00:02
Okay, so in this problem we are asked to evaluate the line integral along c of 3y dx plus 4x dy where c is the straight path or straight line from the point 1 comma 3 to the point 5 comma 9.
00:25
So for c for the parameter station of c i'm going to use the form of formula ro plus t times vector v, where t here will range from 0 to 1.
00:38
Ro here is the initial point, so it will be 1 ,3, and the vector v will be the difference of the vector 5, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9 ,000, 9 ,000, 3, 3.
01:00
Being used to represent c is going to equal to 1 plus 4t, comma 3 plus 6t.
01:12
And once again, the range values of t will be from 0 to 1.
01:19
Our prime of t is just 4 .6.
01:24
And the vector field written in terms of t will equal to the vector field, by the way, can be obtained from these differential form to just be these terms.
01:39
So we're going to have 3y for the first component of the vector field and 4x for the second component.
01:47
So writing these in terms of t is simple by using r of t.
01:53
So 3y would be 3 times 3 plus 6t comma 4x so 4 times 1 plus 4 t.
02:02
Let me just...