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) = $(\frac{t^{2}-256}{48} ,\frac{t^{2}-256}{48} )$, 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

Step-by-step 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) = $(\frac{t^{2}-256}{48}, \frac{t^{2}-256}{48} )$:

F(C(t)) = $\frac{t^{2}-256}{16}-\frac{t^{2}-256}{48}i+ \frac{t^{2}-256}{48}j$

F(C(t)) = $\frac{2t^{2}-512}{48}i+ \frac{t^{2}-256}{48} j$

C'(t) = $(\frac{t}{24}, \frac{t}{24} )$

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

$\int\limits_L {F} .\, dr = \int\limits^a_b {\frac{t^{3}}{384}- \frac{768t}{1152} } \, dt$

$\int\limits_L {F}. \, dr = \frac{t^{4}}{1536} - \frac{768t^{2}}{2304}$

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.