1082 ®201-10íÊ!D01-02í -KëLsãÊU
1. (15 pts) Consider the function
f(x, y) =⎧⎪⎪
⎨⎪⎪⎩
x2+ y2
4 ⋅ ln(x2+ y2) ,if (x, y) ≠ (0, 0), 0 ,if (x, y) = (0, 0).
(a) (3 pts) Is f(x, y) continuous at (0, 0)?
(b) (4 pts) Find fx(0, 0), fy(0, 0), and fx(x, y), fy(x, y) for (x, y) ≠ (0, 0).
(c) (4 pts) Find fxy(0, 0) and fxy(x, y) for (x, y) ≠ (0, 0).
(d) (4 pts) Is fxy(x, y) continuous at (0, 0)?
Solution:
(a) Set x= r cos θ and r sin θ. Then lim
(x,y)→(0,0)
x2+ y2
4 ln(x2+ y2) = lim
r→0+
r2
4 ln(r2) = 1 2 lim
r→0+r2ln r=1 2 lim
r→0+
ln r r−2
= 1 2 lim
r→0+
r−1
−2 r−3 =1 2 lim
r→0+
r2
−2 =0= f(0, 0).
Therefore, f(x, y) is continuous at (0, 0).
f(0, 0) = 0 (1%), Computation of the limit (2%).
(b) For (x, y) ≠ (0, 0), fx(x, y) = x
2 ⋅ ln(x2+ y2) +x2+ y2
4 ⋅ 1
x2+ y2 ⋅ 2x = x
2(ln(x2+ y2) + 1). (1%) fy(x, y) = y
2 ⋅ ln(x2+ y2) +x2+ y2
4 ⋅ 1
x2+ y2 ⋅ 2y = y
2(ln(x2+ y2) + 1). (1%)
fx(0, 0) = lim
h→0
f(h, 0) − f(0, 0)
h = lim
h→0
h2 4 ln h2
h = 1
4lim
h→0h⋅ ln h2
= 1 2lim
h→0
ln∣h∣
h−1 =1 2lim
h→0
h−1
−h−2 =1 2lim
h→0−h = 0 (1%) fy(0, 0) = lim
h→0
f(0, h) − f(0, 0)
h = lim
h→0
h2 4 ln h2
h = 0 (1%).
(c) For (x, y) ≠ (0, 0),
fxy(x, y) = x 2 ⋅ 2y
x2+ y2 = xy
x2+ y2. (2%)
fxy(0, 0) = lim
h→0
fx(0, h) − fx(0, 0)
h = lim
h→0
0
2(ln(h2) + 1) − 0
h = 0 (2%)
(d) Solution 1.
Set x= r cos θ and r sin θ. Then lim
(x,y)→(0,0)fxy(x, y) = lim
r→0+
r2cos θ sin θ r2 = lim
r→0+cos θ sin θ. (2%)
For different θ, the limit value is different. So the limit does not exist. (1%) Therefore, fxy(x, y) is not continuous at (0, 0). (1%)
Solution 2.
First, let’s approach (0, 0) along the y = x. Then y = x gives fxy(x, x) = 1/2 for all x ≠ 0, so fxy(x, y) → 1/2 as (x, y) → (0, 0) along the line y = x. (1%)
Next, we approach (0, 0) along the y = −x. Then y = −x gives fxy(x, −x) = −1/2 for all x≠ 0, so fxy(x, y) → −1/2 as (x, y) → (0, 0) along the line y = −x. (1%)
So lim
(x,y)→(0,0)fxy(x, y) does not exist. (1%)
Therefore, fxy(x, y) is not continuous at (0, 0). (1%)
2. (8 pts) f(x, y) is a differentiable function on R2. Consider two points P0 = (x0, y0) ≠ P1 = (x1, y1) and define a function g(t) = f(x0+ t(x1− x0), y0+ t(y1− y0)).
(a) (2 pts) Compute g′(t) by the chain rule.
(b) (6 pts) Suppose that f(x0, y0) = f(x1, y1) and ∇f ≠ ⃗0. Prove that the line segment P0P1 is tangent to at least one level curve f(x, y) = c for some c.
Solution:
(a) Define g(t) = f(x0+ t(x1− x0), y0+ t(y1− y0)). Then (1%) g′(t) = fx(x0+ t(x1− x0), y0+ t(y1− y0)) xt
+ fy(x0+ t(x1− x0), y0+ t(y1− y0)) yt
(1%) = fx(x0+ t(x1− x0), y0+ t(y1− y0)) (x1− x0) + fy(x0+ t(x1− x0), y0+ t(y1− y0)) (y1− y0) (b) P0≠ P1 and f(x0, y0) = f(x1, y1) ⇒
(2%) g(0) = f(x0, y0) = f(x1, y1) = g(1) ⇒
There exists a 0≤ t∗≤ 1 such that g′(t∗) = g(1) − g(0) 1− 0 = 0 (2%) There exists a point P∗(x0+ t∗(x1− x0), y0+ t∗(y1− y0))
lying on P0P1 with f(x, y) = c = g(t∗)
(1%) At this point P∗, ∇f ⋅ (x1− x0, y1− y0) = g′(t∗) = 0 (1%) ⇒ P0P1 tangent to the level curve f(x, y) = c = g(t∗)
3. (10 pts) Let C be the hyperbola formed by the intersection of the cone x2+3z2= 4y2 and the plane 2x+ y = 5. Find the maximum and the minimum distance between the origin and the point on C (if exist) by the method of Lagrange multipliers.
Solution:
Let f(x, y, z) = x2+y2+z2, g(x, y, z) = x2−4y2+3z2, h(x, y, z) = 2x+y. We want to find extreme values of f under constraints g= 0 and h = 5. By the method of Lagrange multiplizers, we solve the system of equations:
⎧⎪⎪⎪⎪
⎪⎪⎪⎪⎪
⎪⎨⎪⎪⎪⎪⎪
⎪⎪⎪⎪⎪
⎩
fx= λgx+ µhx fy= λgy+ µhy fz= λgz+ µhz g= 0
h= 5
⇒
⎧⎪⎪⎪⎪
⎪⎪⎪⎪⎪
⎪⎨⎪⎪⎪⎪⎪
⎪⎪⎪⎪⎪
⎩
2x= λ2x + 2µ⋯ ⋅ 1O 2y= −λ8y + µ⋯⋯ 2O 2z= λ6z⋯⋯ 3O
x2− 4y2+ 3z2= 0⋯⋯ 4O 2x+ y = 5⋯⋯ 5O
(3pts for correct setting and equations.) O3⇒ λ = 1
3 or z= 0.
Case 1: z= 0, 4O⇒ x = ±2y. If x = 2y, 5O⇒ y = 1, x = 2 ⇒ λ = 0, µ = 2.
There is one solution (x, y, z) = (2, 1, 0), (λ, µ) = (0, 2).
If x= −2y 5O⇒ y = −5
3, x= 10
3 ⇒ λ = −2
3, µ= 50 9 . There is another solution (x, y, z) = (10
3 ,−5
3 , 0), (λ, µ) = (−2 3 ,50
9 ). 2pts.
Case 2: λ=1
3 but z≠ 0, 1O⇒ x = 3
2µ, 2O⇒ y = 3 14µ However, 4O⇒ (3
2µ)2− ( 3
14µ)2+ 3z2= 0
⇒ 3z2 = − (9 4− ( 3
14)2) µ2< 0 . . . (→←) 2pts.
Hence the only solutions are (x, y, z) = (2, 1, 0), (x, y, z) = (10 3 ,−5
3, 0)
∵f(2, 1, 0) = 5 < f(10 3 ,−5
3, 0) = 125
9 ∴f obtains minimum value at (2, 1, 0). 1pt i.e the minimum distance between (0, 0, 0) and C is √
5. 1pt
The C is unbounded and the maximum distance doesn’t exist. 1pt Sol 2:
⎧⎪⎪⎨⎪⎪
⎩
x2+ 3z2 = 4y2
2x+ y = 5 ⇒ Let x = t, y = 5 − 2x = 5 − 2t, 3z2 = 4y2− x2 = 4(5 − 2t)2− t2 x2+ y2+ z2 = t2+ (5 − 2t)2+4
3(5 − 2t)2−1
3t2= f(t).
Find t such that f(t) is minimized.
4. (12 pts) Find the center of mass of a lamina
D= {(x, y) ∈ R2∣x ≥ 0, y ≥ 0, x2+ 9y2 ≤ 1}
whose density function at any point is proportional to the square of its distance from the y-axis.
Solution:
The density function is ρ(x, y) = x2(1 point). Let x= r cos θ and y = r
3sin θ where 0≤ r ≤ 1 and 0≤ θ ≤ π
2(1 point). Thus
∂(x, y)
∂(r, θ) =RRRRRRRRRRRR
cos θ −r sin θ 1
3sin θ 1
3r cos θ RRRRRRRRRRRR= r
3.(1 points)
m = ∬Dρ(x, y)dA = ∬Dx2dA
= ∫0π/2∫01r2cos2θr 3drdθ
= 1
3[∫0π/2cos2θdθ][∫01r3dr]
= 1 3⋅ π
4 ⋅ 1 4= π
48(2 points).
m ¯X = ∬Dxρ(x, y)dA
= ∬Dx3dA(1 point)
= ∫0π/2∫01 1
3r4cos3θdrdθ
= 1
3⋅ [∫0π/2cos3θdθ][∫01r4dr]
= 1
3[∫0π/2(1 − sin2θ)d(sin θ)] ⋅ [1 5]
= 2
45(2 points).
m ¯Y = ∬Dyρ(x, y)dA
= ∬Dx2ydA(1 point)
= 1 9 ∫
π/2
0 ∫01r4cos2θ sin θdrdθ
= 1
9[∫0π/2cos2θd(− cos θ)][∫01r4dr]
= 1 9⋅ [−1
3 cos3θ∣π/20 ] ⋅1 5
= 1
135(2 points).
Then the center of mass is ( ¯X, ¯Y) = ( 32 15π, 16
45π)(1 point).
5. (10 pts) Evaluate the integral
∭U 1
x2+ y2+ z2dV where U = {(x, y, z) ∈ R3 ∶ x2+ y2+ z2≤ 36 and z ≥ 3}.
Solution:
Use spherical coordinate, z ≥ 3 can be expressed as ρ cos φ ≥ 3. We can obtain ρ ≥ 3 sec φ. The bound for ρ is given by 3 sec φ ≤ ρ ≤ 6. Consider the φ value when the upper bound and the lower bound meet, we have sec φ= 2, φ = π
3. The region is rotationally symmetric with respect to z axis, we have 0≤ θ ≤ 2π.
The integral can be expressed in spherical coordinates.
∭U 1
x2+ y2+ z2dV = ∫02π∫
π 3
0 ∫3 sec φ6 1
ρ2 ⋅ ρ2sin φdρdφdθ= 2π ⋅ ∫
π 3
0
ρ sin φ∣63 sec φdφ
= 2π ∫
π 3
0 sin φ(6 − 3 sec φ)dφ = 6π ∫
π 3
0 2 sin φ− tan φdφ
= 6π(−2 cos φ + ln cos φ)0π3 = 6π[(−1 − ln 2) + 2] = 6π lne 2. Grading policies:
Knowing the formula dV = ρ2sin φdρdφdθ for spherical coordinates. No partial credit is allowed for this part. 3pts.
Writing the domain in spherical coordinates. 4pts – Correct upper bound for ρ. 1pt
– Correct lower bound for ρ. 1pt
– Correct upper and lower bound for φ. 1pt – Correct upper and lower bound for θ. 1pt
Evaluating the integral. Partial credits is allowed in this part.3pts
It is possible to use cylindrical coordinate. The equation would be like this
∭U 1
x2+ y2+ z2dV = ∫02π∫36∫
√ 36−z2 0
1
r2+ z2 ⋅ rdrdzdθ
= 2π ∫36∫
√ 36−z2 0
1 2
1
r2+ z2 ⋅ dr2dz= π ∫36ln(r2+ z2)√036−z2dz
= π ∫36(ln(36) − 2 ln z)dz = π[ln(36)z − 2z ln z + 2z]63 = 6π lne 2 The grading policies are the same:
Knowing the formula dV = rdrdzdθ. No partial credit is allowed for this part. 3pts.
Writing the domain in cylindrical coordinates. 4pts
Evaluating the integral. Partial credits is allowed in this part.3pts
