Question

for the problem in session 59 i can't see why t goes from 0 to pi/4 in the second part of part b..could anyone explain this for me please?

Answer 1

That part of the problem is computing the line integral of the vector over the arc of the circle. The path along the arc has been parametrized in terms of theta (solutions call it t), the angle of the position vector with respect to the x-axis.

The position vector begins at (5sqrt(2), 0) which corresponds to theta = 0, and travels along the arc to (5, 5) which corresponds to theta = pi/4. (We know the angle is pi/4 because the acute angle of an isosceles right triangle is pi/4.)

Answer 2

Thank you very much!