Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 2 Winter 2012 Math 255 Review Sheet for Second Midterm Exam Solutions 1. Let r(t)= 〈cos t, sin t, t〉. Find T(π), N(π), and B(π). .......................................................................... r ′ (t) = 〈− sin t, cos t, 1〉, T(t) = r ′ |r ′ | = 1 √ 2 〈− sin t, cos t, 1〉, N(t) = T ′ |T ′ | = 〈− cos t, − sin t, 0〉, B(t) = T × N = 1 √ 2 〈sin t, − cos t, 1〉. We obtain T(π)= 0, −1 √ 2 , 1 √ 2 , N(π)= 〈1, 0, 0〉, B(π)= 0, 1 √ 2 , 1 √ 2 . 2. At what point do the curves r 1 (t)= 〈t, t 2 ,t 3 〉 and r 2 (s)= 〈2s +1,s 2 + 1,e s 〉 intersect? Find cos θ, where θ is the angle of intersection. .......................................................................... By solving t =2s +1, t 2 = s 2 +1, t 3 = e s , we obtain t =1, s =0. (1) Therefore, the point of intersection is (1, 1, 1). We obtain r ′ 1 (t = 1) = 〈1, 2, 3〉, r ′ 2 (s = 0) = 〈2, 0, 1〉. The angle of intersection is the angle θ between these two vectors. We obtain cos θ = 〈1, 2, 3〉·〈2, 0, 1〉 |〈1, 2, 3〉| |〈2, 0, 1〉| = 5 14 . 1
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Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
Note that x and y are independent variables, and z is a function of xand y. By differentiating the equation implicitly with respect to x treatingy as a constant, we obtain
y∂z
∂x= ln(y + z) +
x
y + z
∂z
∂x. (3)
Therefore, we obtain∂z
∂x=
(y + z) ln(y + z)
y(y + z)− x.
By differentiating the equation implicitly with respect to y treating x as aconstant, we obtain
Let us define f(x, y) = e2−x2−y2 ln(1 + x2). The tangent plane is ex-pressed as
z − ln 2 = fx(1,−1)(x− 1) + fy(1,−1)(y + 1).
Since
fx(x, y) = e2−x2−y2[
−2x ln(1 + x2) +2x
1 + x2
]
,
fy(x, y) = e2−x2−y2(−2y) ln(1 + x2),
the equation is obtained as
(1− 2 ln 2)x+ (2 ln 2)y − z + 5 ln 2− 1 = 0.
12. The dimensions of a quarter (United States coin) are measured to be24.4mm (diameter) and 1.8mm (thickness). Each measurement is correct towithin 0.2mm. Use differentials to estimate the largest possible error whenthe volume of the quarter is calculated from these measurements. (Useπ ≃ 3.14.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Let d and w be the diameter and thickness. The volume is V = πd2w/4.The differential is calculated as
dV =∂V
∂ddd+
∂V
∂wdw =
π
2dw dd+
π
4d2 dw. (4)
We are given that |∆d| ≤ 0.2 and |∆w| ≤ 0.2. To find the largest error inthe volume, we therefore use dd = 0.2 and dw = 0.2 together with d = 24.4and w = 1.8.
∆V ≃ dV =3.14
2(24.4)(1.8)(0.2) +
3.14
4(24.4)2(0.2) = (3.14)(24.4)(1.4)
= 107.2624 ≃ 107.
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Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
17. Suppose that the equation F (x, y, z) = 0 implicitly defines each ofthe three variables x, y, and z as functions of the order two: z = f(x, y),y = g(x, z), x = h(y, z). If F is differentiable and Fx, Fy, and Fz are all
∂z∂x = 0. By differentiating F (x, g(x, z), z)) = 0 with respect to z, we
obtain ∂F∂y
∂y∂z + ∂F
∂z = 0. By differentiating F (h(y, z), y, z) = 0 with respect
to y, we obtain ∂F∂x
∂x∂y +
∂F∂y = 0. Thus, by the implicit function theorem, we
obtain∂z
∂x
∂x
∂y
∂y
∂z=
(
−Fx
Fz
)(
−Fy
Fx
)(
−Fz
Fy
)
= −1.
18. Near a buoy, the depth of a lake at the point with coordinates (x, y)is z = 10 + x2 − y3 + 20y, where x, y, and z are measured in meters. Afisherman in a small boat starts at the point (4, 3) and moves toward thebuoy, which is located at (0, 0). Is the water under the boat getting deeperor shallower when he departs? Explain.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Let us define f(x, y) = 10 + x2 − y3 + 20y. Note that we have fx = 2xand fy = −3y2 + 20. We obtain the directional derivative in the direction
of u =
⟨
−4
5,−3
5
⟩
∝ 〈−4,−3〉 as
Duf(x, y) = fx(x, y)
(
−4
5
)
+ fy(x, y)
(
−3
5
)
= −1.6x+ 1.8y2 − 12.
When we start at (x, y) = (4, 3), we have Duf < 0. Therefore, the waterunder the boat is getting shallower.
19. If g(x, y) = x2+y2+2xy+2x−2y, find the gradient vector ∇g(1,−1)and use it to find the tangent line to the level curve g(x, y) = 4 at the point(1,−1). Sketch the level curve, the tangent line, and the gradient vector.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The gradient vector is calculated as ∇g = 〈2x+2y+2, 2y+2x− 2〉 andso ∇g(1,−1) = 〈2,−2〉. Noting that the tangent line is perpendicular to thegradient vector, we write a vector equation of the tangent line as
〈x− 1, y + 1〉 · 〈2,−2〉 = 0.
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Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
We obtainy = x− 2.
To sketch the level curve, let us first understand what g(x, y) = 4 lookslike by defining X = (x + y)/
√2 and Y = (x − y)/
√2. We have Y =
−(1/√2)X2 +
√2, so g(x, y) = 4 is a parabola. The level curve, tangent
line, and gradient vector are sketched as follows.
20. Show that the function f(x, y) = (xy)1/3 is continuous and the partialderivatives fx and fy exist at the origin but the directional derivatives in allother directions do not exist.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The function f(x, y) is continuous at (0, 0) if
lim(x,y)→(0,0)
f(x, y) = f(0, 0) = 0.
That is, (xy)1/3 is continuous at the origin if for every ε > 0 there existsδ > 0 such that
∣
∣
∣(xy)1/3
∣
∣
∣< ε whenever 0 <
√
x2 + y2 < δ.
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Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
Since 0 ≤ (x− y)2, we have 2xy ≤ x2 + y2 < δ2. Thus,
∣
∣
∣(xy)1/3
∣
∣
∣<
∣
∣
∣
∣
∣
(
δ2
2
)1/3∣
∣
∣
∣
∣
.
For every ε, we can choose δ such that ε = (δ2/2)1/3. Therefore, (xy)1/3 iscontinuous at the origin.
The partial derivatives are calculated as
fx(0, 0) = fy(0, 0) = 0.
Let us calculate the directional derivative in the direction of u = 〈a, b〉(a 6= 0, b 6= 0).
Duf(0, 0) = limh→0
[ha(hb)]1/3 − [0(0)]1/3
h= lim
h→0
(ab)1/3
h1/3= ∞.
Therefore, the directional derivatives in the direction of such u do not exist.
Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
Note that
fxx
(
1,1
2
)
= −1
2e−2.
By the second derivatives test, we see that the point (0, 0) is a saddle point
and the point
(
1,1
2
)
is a local maximum.
22. A rectangular house is designed to minimize heat loss. The east andwest walls lose heat at a rate of 9 units/m2 per day, the north and southwalls at a rate of 8 units/m2 per day, the floor at a rate of 1 units/m2 perday, and the roof at a rate of 5 units/m2 per day. Each wall must be atleast 20m long, the height must be at least 2m, and the volume must beexactly 1000m3. Find the dimensions that minimize heat loss. Is it possibleto design a house with even less heat loss if the restrictions on the lengthsof the walls were removed?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Let L denote the heat loss.
L = (9 + 9)xz + (8 + 8)yz + (1 + 5)xy.
We have the following conditions.
x ≥ 20, y ≥ 20, z ≥ 4, xyz = 1000.
Since the volume is fixed, we can eliminate z.
L(x, y) = 6xy +16000
x+
18000
y.
Let us find the critical point(s).
∂L
∂x= 6y − 16000
x2= 0,
∂L
∂y= 6x− 18000
y2= 0.
Using the above two equations and noting that 1 = 4000/xy, we obtain
x =40
3, y = 15, z = 5.
The obtained x and y are less than 20. Let us find the extreme values of Lon the boundary.
L(x, 20) = 120x+16000
x+ 900, x ≥ 20.
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Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
This function L(x, 20) (x ≥ 20) has the minimum value at x = 20.
L(20, y) = 120y + 800 +18000
y, y ≥ 20.
This function L(20, y) (y ≥ 20) has the minimum value at y = 20. Whenx = y = 20, we have z = 2.5. Therefore, the heat loss is minimized whenx = y = 20m, z = 2.5m, and the minimum heat loss is L(20, 20) = 4100.
Math 255-003 -- Math 255-003 -- Math 255-003 -- Math 255-003 --
39. Find the mass m and center of mass (x, y) of the lamina that occupiesthe Cardioid D = {(r, θ)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 + sin θ} when the densityfunction is ρ(x, y) =