Math 241, Quiz 6. 10/3/11. Name:
•Read problems carefully. Show all work. No notes, calculator, or text.
•There are 15 points total.
1. §14.3, #65 (5 points): Let u=erθ sin θ. Compute
∂3u
∂r2∂ θ .
Solution: We have
∂
∂θ (erθ sin θ) = er θ ∂
∂θ (sin θ) + sin θ∂
∂θ (erθ ) = erθ cos θ+ sin θ(rerθ ) = erθ (cos θ+rsin θ).
∂
∂r (erθ (cos θ+rsin θ)) = erθ ∂
∂r (cos θ+rsin θ) + (cos θ+rsin θ)∂
∂r (erθ )
=erθ sin θ+ (cos θ+rsin θ)θerθ =erθ (sin θ+θcos θ+rθ sin θ).
∂
∂r (erθ (sin θ+θcos θ+rθ sin θ))
=erθ ∂
∂r (sin θ+θcos θ+rθ sin θ) + (sin θ+θcos θ+rθ sin θ)∂
∂r (erθ ))
=erθθsin θ+ (sin θ+θcos θ+rθ sin θ)θerθ =θerθ (2 sin θ+θcos θ+rθ sin θ).
2. §14.3, #87 (5 points): You are told that there is a function fwhose partial derivatives are
fx(x, y) = x+ 4y, fy(x, y ) = 3x−y.
Should you believe it? Why or why not? Briefly explain.
Solution: We have
fxy(x, y ) = 4 6=3=fyx(x, y)
which violates Clairaut’s Theorem. Therefore, no such function can exist.
3. §14.4, #5 (5 points): Find an equation of the tangent plane to the surface
z=ycos(x−y)
at the point (2,2,2).
Solution: We have
fx=y(−sin(x−y)) = −ysin(x−y) =⇒fx(2,2) = 0;
fy=y(sin(x−y)) + cos(x−y) =⇒fy(2,2) = 1.
The normal vector to the tangent plane is h0,1,−1i; hence the tangent plane is
0(x−2) + 1(y−2) + (−1)(z−2) = y−2−z+ 2 = y−z= 0.