#$&* course Mth 277 Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment `ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt. The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi). Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4. Thus our line integral has value 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[-y dx + 3y dy, C] y^2 = x (1,1) to the point (9,3) 2y=dx -2/3*y^3 + 3/2*y^2 1 to 3 -2/3*27 + 3/2*9 - [-2/3 + 3/2] -54/3+27/2-5/9 -324/18+243/18-10/18 -91/18 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ********************************************* Question: Evaluate Int[(x^2 - y^2)dx + x dy, C] where C is the circular path given by x = 2 cos(theta), y = 2sin(theta), 0 <= theta <= 2pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(x^2 - y^2)dx + x dy, C] x = 2 cos(theta) y = 2sin(theta) 0 <= theta <= 2pi dx = -2sin(theta)d(theta) dy = 2cos(theta)d(theta) (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)) d(Theta) Int( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta 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 Int( -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta + 4 cos^2(theta) dTheta 0 to 2 pi Int( sin^3(theta) ) dTheta 0 to 2 pi confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our parameter here is theta, so our derivatives will be with respect to theta. Thus ' will indicate derivative with respect to theta. On this path we have dx = x ' dTheta = -2 sin(theta) dTheta and dy = y ' dTheta = 2 cos(theta) dTheta Thus (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta and x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta Thus our integral becomes integral( 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi) = integral( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta , theta from 0 to 2 pi). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 277 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):OK ********************************************* 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: Int[-y dx + 3y dy, C] y^2 = x (1,1) to the point (9,3) 2y=dx -2/3*y^3 + 3/2*y^2 1 to 3 -2/3*27 + 3/2*9 - [-2/3 + 3/2] -54/3+27/2-5/9 -324/18+243/18-10/18 -91/18 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First verify that (1, 1) and (9, 3) are on the curve. This is easily verified for the first point, since 1^2 = 1, and for the second, since 3^2 = 9. This curve is part of the parabola x = y^2, which has vertex at the origin and opens to the right. The curve C is part of the 'upper half' of this parabola. If x = y^2, then dx = d(y^2) = 2 y dy. On this curve y changes from 1 to 3 (and x = y^2 changes from 1^2 = 1 to 3^2 = 9). Expressing our integrand -y dx + 3 y dy in terms of y we get -y ( 2 y dy) + 3 y dy = (-2 y^2 + 3 y) dy. Our integral is therefore integral((-2 y^2 + 3 y) dy, y from 1 to 3). Our antiderivative is -2/3 y^2 + 3/2 y^2. Between y = 1 and y = 3 this antiderivative changes from -2/3 + 3/2 to -6 + 27/2, a change of ... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Evaluate Int[(x^2 - y^2)dx + x dy, C] where C is the circular path given by x = 2 cos(theta), y = 2sin(theta), 0 <= theta <= 2pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(x^2 - y^2)dx + x dy, C] x = 2 cos(theta) y = 2sin(theta) 0 <= theta <= 2pi dx = -2sin(theta)d(theta) dy = 2cos(theta)d(theta) (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) d(Theta) Int( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta 0 to 2 pi (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * (-2 sin(theta)) dTheta = -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta Int( -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta + 4 cos^2(theta) dTheta 0 to 2 pi Int( sin^3(theta) ) dTheta 0 to 2 pi confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 277 Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment `ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt. The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi). Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4. Thus our line integral has value 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[-y dx + 3y dy, C] y^2 = x (1,1) to the point (9,3) 2y=dx -2/3*y^3 + 3/2*y^2 1 to 3 -2/3*27 + 3/2*9 - [-2/3 + 3/2] -54/3+27/2-5/9 -324/18+243/18-10/18 -91/18 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ********************************************* Question: Evaluate Int[(x^2 - y^2)dx + x dy, C] where C is the circular path given by x = 2 cos(theta), y = 2sin(theta), 0 <= theta <= 2pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int[(x^2 - y^2)dx + x dy, C] x = 2 cos(theta) y = 2sin(theta) 0 <= theta <= 2pi dx = -2sin(theta)d(theta) dy = 2cos(theta)d(theta) (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)) d(Theta) Int( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta 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 Int( -4 (cos^2(theta) sin(theta) + sin^3(theta)) dTheta + 4 cos^2(theta) dTheta 0 to 2 pi Int( sin^3(theta) ) dTheta 0 to 2 pi confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!