Solutions to Homework from Sections 11.2-11.4
Math 234, Spring 2008
Section 11.2
10. We will show that lim(x,y)→(0,0) 6x3y
2x4+y4does not exist. First, consider the limit as we approach (0,0) along the x-axis
(so y= 0). On this path, 6x3y
2x4+y4=6x3(0)
2x4+04=0
2x4= 0 (for x6= 0). So the limit as x→0 is 0.
However, now assume we approach the origin along the line y=x. Then 6x3y
2x4+y4=6x3x
2x4+x4=6x4
3x4= 2 (for x6= 0), so the
limit as x→0 is 2. Since 2 6= 0, the limits along these two paths do not agree, so lim(x,y)→(0,0) 6x3y
2x4+y4does not exist.
22. h(x, y) = g(f(x, y)) = √x2−y−1
√x2−y+1 . This is a rational function, so it is continuous wherever it is defined, which is where
x2≥y. (As long as px2−yis defined, the denominator of the rational function is non-zero.) So h(x, y ) is continuous
on the set {(x, y) : x2≥y}.
34. We want to find lim(x,y)→(0,0)(x2+y2) ln(x2+y2). In polar coordinates, r2=x2+y2, so this limit is limr→0+r2ln(r2).
We can find this limit using L’Hospital’s Rule:
lim
(x,y)→(0,0)(x2+y2) ln(x2+y2) = lim
r→0+r2ln(r2) = lim
r→0+r22 ln(r) = lim
r→0+
2 ln(r)
1/r2= lim
r→0+
2/r
−2/r3= lim
r→0+−r2= 0
Section 11.3
4. (a) ∂ h
∂v measures the rate of change of the wave height with respect to the wind speed, and ∂ h
∂t measures the rate of
change of the wave height with respect to time.
(b) fv(40,15) ≈f(50,15)−f(30,15)
20 =36−16
20 = 1 feet/knot. ft(40,15) ≈f(40,20)−f(40,10)
10 =28−21
10 = 0.7 feet/hour. This
means that when the wind has been blowing at 40 knots for 15 hours, the wave heights are about 1 foot higher
than if the wind had been blowing at 39 knots, and if the wind continues blowing at 40 knots for another hour,
the wave heights will increase about 0.7 feet.
(c) It looks like limt→∞
∂h
∂t = 0. In other words, as time goes on, the waves reach a maximum height and stay there.
32. f(x, y, z, t) = xy2
t+2z. Then:
fx(x, y, z, t) = y2
t+ 2zfy(x, y, z, t) = 2xy
t+ 2zfz(x, y, z, t) = −2xy2
(t+ 2z)2ft(x, y, z, t) = −xy2
(t+ 2z)2
80. T(x, t) = T0+T1e−λx sin(ωt −λx).
(a) ∂T
∂x =−λT1e−λx sin(ωt −λx)−λT1e−λx cos(ω t −λx) = −λT1e−λx(sin(ωt −λx) + cos(ωt −λx)). This tells us how
quickly the temperature increases (or decreases) as we penetrate further into the ground.
(b) ∂T
∂t =ωT1e−λx cos(ωt −λx). This tells us how quickly the temperature increases (or decreases) over time.
(c) Txx =λ2T1e−λx(sin(ωt −λx) + cos(ω t −λx)) + λ2T1e−λx(cos(ωt −λx)−sin(ω t −λx)) = 2λ2T1e−λx cos(ωt −λx) =
2λ2
ωTt. So k=2λ2
ω.
(d) Notice the range: 0 ≤t≤730, so the graph covers two years of time, in order to show the periodic behavior.
0
10
x
0
730
t
-10
10
T
