from 0 to 2 pi). By symmetry this integral is easily seen to be zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I spy some errors I think, In your dx you say its -2 sin(theta), well in your equation (x^2 - y^2) dx; you fill in dx with a positive 2sin(theta), and I think you should end up with -8(cos^2(theta)sin(theta) - sin^3(theta))*d(theta) + 4cos^2(theta) * d(theta) working this from 0 to 2pi to get an answer of 0, because you end up with -sin^3(theta) - sin^2(theta), from 0 to 2pi, the sin of both are 0.