Question

Consider the oriented path which is a straight line segment L running from (0,0) to (16, 16 (a) Calculate the line integral of the vector field F = (3x-y) i +j along L using the parameterization B (t) = (2,20, 0 Enter an exact answer. t 8. 256 48 , 48 256). (b) Consider the line integral of the vector field F = (3r-y) i +j along L using the parameterization C(1)-( ,16 3t 32 16$1532 . The line integral calculated in (a) is the line integral of the parameterization given in (b).

Answer 1

This question is missing some parts. Here is the complete question.

Consider the oriented path which is a straight line segment L running from (0,0) to (16,16).

(a) Calculate the line inetrgal of the vector field F = (3x-y)i + xj along line L using the parameterization B(t) = (2t,2t), 0 ≤ t ≤ 8.

Enter an exact answer.

`\int\limits_L {F} .\, dr =`

(b) Consider the line integral of the vector field F = (3x-y)i + xj along L using the parameterization C(t) =
`((t^(2)-256)/(48) ,(t^(2)-256)/(48) )`, 16 ≤ t ≤ 32.

The line integral calculated in (a) is ____________ the line integral of the parameterization given in (b).

Answer: (a)
`\int\limits_L {F} .\, dr =` 384

(b) the same as

Explanation: Line Integral is the integral of a function along a curve. It has many applications in Engineering and Physics.

It is calculated as the following:

`\int\limits_C {F}. \, dr = \int\limits^a_b {F(r(t)) . r'(t)} \, dt`

in which (.) is the dot product and r(t) is the given line.

In this question:

(a) F = (3x-y)i + xj

r(t) = B(t) = (2t,2t)

interval [0,8] are the limits of the integral

To calculate the line integral, first substitute the values of x and y for 2t and 2t, respectively or

F(B(t)) = 3(2t)-2ti + 2tj

F(B(t)) = 4ti + 2tj

Second, first derivative of B(t):

B'(t) = (2,2)

Then, dot product between F(B(t)) and B'(t):

F(B(t))·B'(t) = 4t(2) + 8t(2)

F(B(t))·B'(t) = 12t

Now, line integral will be:

`\int\limits_C {F}. \, dr = \int\limits^8_0 {12t} \, dt`

`\int\limits_L {F}. \, dr = 6t^(2)`

`\int\limits_L {F.} \, dr = 6(8)^(2) - 0`

`\int\limits_L {F}. \, dr = 384`

Line integral for the conditions in (a) is 384

(b) same function but parameterization is C(t) =
`((t^(2)-256)/(48), (t^(2)-256)/(48) )`:

F(C(t)) =
`(t^(2)-256)/(16)-(t^(2)-256)/(48)i+ (t^(2)-256)/(48)j`

F(C(t)) =
`(2t^(2)-512)/(48)i+ (t^(2)-256)/(48) j`

C'(t) =
`((t)/(24), (t)/(24) )`

`\int\limits_L {F}. \, dr = \int\limits {((t)/(24))((2t^(2)-512)/(48))+ ((t)/(24) )((t^(2)-256)/(48)) } \, dt`

`\int\limits_L {F} .\, dr = \int\limits^a_b {(t^(3))/(384)- (768t)/(1152) } \, dt`

`\int\limits_L {F}. \, dr = (t^(4))/(1536) - (768t^(2))/(2304)`

Limits are 16 and 32, so line integral will be:

`\int\limits_L {F} \, dr = 384`

With the same function but different parameterization, line integral is the same.