MATH 265 Midterm: MATH 265 Iowa State m4Practice Solutions

• School: Iowa State University
• Department: Mathematics
• Course: MATH 265
• Professor: All

Document Summary

First practice midterm solutions: let c be the curve de ned as follows: cos(2 t) x(t) = (cid:12) t2 y(t) = (cid:12) (t 2)et + 3 t + 1. 2 6 t 6 3 z(t) = (cid:12) 10. Zc (2x + y cos(xy)) dx + (x cos(xy)) dy + 2 dz. Solution: this is a complicated curve that starts at (0, 1, 10) and goes to (1, 4, 1). Our best hope is that this is a path independent integral, i. e. , rc f r ds. Integrating the rst term with respect to x we can conclude f = x2 + sin(xy) + c(y, z). Taking the derivative of this with respect to y and comparing to the second term we would want. = x cos(xy) from which we can conclude that c. Y = 0 so that f = x2 + sin(xy) + d(z).