Help with evaluating the line integral ∫c y dx +x^2 dy

[Tonia]

Help with evaluating the line integral ∫c y dx +x^2 dy?

Someone on a different internet website helped me with this line integration problem but I can't see how they got some of the steps and my Calculus book doesn't explain it well: The answers are supposed to be: a) 100/3, b) 60
I did read the post "Read Before Posting". My problem is that I am missing some important steps for how to solve this problem, but I can't solve it if I don't know what to do first. I know about derivatives but maybe I am not using them correctly.

Evaluate the line integral ∫c y dx + x^2 dy a) along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A(0, -1) to the point B (4, 3) along the line segments AC and CB, where C is the point (4, -1) Hint: Along AC, dy = 0, along CB, dx = 0
a) = [(t^2 - 1) * 2 + (2t)^2 * 2t] dt, since x = 2t and y = t^2 - 1 = (8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) {for t = 0 to 2} = 100/3

I don't understand how they got the 8t^3 + 2t^2 – 2)dt and why it equals 2t^4 + (2/3)t^3 - 2t and why that equals 100/3:
(8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) = 100/3

Also: Does dx = 0? Because the derivative of 4 = 0?
Does dy = 0? Because the derivative of -2 = 0?
What does the derivative of dt = ?? I don't know what this would be.

(Do this one segment at a time and add the results):

b) For AC, we have y = -1 and so dy = 0. ∫AC (y dx + x^2 dy) = ∫(x = 0 to 4) [-1 * 1 + x^2 * 0] dx = -4. Why and How does [-1 * 1 + x^2 * 0] dx = -4? The person who helped me didn't explain how they got this and I can't figure it out.
For CB, we have x = 4 and so dx = 0. ∫CB (y dx + x^2 dy) = ∫(y = -1 to 3) (y * 0 + 4^2 * 1) dy = 64. Why and how does (y * 0 + 4^2 * 1) dy = 64? The person who helped me did not explain how they got this either. For b) ∫c (y dx + x^2 dy) = -4 + 64 = 60.

 

[Deleted member 4993]

Help with evaluating the line integral ∫c y dx +x^2 dy?

Someone on a different internet website helped me with this line integration problem but I can't see how they got some of the steps and my Calculus book doesn't explain it well: The answers are supposed to be: a) 100/3, b) 60

Evaluate the line integral ∫c y dx + x^2 dy a) along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A0, -1) to the point B4, 3) along the line segments AC and CB, where C is the point (4, -1) Hint: Along AC, dy = 0, along CB, dx = 0
a) = [(t^2 - 1) * 2 + (2t)^2 * 2t] dt, since x = 2t and y = t^2 - 1 = (8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) {for t = 0 to 2} = 100/3

I don't understand how they got the 8t^3 + 2t^2 – 2)dt and why it equals 2t^4 + (2/3)t^3 - 2t and why that equals 100/3:
(8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) = 100/3

Also: Does dx = 0? Because the derivative of 4 = 0?
Does dy = 0? Because the derivative of -2 = 0?
What does the derivative of dt = ?? I don't know what this would be.

(Do this one segment at a time and add the results):

b) For AC, we have y = -1 and so dy = 0. ∫AC (y dx + x^2 dy) = ∫(x = 0 to 4) [-1 * 1 + x^2 * 0] dx = -4. Why and How does [-1 * 1 + x^2 * 0] dx = -4? The person who helped me didn't explain how they got this and I can't figure it out.
For CB, we have x = 4 and so dx = 0. ∫CB (y dx + x^2 dy) = ∫(y = -1 to 3) (y * 0 + 4^2 * 1) dy = 64. Why and how does (y * 0 + 4^2 * 1) dy = 64? The person who helped me did not explain how they got this either. For b) ∫c (y dx + x^2 dy) = -4 + 64 = 60.

It seems to me - you are staring at the problemon the screen - instead of doing the problem with pencil and paper!!

along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A:sad:0, -1) to the point B:sad:4, 3)

Do you understand the process of getting co-ordinates of A(0,-1) & B(4,3)?

Hint: Along AC, dy = 0, along CB, dx = 0

Do you understand why the statement above is true? If not - get a piece of graph paper and plot A, B and C and see the route of integration.

Your questions almost tells me that either you do not know definite integration of simple polynomials or you are not willing to work it out!!

Come back and tell us what you see!!

 

[Tonia]

It seems to me - you are staring at the problemon the screen - instead of doing the problem with pencil and paper!!



Do you understand the process of getting co-ordinates of A(0,-1) & B(4,3)?



Do you understand why the statement above is true? If not - get a piece of graph paper and plot A, B and C and see the route of integration.

Your questions almost tells me that either you do not know definite integration of simple polynomials or you are not willing to work it out!!

Come back and tell us what you see!!

I am willing to work the problem out, I just do not know how to get the co-ordinates and plot A, B, and C. I am trying to solve a Calculus III problem and it is apparently a little bit too advanced for me. I don't know definite integration of simple polynomials yet.

 

[Tonia]

It seems to me - you are staring at the problemon the screen - instead of doing the problem with pencil and paper!!



Do you understand the process of getting co-ordinates of A(0,-1) & B(4,3)?



Do you understand why the statement above is true? If not - get a piece of graph paper and plot A, B and C and see the route of integration.

Your questions almost tells me that either you do not know definite integration of simple polynomials or you are not willing to work it out!!

Come back and tell us what you see!!

If I wasn't willing to do the work, I wouldn't be attempting to solve a Calculus III problem. I am self-teaching myself advanced Calculus. But it looks like I am in over my head, as it isn't good to try to solve a problem that I need previous knowledge of how to do integration.

 

[Tonia]

It seems to me - you are staring at the problemon the screen - instead of doing the problem with pencil and paper!!



Do you understand the process of getting co-ordinates of A(0,-1) & B(4,3)?



Do you understand why the statement above is true? If not - get a piece of graph paper and plot A, B and C and see the route of integration.

Your questions almost tells me that either you do not know definite integration of simple polynomials or you are not willing to work it out!!

Come back and tell us what you see!!

I will start with more basic Calculus

 

[Deleted member 4993]

Help with evaluating the line integral ∫c y dx +x^2 dy?

Someone on a different internet website helped me with this line integration problem but I can't see how they got some of the steps and my Calculus book doesn't explain it well: The answers are supposed to be: a) 100/3, b) 60
I did read the post "Read Before Posting". My problem is that I am missing some important steps for how to solve this problem, but I can't solve it if I don't know what to do first. I know about derivatives but maybe I am not using them correctly.

Evaluate the line integral ∫c y dx + x^2 dy a) along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A(0, -1) to the point B (4, 3)

At t= 0, what are the values of x & y. Those are the co-ordinates of A.

At t= 2, what are the values of x & y. Those are the co-ordinates of B.
along the line segments AC and CB, where C is the point (4, -1) Hint: Along AC, dy = 0, along CB, dx = 0

Plot A, B and C on a graph paper and try to figure out the logic for co-ordinates of C, using the given hint.
a) = [(t^2 - 1) * 2 + (2t)^2 * 2t] dt, since x = 2t and y = t^2 - 1 = (8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) {for t = 0 to 2} = 100/3

I don't understand how they got the 8t^3 + 2t^2 – 2)dt and why it equals 2t^4 + (2/3)t^3 - 2t and why that equals 100/3:
(8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) = 100/3

Also: Does dx = 0? Because the derivative of 4 = 0?
Does dy = 0? Because the derivative of -2 = 0?
What does the derivative of dt = ?? I don't know what this would be.

(Do this one segment at a time and add the results):

b) For AC, we have y = -1 and so dy = 0. ∫AC (y dx + x^2 dy) = ∫(x = 0 to 4) [-1 * 1 + x^2 * 0] dx = -4. Why and How does [-1 * 1 + x^2 * 0] dx = -4? The person who helped me didn't explain how they got this and I can't figure it out.
For CB, we have x = 4 and so dx = 0. ∫CB (y dx + x^2 dy) = ∫(y = -1 to 3) (y * 0 + 4^2 * 1) dy = 64. Why and how does (y * 0 + 4^2 * 1) dy = 64? The person who helped me did not explain how they got this either. For b) ∫c (y dx + x^2 dy) = -4 + 64 = 60.

.

 

[dasinger]

Help with evaluating the line integral ∫c y dx +x^2 dy?

Someone on a different internet website helped me with this line integration problem but I can't see how they got some of the steps and my Calculus book doesn't explain it well: The answers are supposed to be: a) 100/3, b) 60
I did read the post "Read Before Posting". My problem is that I am missing some important steps for how to solve this problem, but I can't solve it if I don't know what to do first. I know about derivatives but maybe I am not using them correctly.

Evaluate the line integral ∫c y dx + x^2 dy a) along the curve x = 2t, y = t^2 - 1 from t = 0 to t = 2 b) from the point A(0, -1) to the point B (4, 3) along the line segments AC and CB, where C is the point (4, -1) Hint: Along AC, dy = 0, along CB, dx = 0
a) = [(t^2 - 1) * 2 + (2t)^2 * 2t] dt, since x = 2t and y = t^2 - 1 = (8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) {for t = 0 to 2} = 100/3

I don't understand how they got the 8t^3 + 2t^2 – 2)dt and why it equals 2t^4 + (2/3)t^3 - 2t and why that equals 100/3:
(8t^3 +2t^2 – 2)dt = (2t^4 + (2/3)t^3 - 2t) = 100/3






Also: Does dx = 0? Because the derivative of 4 = 0?
Does dy = 0? Because the derivative of -2 = 0?
What does the derivative of dt = ?? I don't know what this would be.

(Do this one segment at a time and add the results):

b) For AC, we have y = -1 and so dy = 0. ∫AC (y dx + x^2 dy) = ∫(x = 0 to 4) [-1 * 1 + x^2 * 0] dx = -4. Why and How does [-1 * 1 + x^2 * 0] dx = -4? The person who helped me didn't explain how they got this and I can't figure it out.
For CB, we have x = 4 and so dx = 0. ∫CB (y dx + x^2 dy) = ∫(y = -1 to 3) (y * 0 + 4^2 * 1) dy = 64. Why and how does (y * 0 + 4^2 * 1) dy = 64? The person who helped me did not explain how they got this either. For b) ∫c (y dx + x^2 dy) = -4 + 64 = 60.

Let's start by looking at what we mean by the line integral along the curve x = 2t, y = t^2 - 1. This can be written as

∫ y(dx/dt)+x^2(dy/dt) dt, where the integration goes from t=0 to t=2. Now plug in dx/dt=2 and dy/dt=2t to get

∫(t^2-1)(2)+(4t^2)(2t)dt=∫2t^2-2+8t^3 dt. This is just a one-variable integral problem. You integrate to get 2t^4+(2/3)t^3-2t, and plug in the limits t=2 and t=0 to get (32+16/3-4)-(0)=100/3.



For the second part of the problem, you have to do the line integral as the sum of two integrals, one along the line from A(0,-1) to C(4,-1) and the other along the line from C to B(4,3). You can parametrize the line from A to C by using the variable x from 0 to 4 while y=-1. In other words, we have the parametrized curve x=t,y=-1, so dx/dt=1 and dy/dt=0. So plugging in
∫ y(dx/dt)+x^2(dy/dt) dt = ∫ (-1)(1)+t^2(0) dt=∫ (-1) dt with t going from 0 to 4. This integrates to -4. Along this path dy=0 because y is constant.
I have used the letter t for the parameter along the curve, but the author used x, since x=t along this line.
On the second path dx=0 because x is constant (and equal to 4).
I hope this explanation help!

 

[Tonia]

thank you

thanks for your help, Subhotosh Khan and Dasinger!

.