Math 103, Practice for the Second Midterm – SolutionsIvan Matic

1. Find the equation of the tangent plane to the surface z2 +4z = x2 + y2 at the point (1,2,1).Solution. The equation of the tangent plane at the point (x0,y0,z0) to the surface given by F(x,y,z) = 0 isgiven by the equation

Fx(x0,y0,z0)(x− x0)+Fy(x0,y0,z0)(y− y0)+Fz(x0,y0,z0)(z− z0) = 0.

Therefore the required tangent plane is

−2(x−1)−4(y−2)+6(z−1) = 0.

2. If z = xy+ xey/x, prove that x ∂ z∂x + y ∂ z

∂y = xy+ z.

Solution. From the equality z = xy + xey/x we get ∂ z∂x = y + ey/x + xey/x ·

�− y

x2

�= y + ey/x− y

xey/x and∂ z∂y = x+ ey/x. Therefore

x∂ z∂x

+ y∂ z∂y

= xy+ xeyx− yey/x + xy+ yey/x = z+ xy.

3. Find absolute minimum and absolute maximum of the function f (x,y,z) = xyz if the numbers x, y, z satisfy:x2 + y2 + z2 ≤ 9.Solution. fx = yz, fy = zx, fz = xy. To have local minimum or maximum at point (x0,y0,z0) we must havex0y0 = 0, y0z0 = 0, z0x0 = 0. In this case two of the numbers x0, y0, z0 must be zero hence f (x0,y0,z0) = 0.Let us now find critical points of f on x2 + y2 + z2 = 9. Let g(x,y,z) = x2 + y2 + z2− 9. According to thetheory of Lagrange multipliers, all critical points of f that satisfy g(x,y,z) = 0 must satisfy the system:

yz = 2λxzx = 2λyxy = 2λ z

x2 + y2 + z2 = 9.

Multiplying the first equation by x it becomes xyz = 2λx2. Similarly, from the second equation we getxyz = 2λy2, and from the third xyz = 2λ z2. This implies that x2 = y2 = z2 and from the last equation we getx2 = y2 = z2 = 3. Therefore all critical points belong to the set

S = {(−√

3,−√

3,−√

3),(−√

3,−√

3,√

3),(−√

3,√

3,−√

3),(−√

3,√

3,√

3),(√

3,−√

3,−√

3),(√

3,−√

3,√

3),(√

3,√

3,−√

3),(√

3,√

3,√

3)}.

Evaluating f at each of these points we get that max f =√

33 and min f =−√

33.

4. Evaluate the integral� 1

−1

� z2

0

� y

0

z+ tan zx2+y2+1

x2 + y2 +1dxdydz.

Solution. Let f (x,y,z) =z+tan z

x2+y2+1x2+y2+1 . Since f (x,y,−z) = − f (x,y,z) and the domain of integration is sym-

metric with respect to xy-plane we get that the required integral is equal to 0.

Solution. Let A be the triangle with vertices (0,0), (1,1), and (0,1). Let B = {(x,y) : x2 +y2 ≤ 2,0≤ x,y≤x}. Since the integrand is symmetric with respect to x and y axis, we have:

� �

D(x2 + y2)dA = 2

�� �

A(x2 + y2)dA+

� �

B(x2 + y2)dA

�.

Let us calculate the first integral over A:� �

A(x2 + y2)dA =

� 1

0

� 1

x(x2 + y2)dydx =

� 1

0

�x2(1− x)+

13(1− x3)

�dx =

13.

On the other hand, the integral over B can be calculated by turning it into polar coordinates:� �

B(x2 + y2)dA =

� π/4

−π/2

� √2

0ρ3 dρdθ =

3π4

· 44

=3π4

.

Therefore� �

D(x2 + y2)dA = 2 ·�1

3 + 3π4

�.

8. Find the center of mass of the solid of uniform density that is bounded by the planes z = 1, z = 0 and thesphere x2 + y2 + z2 = 2.Solution. Let S be the given solid. Draw the picture of the solid S! Clearly, x and y coordinates of the centerof mass are 0, because the solid is symmetric with respect to xz and yz planes. Let K be the density of thesolid. The total mass is

� � �S K dV = K

� � �S dV . Using the cylindrical coordinates we get

M = K� � �

SdV = K

� 1

0

� √2−z2

0

� 2π

0r dθdrdz = 2πK

� 1

0

12(2− z2)dz =

5πK3

.

To calculate the z-coordinate of the center of mass we need to do:

Mz =1M

� � �

SzK dV =

KM

� 1

0

� √2−z2

0

� 2π

0zr dθdrdz =

2πKM

� 1

0

z(2− z2)2

dz =3πK4M

=9

20.

9. (a) If a, b, c are positive real numbers, prove that a3 +b3 + c3 ≥ 3abc.(b) Find the minimal value of the function f (x,y,z) = x3 + y3 + z3 if x, y, z satisfy x2y+ y2z+ z2x = 3.

Solution.

(a) Let f (a,b,c) = a3 + b3 + c3. It is enough to prove the following statement: If a, b, c are positivereal numbers such that abc = k then min f (a,b,c) = 3k. We use Largrange multipliers to prove thelast statement. Let g(a,b,c) = abc− k. Then the minimizer of f under the constraints g(a,b,c) = 0satisfies:

3a2 = λbc3b2 = λca3c2 = λababc = k.

Multiplying the first equation by a, the second by b and the third by c we get a3 = b3 = c3, or, equiv-alently a = b = c. This implies that a = b = c = 3√k and a3 + b3 + c3 = 3k in this case. All we knowis that a3 +b3 + c3 is a critical point. Since at least one of a, b, c must be bigger then 3√k (in order forabc = k to be satisfied) we conclude that f (a,b,c) ≥ k. When one of a, b, c tend to infinity we havef (a,b,c)→ ∞, hence f attains minimum at ( 3√k, 3√k, 3√k) therefore a3 +b3 + c3 ≥ 3k.

(b) Applying the inequality from the previous problem to a = x, b = x, c = y we get 2x3 + y3 ≥ 3x2y.Similarly we get 2y3 + z3 ≥ 3y2z and 2z3 + x3 ≥ 2z2x. Adding the last three inequalities yields 3(x3 +y3 + z3)≥ 3(x2y+ y2z+ z2x) = 9. The minimum can be attained for x = y = z = 1.

10. If the ellipse x2

a2 + y2

b2 = 1 is to enclose the circle x2 + y2 = 2y, what values of a and b minimize the area ofthe ellipse?

Solution. We may assume that a > 0 and b > 0. The volume of the ellipse is abπ . Hence we want to find theminimum of the function f (a,b) = abπ under the condition that the ellipse x2

a2 + y2

b2 = 1 encloses the circlex2 + y2 = 2y. This is a problem where we will use Lagrange multipliers, but we first have to transform theconstraints in the form g(a,b) = 0. So far the constraints are in a crazy geometrical form. There is somework to be done.

Hence, our first goal is to understand what does it mean for the ellipse x2

a2 + y2

b2 = 1 to enclose the circle

x2 + y2 = 2y. This can be seen in the following way: Each point on the ellipse x2

a2 + y2

b2 = 1 is outside of thecircle x2 + y2 = 2y. This is still not in the satisfactory form g(a,b) = 0 but we are slowly getting there.

Our constraint can be further understood as Each point of the ellipse x2

a2 + y2

b2 = 1 satisfies x2 + y2−2y ≥ 0,

or If (x,y) satisfies x2

a2 + y2

b2 = 1 then x2 + y2−2y≥ 0, or

min{x2 + y2−2y}≥ 0 under the condition x2

a2 + y2

b2 = 1.

We will now use Lagrange multipliers to find the minimum of M(x,y) = x2 + y2− 2y under the conditionx2

a2 + y2

b2 = 1. This minimum will be a function of a and b. Then we will require that minimum to be 0 inorder to find the conditions on a and b that are necessary for the ellipse to enclose the circle.

Now we have the system:

2x =2λxa2

2y−2 =2λyb2

x2

a2 +y2

b2 = 1.

From the first equation we arrive to the following two cases x = 0 and x �= 0:

1◦ x = 0. From the last equation (the equation of the ellipse) we conclude that y = b or y = −b. Ourcandidates are (0,b) and (0,−b).

2◦ x �= 0. The first equation now becomes λ = a2. We can now place this value of λ in the second equation.The equation becomes y · (1− a2

b2 ) = 1. We now consider two cases:

2.1◦ a2 = b2. Then this equation doesn’t have a solution for y.2.2◦ a2 �= b2. We can solve for y: y = b2

b2−a2 . Solving for x from the third equation gives us: x2 =a2((b2−a2)2−b2)

(b2−a2)2 . Therefore the critical points are

(x,y) ∈��

+

�a2((b2−a2)2−b2)

(b2−a2)2 ,b2

b2−a2

�,

�−

�a2((b2−a2)2−b2)

(b2−a2)2 ,b2

b2−a2

��.

All critical points are

(x,y) ∈��

+

�a2((b2−a2)2−b2)

(b2−a2)2 ,b2

b2−a2

�,

�−

�a2((b2−a2)2−b2)

(b2−a2)2 ,b2

b2−a2

�,(0,b),(0,−b)

�.

Now we have M(0,±b) = b2−2b, and in the case a2 �= b2:

M

�±

�a2((b2−a2)2−b2)

(b2−a2)2 ,b2

b2−a2

�=

a2(b2−a2)2−a2b2 +b4−2b4 +2b2a2

(b2−a2)2

=a2(b2−a2)2−b2(b2−a2)

(b2−a2)2

=a2(b2−a2)−b2

b2−a2 =a2b2−a4−b2

b2−a2 .

The minimum of M is the smaller of the numbers b2−2b and a2b2−a4−b2

b2−a2 .The smaller of these two numbers has to be ≥ 0. This means that both of the numbers have to be ≥ 0. Inparticular we must have b2−2b≥ 0, or, equivalently b≥ 2 (we have assumed that b > 0). Now consider thetwo cases:

1◦ a ≥ b. Then a2b2−a4−b2

b2−a2 ≥ 0 because the denominator is negative, and the numerator satisfies a2b2−a4−b2 < a2 ·a2−a4−b2 = −b2 < 0. Now the only condition is b ≥ 2. Turning back to the originalproblem and using the fact a ≥ b we see that the minimal ellipse is one for which a = b = 2. Thisellipse happens to enclose the circle and its area is 4π .

2◦ a < b. Then a2b2−a4−b2

b2−a2 ≥ 0 is equivalent to a2b2−a4−b2 ≥ 0. Now the original problem is asking usto minimize the function f (a,b) = ab under the condition g(a,b) = 0 where g(a,b) = a2b2−a4−b2.The function f attains its minimum at the same points as h(a,b) = ( f (a,b))2. It is more convenient tominimize h(a,b) = a2b2 under the condition g(a,b) = 0 because we can do the substitution α = a2 andβ = b2 to transform the problem into minimizing h(α,β ) = α ·β under the condition αβ−α2−β = 0.We use the Lagrange multipliers again:

β = µ(β −2α)α = µ(α−1)

αβ −α2−β = 0.

From the second equation we get that α(1−µ) =−1. Consider the following two cases:2.1◦ µ = 1. In this case the second equation becomes the impossible one. There are now critical points

in this case.2.2◦ µ �= 1. Then α = 1

µ−1 . From the first equation we get β = 2µαµ−1 = 2µ2

(µ−1)2 . It is now convenientto notice that β = 2α2 and substitute this into the third equation to get: 2α3−α2−2α2 = 0 andα2(2α−3) = 0. We get α = 0 or α = 3

2 . Then our candidates are (α,β ) ∈�(0,0),

�32 , 9

2��

.

Using our assumption a,b > 0 we get that the only critical point of M is (a,b) =��

32 ,

�92

�. The area

of the ellipse is now equal to π · 3√

32 . Since π 3

√3

2 < 4π we get that the minimal area of the ellipse is

π · 3√

32 and it is attained when a =

�32 and b = 3√

2.