21-256 Solutions to Sample Exam 2 Section A

21-256 Solutions to Sample Exam 2
Section A
A1. Find the partial derivatives of
p
h(x, y, z) = x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2 + x3 y 3 z 3 sin(x2 y + y 2 z + z 2 x)
with respect to x, y and z.
Solution. Substituting u = xyz and v = x2 y + y 2 z + z 2 x gives h =
√
u2 + v 2 + u3 sin v). So
∂h
∂h ∂u ∂h ∂v
=
+
∂x ∂u ∂x ∂v ∂x
u
v
2
3
= √
+ 3u sin v · yz + √
+ u cos v · (2xy + z 2 )
u2 + v 2
u2 + v 2
!
xyz
+ 3x2 y 2 z 2 sin(x2 y + y 2 z + z 2 x) yz
= p
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
x2 y + y 2 z + z 2 x
!
p
+ x y z cos(x y + y z + z x) (2xy + z 2 )
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
∂h
∂h ∂u ∂h ∂v
=
+
∂y
∂u ∂y
∂v ∂y
u
v
2
3
= √
+ 3u sin v · xz + √
+ u cos v · (2yz + x2 )
u2 + v 2
u2 + v 2
!
xyz
= p
+ 3x2 y 2 z 2 sin(x2 y + y 2 z + z 2 x) xz
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
!
x2 y + y 2 z + z 2 x
3 3 3
2
2
2
+ p
+ x y z cos(x y + y z + z x) (2yz + x2 )
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
∂h
∂h ∂u ∂h ∂v
=
+
∂x ∂u ∂x ∂v ∂z
u
v
2
3
= √
+ 3u sin v · xy + √
+ u cos v · (2zx + y 2 )
u2 + v 2
u2 + v 2
!
xyz
2 2 2
2
2
2
+ 3x y z sin(x y + y z + z x) xy
= p
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
!
x2 y + y 2 z + z 2 x
+ p
+ x3 y 3 z 3 cos(x2 y + y 2 z + z 2 x) (2zx + y 2 )
x2 y 2 z 2 + (x2 y + y 2 z + z 2 x)2
+
3 3 3
2
2
2
In this case, since the expressions are so long, I’d accept the answer mixed in terms of u, v
and x, y, z. If the substitutions were simpler I’d want the answer put in terms of x, y, z.
A2. Find a function g of variables x, y such that
∂g
∂g
= −1,
= 3 and g(1, 1) = 4.
∂x
∂y
∂g
Solution. Since
= −1, integrating gives g(x) = −x + A(y) for some function A of y.
∂x
Partially differentiating this expression with respect to y gives
3=
∂g
= A0 (y)
∂y
so A(y) = 3y + C for some constant C. So g(x, y) = −x + 3y + C. Since g(1, 1) = 4, we
have
4 = −1 + 3 + C ⇒ C = 2
so g(x, y) = −x + 3y + 2 will do just fine.
A3. A bug is crawling on the xy-plane in such a way that, after t seconds, its position is
√
∂T
( 1 + t, 2 + 31 t). The temperature at the point (x, y) is T (x, y). If
(2, 3) = 4 and
∂x
∂T
(2, 3) = 3, how fast is the temperature increasing on the bug’s path after 3 seconds?
∂y
√
dT
when t = 3. Putting x = 1 + t and y = 2 + 13 t and using
Solution. We want to find
dt
the chain rule gives
dT
∂T dx ∂T dy
=
+
dt
∂x dt
∂y dt
∂T 1
∂T
1
+
=
· √
·
∂x 2 1 + t
∂y 3
So when t = 3 we have
dT
1
1
=4· +3· =2
dt
4
3
A4. Let z = f (x, y), where x = s + t and y = s − t. Show that
∂z
∂x
2
−
∂z
∂y
Solution. Using the chain rule, we have
∂z
∂z ∂x ∂z ∂y
=
+
∂s
∂x ∂s ∂y ∂s
∂z
∂z
=
+
∂x ∂y
∂z
∂z ∂x ∂z ∂y
=
+
∂t
∂x ∂t
∂y ∂t
∂z
∂z
=
−
∂x ∂y
Multiplying them together yields
2 2
∂z ∂z
∂z
∂z
∂z
∂z
∂z
∂z
=
+
−
=
−
∂s ∂t
∂x ∂y
∂x ∂y
∂x
∂y
as desired.
2
=
∂z ∂z
.
∂s ∂t
Section B
B1. Given that ln(π) ≈ 1.14, use linear approximation to estimate the value of sin(0.24e1.14 ) +
cos(0.24e1.14 ).
Solution. Let f (x, y) = sin(xey ) + cos(xey ) and let L(x, y) be the linearization of f at
( 41 , ln(π)). Then f (0.24, 1.14) ≈ L(0.24, 1.14), since 0.24 ≈ 14 and ln(π) ≈ 1.14.
The partial derivatives of f are:
∂f
= xey cos(xey ) − xey sin(xey )
∂y
∂f
= ey cos(xey ) − ey sin(xey )
∂x
Now, when x =
1
4
and y = ln(π), we have xey = 41 eln(π) =
π
4.
But sin( π4 ) = cos( π4 ) =
√1 ,
2
which means that at the point ( 14 , ln(π)) we have
∂f
∂f
=
=0
∂x
∂y
So quite simply
√
1
π
π
1
L(x, y) = f ( , ln(π)) = sin( ) + cos( ) = 2 · √ = 2
4
4
4
2
(Yes, the linearization is constant!)
So sin(0.24e1.14 ) + cos(0.24e1.14 ) = f (0.24, 1.14) ≈ L(0.24, 1.14) =
√
2.
B2. Show that the functions f (x, y) = 2e2x cos(3y) and g(x, y) = (1 + x)2 (2 − y 2 ) have the same
linearization at the point (0, 0).
Solution. The functions will have the same linearization so long as f (0, 0) = g(0, 0),
fx (0, 0) = gx (0, 0), and fy (0, 0) = gy (0, 0). Well
– f (0, 0) = 2e0 cos(0) = 2 and g(0, 0) = 12 (2 − 02 ) = 2;
– fx = 4e2x cos(3y) and gx = 2(1 + x)(2 − y 2 ), so
fx (0, 0) = 4e0 cos(0) = 4
and
gx (0, 0) = 2(1 + 0)(2 − 02 ) = 4
– fy = −6e2x sin(3y) and gy = (1 + x)2 (−2y) so
fy (0, 0) = −6e0 sin(0) = 0
and
gy (0, 0) = (1 + 0)2 (−0) = 0
So f and g have the same linearizations at (0, 0), namely
L(x, y) = 2 + 4x
B3. Find the direction of the line of intersection of the tangent planes to the surfaces
z=
x cos(y) + y cos(x)
√
3
and
2 sin(x) + 2 sin(y) + 2 sin(z) = 3
at the point ( π6 , π6 , π6 ).
Solution. Let f (x, y) =
x cos y+y
√ cos x .
3
Then
√
π π
cos y − y sin x √
fx ( , ) =
=
6 6
3
x=y= π6
3
2
− π6 ·
√
3
− π6 · 12 +
π π
−x sin y + cos x √
√
fy ( , ) =
=
6 6
3
3
x=y= π6
1
2
=
1
π
− √
2 12 3
√
3
2
=
1
π
− √
2 12 3
so the equation to the tangent line of the surface described by the first equation is
π
1
π
π
1
π
π
z− =
− √
x−
+
− √
y−
6
2 12 3
6
2 12 3
6
So its normal vector is
1
2
1
2

− 12π√3

− 12π√3 
−1
Now let g(x, y, z) = 2 sin(x) + 2 sin(y) + 2 sin(z). Then
√
π π π
3 √
gx ( , , ) = 2 cos x
=2·
= 3
6 6 6
2
x= π6
√
and likewise gy ( π6 , π6 , π6 ) = gz ( π6 , π6 , π6 ) = 3. So the equation of the tangent line of the
surface described by the second equation is
√
√
√
π
π
π
3(x − ) + 3(y − ) + 3(z − ) = 0
6
6
6
So its normal vector is
√ 
 
1
√3
 3
1
or
equivalently
√
1
3
So the line of intersection has direction
1
    1

π√
π√
1
2 − 12 3
2 − 12 3 + 1
1



 2 − 12π√3  × 1 = − 12 + 12π√3 − 1
1
−1
0
On the actual test I won’t give you anything this messy.
B4. Find the angle between the tangent planes of the surfaces
x2 y + y 2 z + z 2 x = 1
and
xyz + xy + yz + zx = −1
at the point (1, −1, 1).
Solution. Let f (x, y, z) = x2 y + y 2 z + z 2 x. Then when x = 1, y = −1, z = 1 we have
– fx = 2xy + z 2 = −2 + 1 = −1;
– fy = x2 + 2yz = 1 − 2 = −1;
– fz = y 2 + 2zx = 1 + 2 = 3.
So the tangent plane to the first surface has equation
(−1)(x − 1) + (−1)(y + 1) + 3(z − 1) = 0
 
−1
and its normal vector is n1 = −1.
3
Let g(x, y, z) = xyz + xy + yz + zx. Then when x = 1, y = −1, z = 1 we have
– gx = yz + y + z = −1 − 1 + 1 = −1;
– gy = xz + x + z = 1 + 1 + 1 = 3;
– gz = zy + y + x = −1 − 1 + 1 = −1.
So the tangent plane to the second surface has equation
(−1)(x − 1) + 3(y + 1) + (−1)(z − 1) = 0
 
−1
and its normal vector is n2 =  3 .
−1
So the angle between the two planes is
!
n
·
n
1
−
3
−
3
5
1
2
−1
−1
−1
p
p
cos
= cos
= cos
−
kn1 kkn2 k
11
(−1)2 + (−1)2 + 32 (−1)2 + 32 + (−1)2
Section C
C1. Find and classify the local extrema of the function h(x, y) = xy + x−1 + y −1 .
Solution. To find the critical points:
y − x−2
0
∇h(x, y) =
=
x − y −2
0
The first component gives y = x−2 , and substituting in the second gives x = x−4 , or
equivalently x5 = 1. Hence x = 1, and y = 1−2 = 1. So the only critical point is (1, 1).
(The points at which ∇h doesn’t exist is when x = 0 or y = 0, where h isn’t defined either.)
The Hessian matrix at (x, y) = (1, 1) is
−3
2x
1
2 1
Hh =
=
1
2y −3
1 2
Now 2 > 0 and det(Hh ) = 2 · 2 − 1 · 1 = 3 > 0, so the point (1, 1) is a local minimum of h.


2 −1 0
C2. Find values of a such that the matrix −1 2 a is positive definite. Deduce that the
0
a 8
2
2
2
function g(x, y, z) = x + y + 4z − xy − yz + 11z − 2 has a local minimum at (− 12 , −1, − 32 ).
2 −1
Solution. Since 2 > 0 and det
= 3 > 0, it suffices to find the values of a making
−1 2
the determinant of the whole matrix positive. Now


2 −1 0
2
a
−1
a
− (−1) det
det −1 2 a = 2 det
a 8
0 8
0
a 8
= 2(16 − a2 ) + (−8)
= 24 − 2a2
√
√
So the matrix is positive definite when 24 − 2a2 > 0, i.e. when − 12 < a < 12.
Now, when x = − 21 , y = −1, z = − 23 , we have
 
  
−1 + 1
0
2x − y
∇g =  2y − x − z  =  −2 + 21 + 32  = 0
0
8z − y + 11
−12 + 1 + 11

so (− 21 , −1, − 32 ) is a critical point. And


2 −1 0
Hg = −1 2 −1
0 −1 8
√
√
which is our matrix with a = −1. Since − 12 < −1 < 12, the matrix is positive definite,
so the critical point is a local minimum.
C3. Find the dimensions of the (closed) box of minimal surface area that has volume 1000 m3 .
Solution. Suppose the dimensions of the box are x by y by z metres. The surface area is
thus 2xy + 2yz + 2zx and the volume is xyz. We’re told that xyz = 1000, so z = 1000
xy .
Substituting this into the formula for the surface area gives
S(x, y) = 2xy +
2000 2000
+
x
y
We wish to minimize this function, bearing in mind that for the problem to make sense we
must take x, y, z > 0. (If any dimension is equal to zero the volume is zero, not 1000!)
Now solving
∇S =
gives y =
1000
x2
2y −
2x −
2000 x2
2000
y2
0
=
0
and hence x3 = 1000. So x = 10 and y =
1000
102
= 10, and z =
1000
10·10
= 10.
Moreover when x = y = 10 we have
4000
HS =
x3
2
2
4000
y3
4 2
=
2 4
Since 4 > 0 and det(HS ) = 4 · 4 − 2 · 2 = 16 − 4 = 12 > 0, this matrix is positive definite,
so (x, y) = (10, 10) is a minimum.
So in conclusion, the box is a cube of side length 10 m.
C4. Find the global extrema of the function f (x, y) = x3 + y 3 on the compact set x2 + y 2 6 1.
First we find the critical points. Well
2 0
3x
=
∇f =
0
3y 2
⇒
x=y=0
so the only critical point is (0, 0); and ∇f exists everywhere, so definedness is not an issue.
Now f (0, 0) = 0, so note this value.
Now √
we look at the boundary,
which is the circle x2 + y 2 = 1. On the boundary, either
√
y = 1 − x2 or y = − 1 − x2 , with x varying between −1 and 1, so we must split into
these cases.
√
√
3
Case 1 (y = 1 − x2 ). Let g1 (x) = f (x, 1 − x2 ) = x3 + (1 − x2 ) 2 . Then
p
1
g10 (x) = 3x2 − 3x(1 − x2 ) 2 = 3x(x − 1 − x2 )
This is equal to 0 when. . .
– x = 0 and y = 1, in which case f (0, 1) = 1;
√
– x = 1 − x2 , i.e. when x = ± √12 and y = √12 , in which cases we have f ( √12 , √12 ) =
and f (− √12 , √12 ) = 0.
1
√
3
2
Checking the end-points (x = −1 and x = 1) we get f (1, 0) = 1 and f (−1, 0) = −1.
√
So when y = − 1 − x2 the maximum is f (0, 1) = f (1, 0) = 1, and the minimum is 0 (which
is already covered).
√
√
Case 2 (y = − 1 − x2 ). Let g2 (x) = f (x, − 1 − x2 ). In a similar fashion
√ to with g1 we
find f (0, −1) = f (−1, 0) = −1 is the smallest value f takes when y = − 1 − x2 , and the
greatest value is 0.
In summary:
– The global minimum of f on the closed disc x2 + y 2 6 1 is 1, at (0, 1) and (1, 0);
– The global maximum of f on the closed disc x2 + y 2 6 1 is −1, at (0, −1) and (−1, 0).