#$&* course Mth 277 5/2 1 Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment `ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt. The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi). Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4. Thus our line integral has value 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not quite understanding approach looking at solution for pointer Ok well we look to see if we can find what curve we are integrating over and by looking at the coordinates alone we see x = y^2(1^2 = 1 and 3^2 = 9, which are points given) not to mention it is given is question(which I over looked somehow first time thru). For pts given we see we are dealing with top half of parabolia.We have dx in our integral so we need to know what ‘dx is x = y^2 so `dx =2y Plugging into our integral we now have our integral parameterized wrt y Int(-y dx + 3y dy, C] = Int(-y (2y) + 3y dy, C) = Int(-2y^2 + 3y dy,1,3) = (-2/3)y^3 + (3/2)y^2, evaluated [(-2/3)3^3 + (3/2)3^2] - [(-2/3)1^3 + (3/2)1^2] = -16/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Would our integral not be -2/3 y^3 + 3/2 y^2, instead of the stated “-2/3 y^2 + 3/2 y^2” ********************************************* 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] = Int( [(2cos(theta))^2 - (2sin(theta))^2]*-2sin(`theta) + 2cos(`theta)*2sin(`theta) d`theta) =Int((4cos(`theta)^2 - 4sin(`theta)^2)*-2sin(`theta) + 2cos(`theta)*2sin(`theta) d`theta) =Int((-8cos^2sin(`theta) + 8sin(`theta)^3) + 4sincos(`theta) d`theta) =Int(-8cos(`theta)^2sin(`theta) d`theta) + Int( 8 sin(`theta)^3 d`theta) + Int(4sin(`theta)cos(`theta) d`theta -8*Int( cos(`theta)^2sin(`theta) d`theta) with u sub u = cos(`theta) du = -sin(`theta) -8*Int(-u^2) = 8*cos(`theta)^3/3 8*Int(sin(`theta)^3 d`theta) = Int(sin(`theta)*(1-cos(`theta)^2) `dtheta) = Int(sin(`theta) d`theta) - Int(sin*cos(`theta)^2 d`theta), using u sub again we get =8*(-cos(`theta) - (-cos(`theta)^3/3)) = 8(-cos(`theta) + cos(`theta)^3/3) Int(4sin(`theta)cos(`theta) d`theta) with u = cos(`theta) du = -sin(`theta) 4Int(-u du) = 4(u^2/2) = 4(cos(`theta)^2/2) So that we have Int(-8cos(`theta)^2sin(`theta) d`theta) + Int( 8 sin(`theta)^3 d`theta) + Int(4sin(`theta)cos(`theta) d`theta = 8*cos(`theta)^3/3 + 8(-cos(`theta) + cos(`theta)^3/3) + 4(cos(`theta)^2/2), evaluated [8*cos(2`pi)^3/3 + 8(-cos(2`pi) + cos(2`pi)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] = 0, so would this be one of those situations where we integrate from0 to `pi/2 then mult result by 4 to find true value???? [8*cos(2`pi)^3/3 + 8(-cos(2`pi) + cos(2`pi)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] If so [8*cos(`pi/2)^3/3 + 8(-cos(`pi/2) + cos(`pi/2)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] =-(2/3), to account for total area we mult by 4 and take absolute value =8/3 ?????Not sure this is correct?????????????? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our parameter here is theta, so our derivatives will be with respect to theta. Thus ' will indicate derivative with respect to theta. On this path we have dx = x ' dTheta = -2 sin(theta) dTheta and dy = y ' dTheta = 2 cos(theta) dTheta Thus (x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta and x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta Thus our integral becomes integral( 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi) = integral( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta , theta from 0 to 2 pi). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not sure but I believe that “( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta” , should be( 4 cos^2(theta) - 4 sin^2(theta) ) * -2 sin(theta) dTheta = -8 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta. But not sure I am calculating integral correct??????? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 277 5/2 1 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: Not quite understanding approach looking at solution for pointer Ok well we look to see if we can find what curve we are integrating over and by looking at the coordinates alone we see x = y^2(1^2 = 1 and 3^2 = 9, which are points given) not to mention it is given is question(which I over looked somehow first time thru). For pts given we see we are dealing with top half of parabolia.We have dx in our integral so we need to know what ‘dx is x = y^2 so `dx =2y Plugging into our integral we now have our integral parameterized wrt y Int(-y dx + 3y dy, C] = Int(-y (2y) + 3y dy, C) = Int(-2y^2 + 3y dy,1,3) = (-2/3)y^3 + (3/2)y^2, evaluated [(-2/3)3^3 + (3/2)3^2] - [(-2/3)1^3 + (3/2)1^2] = -16/3 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): Would our integral not be -2/3 y^3 + 3/2 y^2, instead of the stated “-2/3 y^2 + 3/2 y^2” ********************************************* 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] = Int( [(2cos(theta))^2 - (2sin(theta))^2]*-2sin(`theta) + 2cos(`theta)*2sin(`theta) d`theta) =Int((4cos(`theta)^2 - 4sin(`theta)^2)*-2sin(`theta) + 2cos(`theta)*2sin(`theta) d`theta) =Int((-8cos^2sin(`theta) + 8sin(`theta)^3) + 4sincos(`theta) d`theta) =Int(-8cos(`theta)^2sin(`theta) d`theta) + Int( 8 sin(`theta)^3 d`theta) + Int(4sin(`theta)cos(`theta) d`theta -8*Int( cos(`theta)^2sin(`theta) d`theta) with u sub u = cos(`theta) du = -sin(`theta) -8*Int(-u^2) = 8*cos(`theta)^3/3 8*Int(sin(`theta)^3 d`theta) = Int(sin(`theta)*(1-cos(`theta)^2) `dtheta) = Int(sin(`theta) d`theta) - Int(sin*cos(`theta)^2 d`theta), using u sub again we get =8*(-cos(`theta) - (-cos(`theta)^3/3)) = 8(-cos(`theta) + cos(`theta)^3/3) Int(4sin(`theta)cos(`theta) d`theta) with u = cos(`theta) du = -sin(`theta) 4Int(-u du) = 4(u^2/2) = 4(cos(`theta)^2/2) So that we have Int(-8cos(`theta)^2sin(`theta) d`theta) + Int( 8 sin(`theta)^3 d`theta) + Int(4sin(`theta)cos(`theta) d`theta = 8*cos(`theta)^3/3 + 8(-cos(`theta) + cos(`theta)^3/3) + 4(cos(`theta)^2/2), evaluated [8*cos(2`pi)^3/3 + 8(-cos(2`pi) + cos(2`pi)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] = 0, so would this be one of those situations where we integrate from0 to `pi/2 then mult result by 4 to find true value???? [8*cos(2`pi)^3/3 + 8(-cos(2`pi) + cos(2`pi)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] If so [8*cos(`pi/2)^3/3 + 8(-cos(`pi/2) + cos(`pi/2)^3/3) + 4(cos2`pi)^2/2)] - [8*cos(0)^3/3 + 8(-cos(0) + cos(0)^3/3) + 4(cos(0))^2/2)] =-(2/3), to account for total area we mult by 4 and take absolute value =8/3 ?????Not sure this is correct?????????????? 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): I’m not sure but I believe that “( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta” , should be( 4 cos^2(theta) - 4 sin^2(theta) ) * -2 sin(theta) dTheta = -8 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta. But not sure I am calculating integral correct??????? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 277 5/2 1 Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment `ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt. The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi). Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4. Thus our line integral has value 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not quite understanding approach looking at solution for pointer Ok well we look to see if we can find what curve we are integrating over and by looking at the coordinates alone we see x = y^2(1^2 = 1 and 3^2 = 9, which are points given) not to mention it is given is question(which I over looked somehow first time thru). For pts given we see we are dealing with top half of parabolia.We have dx in our integral so we need to know what ‘dx is x = y^2 so `dx =2y Plugging into our integral we now have our integral parameterized wrt y Int(-y dx + 3y dy, C] = Int(-y (2y) + 3y dy, C) = Int(-2y^2 + 3y dy,1,3) = (-2/3)y^3 + (3/2)y^2, evaluated [(-2/3)3^3 + (3/2)3^2] - [(-2/3)1^3 + (3/2)1^2] = -16/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Would our integral not be -2/3 y^3 + 3/2 y^2, instead of the stated “-2/3 y^2 + 3/2 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not sure but I believe that “( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta” , should be( 4 cos^2(theta) - 4 sin^2(theta) ) * -2 sin(theta) dTheta = -8 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta. But not sure I am calculating integral correct??????? "