MATH 116 SUMMER 2008 QUIZ 2

Name : Student Id : Section : Problem: Assume

w = x^{3} + yz + x ln z, x = u + cos v, y = u sin v , and z = u^{2} + 1 Evaluate \frac{∂w}{∂u} and \frac{∂w}{∂v} at (u, v) = (0, π). Solution: By the Chain Rule,

\frac{∂w}{∂u} = \frac{∂w}{∂x} \frac{∂x}{∂u} + \frac{∂w}{∂y} \frac{∂y}{∂u} + \frac{∂w}{∂z} \frac{∂z}{∂u}

• \frac{∂w}{∂x} = 3x^{2} + ln z = 3(u + cosv)^{2} + ln(u^{2} + 1), \frac{∂x}{∂u} = 1

• \frac{∂w}{∂y} = u^{2} + 1, \frac{∂y}{∂u} = sin v

• \frac{∂w}{∂z} = y + \frac{x}{z} = u sin v + \frac{u+cos v}{u^{2}+1} , \frac{∂z}{∂u} = 2u

\frac{∂w}{∂u} = 3(u + cosv)^{2} + ln(u^{2} + 1) + (u^{2} + 1) sin v + (u sin v + \frac{u + cos v}{u^{2} + 1} )2u

\frac{∂w}{∂u} |_{(0,π)} = 3(0 − 1)^{2} + ln 1 + sin π + 0 = 3.

\frac{∂w}{∂v} = \frac{∂w}{∂x} \frac{∂x}{∂v} + \frac{∂w}{∂y} \frac{∂y}{∂v} + \frac{∂w}{∂z} \frac{∂z}{∂v}

\frac{∂x}{∂v} = − sin v, \frac{∂y}{∂v} = u cos v, \frac{∂z}{∂v} = 0.

\frac{∂w}{∂v} = − sin v(3(u + cosv)^{2} + ln(u^{2} + 1)) + u cos v(u^{2} + 1) + 0

\frac{∂w}{∂v} |_{(0,π)} = 0 + 0 = 0.