Calculus III, Exam 2: Review Solutions
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Calculus III Exam 2 Practice Problems - Solutions
1. Compute all first and second partial derivatives for the following functions.
(a) f (x, y) = (x + y)2
Solutions: fx = 2(x + y), fy = 2(x + y), fxx = 2, fxy = fyx = 2, fyy = 2
(b) f (x, y, z) = x2 y + x2 z − yz 2
Solutions: fx = 2xy + 2xz, fy = x2 − z 2 , fz = x2 − 2yz, fxx = 2y + 2z, fxy = fyx = 2x,
fxz = fzx = 2x, fyy = 0, fyz = fzy = −2z, fzz − 2y
(c) f (x, y) = e2x+y
2
2
2
2
2
Solutions: fx = 2e2x+y , fy = 2ye2x+y , fxx = 4e2x+y , fxy = fyx = 4ye2x+y , fyy =
2
2
2e2x+y + 4y 2 e2x+y
(d) f (x, y) = sin(2x + y)
Solutions: fx = 2 cos(2x + y), fy = cos(2x + y), fxx = −4 sin(2x + y), fxy = fyx =
−2 sin(2x + y), fyy = − sin(2x + y)
(e) f (x, y, z) = sin x sin y sin z
Solutions: fx = cos x sin y sin z, fy = sin x cos y sin z, fz = sin x sin y cos z, fxx = − sin x sin y sin z,
fxy = fyx = cos x cos y sin z, fxz = fzx = cos x sin y cos z, fyy = − sin x sin y sin z, fyz = fzy =
sin x cos y cos z, fzz = − sin x sin y sin z
(f) f (x, y) = xeyz
Solutions: fx = eyz , fy = xzeyz , fz = xyeyz , fxx = 0, fxy = fyx = zeyz , fxz = fzx = yeyz ,
fyy = xz 2 eyz , fyz = fzy = xeyz + xz 2 eyz , fzz = xy 2 eyz
2. Let the temperature of a flat plate be given by the function T (x, y) = 3x2 + 2xy. What is the
value of the directional derivative of this function at the point (3, −6) in the direction ~v = 4~ı − 3~?
In what direction is the plate cooling most rapidly?
Solution: The unit vector in the direction of ~v is ~u = h4/5, −3/5i, so D~u T (3, −6) = ∇T (3, −6) ·
h4/5, −3/5i = h6x + 2y, 2xi(3, −6) · h4/5, −3/5i = h6, 6i · h4/5, −3/5i = 24/5 − 18/5 = 6/5.
The direction the in which the plate is cooling most rapidly is −∇T (3, −6) = −h6, 6i.
3. Find and test all critical points of f (x, y) = x3 + 3xy 2 + 3y 2 − 15x + 2.
Solution: fx = 3x2 + 3y 2 − 15 = 0 if and only if x2 √
+ y 2 = 5.√fy = 6xy + 6y = 6y(x + 1) = 0 if
and only if x = −1 or y = 0. Thus (−1, 2), (−1, −2), ( 5, 0), (− 5, 0) are the only possible critical
points (since fx , fy exist everywhere). Now apply SDT: fxx = 6x, fxy = 6y, fyy = 6x + 6, so
∆(−1, 2) = (−6)(0) − (12)2 < 0 =⇒ saddle point
∆(−1, 2) = (−6)(0) − (12)2 < 0 =⇒ saddle point
√
√
√
√
√
∆( 5, 0) = (6 5)(6 5 + 6) − (0)2 > 0 and fxx ( 5, 0) = 6 5 > 0 =⇒ local min
√
√
√
√
√
∆(− 5, 0) = (−6 5)(−6 5 + 6) − (0)2 > 0 and fxx (− 5, 0) = −6 5 < 0 =⇒ local max
4. Assume that f (x, y) = xy + y 2 − 3x − 3 = 0 defines y implicitly as a function of x. Use the chain
dy
rule to find dx
.
Solution: Differentiate with respect to x:
dy
dx
df
∂f ∂f dy
dy
=
+
= (y−3)+(x+2y)
= 0. Therefore
dx
∂x ∂y dx
dx
= −(y − 3)/(x + 2y).
5. Find the point on x + y − z = 3 that is closest to the point (1, −1, 2).
Solution: We want to maximize the function d(x, y, z) = (x − 1)2 + (y + 1)2 + (z − 2)2 subject to
the constraint x + y − z − 3 = 0. So z = x + y − 3 and
d(x, y) = (x − 1)2 + (y + 1)2 + [(x + y − 3) − 2]2 = 2x2 + 2xy + 2y 2 − 12x − 8y + 27.
Calculus III, Exam 2: Review Solutions
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Computing dx = 4x + 2y − 12 and dy = 2x + 4y − 8, we see both exist for all values of x, y, so
the only critical points of d(x, y) occur where dx = dy = 0. Moreover, as d is defined for all (x, y),
there is no boundary to consider, so any extreme points occur at critical points. Now dx = 0
implies y = 6 − 2x. Plugging this in to the equation dy = 0 we obtain 2x + 4(6 − 2x) − 8 = 16 − 6x,
so x = 8/3, y = 2/3, z = 1/3. Hence (8/3, 2/3, 1/3) is the point on the plane closest to (1, −1, 2).
(The second derivative test, with dxx = 4, dxy = 2, dyy = 4, implies that D = 42 − 22 > 0 and
dxx > 0 proves this is a minimum).
6. Evaluate the following expressions.
2+y
2+y
2+0
= lim(0,0) x+cos
y = 0+1 = 2
(x,y)→(0,0) x + cos y
∂
2xy
y2
[tan−1 (x2 y) + tan−1 (xy 2 )] = 1+(x
(b)
2 y)2 + 1+(xy 2 )2
∂x
(c) fxy (0, 1) where f (x, y) = (x3 + x)y 2 : fx = (3x2 + 1)y 2 so fxy = (3x2 + 1)(2y) and fxy (0, 1) =
(1)(2) = 2
(a)
lim
(d) ∇(x3 y 4 ) = h3x2 y 4 , 4x3 y 3 i
(e) fx (2, 4) where f (x, y) = xey/x : fx = ey/x + xey/x (−yx−2 ) so fx (2, 4) = e2 + 2e2 (−1)
√
√
√
(f) D~u (x2 − y 3 ) where ~u = 12~ı − 23 ~: D~u (x2 − y 3 ) = h2x, −3y 2 i · h1/2, 3/2i = x − (3 3/2)y 2
7. Find the local max, local min, and saddle points of f (x, y) = 4xy − x4 − y 4 . Does f have any
absolute extreme values?
Solution: fx = 4y − 4x3 = 0 when y = x3 . fy = 4x − 4y 3 = 0 when x = y 3 . Combining these two,
we need y = (y 3 )3 = y 9 , which only happens if y = 0, 1, −1 and so x = 0, 1, −1 correspondingly.
That is, the only critical points of f are (0, 0), (1, 1), (−1, −1) (since fx , fy are defined for all x, y).
Now apply SDT: fxx = −12x2 , fxy = 4, fyy = −12y 2 , so
∆(0, 0) = (0)(0) − (4)2 < 0 =⇒ saddle point
∆(1, 1) = (−12)(−12) − (4)2 > 0 and fxx (1, 1) = −12 =⇒ local max
∆(−1, −1) = (−12)(−12) − (4)2 > 0 and fxx (−1, −1) = −12 =⇒ local max
Lastly, f has an absolute max at both (1, 1) and (−1, −1), but it doesn’t have an absolute min,
for as x2 + y 2 → ∞, f → −∞.
8. Find the directions of maximum increase and decrease for f (x, y) = x2/3 + xy + y 2/3 at (−1, 1).
What are the corresponding magnitudes of these changes?
Solution: Max increase occurs in the direction of ∇f (−1, 1) = h2/(3x1/3 )+y, x+2/(3y 1/3 )i(−1, 1) =
h−2/3+1,
−1+2/3i = h1/3, −1/3i and the magnitude of change in this direction is ||∇f (−1, 1)|| =
p
2/9. The
p greatest decrease occurs in the direction −∇f (−1, 1) = h−1/3, 1/3i with corresponding
change − 2/9.
∂w
∂w
and
given w = x2 + xy + y 2 with x = 2r + s and y = r − 2s.
∂s
∂t
∂w ∂x
∂w ∂y
∂w
∂w ∂x
Solution: ∂w
∂r = ∂x ∂r + ∂y ∂r = (2x + y)(2) + (x + 2y)(1) and ∂s = ∂x ∂s +
y)(1) + (x + 2y)(−2).
9. Find
∂w ∂y
∂y ∂s
= (2x +
10. Find the extreme values of f (x, y) = xy in the region bounded by the positive x-axis, the positive
y-axis, and the circle x2 + y 2 − 10 = 0.
Solution: First we look for critical points of f : fx = y = 0 and fy = x = 0 only at the origin,
(0, 0), which is not an interior point of the domain, so the book and most people (including myself)
will disagree in calling this a critical point. Technically, it is a boundary point of the domain, and
so is not a critical point, but if you call it one, I won’t hurt you. What this does say, however,
is that the function can only assume its max and min values on the boundary. Let’s consider
Calculus III, Exam 2: Review Solutions
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√
this, then.
is f (0, y) = 0 for√all y. On y √
= 0 and
√ On x = 0, 0 ≤ y ≤ 10, the function
2
2
2 ) = x 10 − x2
0 ≤ x ≤ 10, f √
(x, 0) = 0 for all x. Finally,
on
x
+
y
=
10,
we
have
f
(x,
10
−
x
√
(we know y √
= 10 − x2 and not − 10 − x2 since we
√ are in the first quadrant so that y ≥ 0).
Hence f 0 (x, 10 − x2 ) = (x)[ 12 (10 − x2 )−1/2 (−2x)] + 10 − x2 = (10 − x2 )−1/2 [−x2 + (10 − x2 )] =
√
2
(10 −√x2 )−1/2 [10 −
√2x ]. This does not exist when x = 10 (so y = 0 and f = 0) and is zero when
x = 5 (so y = 5 and f = 5). Hence the absolute minimum value of f is 0, which occurs for
all values of (x, y) on the boundaries
x = 0 and y = 0, and the absolute maximum value of f is 5
√ √
which occurs at the point ( 5, 5).
11. Evaluate the following iterated integrals.
R 1 R x3
R1
R1
3
2
2
(a) 0 0 ey/x dydx = 0 [xey/x ]x0 dx = 0 [xex − xe0 ]dx = [ 21 ex − 12 x]10 = 12 [(e − 1) − (1 − 0)] =
1
2 [e − 2]
R 2 R 4−x2
R2
R2
R2
2
(b) 0 x2 2xdydx = 0 2x[y]4−x
dx = 0 2x[(4 − x2 ) − (x2 )]dx = 0 (8x − 4x3 )dx = [4x2 −
x2
x4 ]20 = 16 − 16 = 0
R2R1
R 1 R 2x 2
R1 2
R 1 x2
2
x2 1
1
0
(c) 0 y/2 ex dxdy = 0 0 ex dydx = 0 ex [y]2x
0 dx = 0 e [2x]dx = [e ]0 = e − e = e − 1
ReRxRz
ReRx
ReRx
Re
Re
(d) 1 1 0 (2y/z 3 )dydzdx = 1 1 (y 2 /z 3 )z0 dzdx = 1 1 (1/z)dzdx = 1 [ln |z|]x1 dx = 1 ln xdx =
[x ln x − x]e1 = (e − e) − (0 − 1) = 1
√
R 3/2 R √9−4y2
R 3/2
R 3/2 p
2 3/2
9−4y 2
3/2
)
√
√
(e) 0
ydxdy
=
y[x]
= 0 2y 9 − 4y 2 dy = [− 14 (9−4y
]0 = − 16 (0)3/2 −
3/2
0
2
2
−
9−4y
−
9−4y
9
(− 16 )(9)3/2 = 27
6 = 2
R 1 R x2
R 1 R x2 R x+y
R 1 R x2
dydx = 0 0 [2x2 +2xy−xy−y 2 −
(f) 0 0 0 (2x−y−z)dzdydx = 0 0 [2xz−yz− 12 z 2 ]x+y
0
R 1 R x2 3 2 3 2
R1 3 2
R1 3 4 1 6
1
1 3 x2
2
2
2 (x + 2xy + y )]dydx = 0 0 [ 2 x − 2 y ]dydx = 0 [ 2 x y − 2 y ]0 dx = 0 [ 2 x − 2 x ]dx =
3 5
1 7 1
3
1
[ 10
x − 14
x ]0 = 10
− 14
= 16
70
12. The two legs of a right triangle are measured as 5 m and 12 m, respectively, with a possible error
in measurement of at most 0.2 cm in each. Use differentials to estimate the maximum error in the
calculated value of (a) the area of the triangle and (b) the length of the hypotenuse.
Solution: (a) A = 12 bh, so dA = Ab db + Ah dh = [ 12 h][db] + [ 12 b][dh]. Thus, ∆A ≈ dA =
1
2 [(12)(0.2) + (5)(0.2)] = 1.7.
√
(b) H = b2 + h2 , so dH = Hb db+Hh dh = 12 (b2 +h2 )−1/2 [2b]db+ 12 (b2 +h2 )−1/2 [2h]dh. Therefore,
∆H ≈ dH = (52 + 122 )−1/2 [(12)(0.2) + (5)(0.2)] = 17
65 .
13. Find the volume of the solid bounded by the cylinder x2 + y 2 = 4 and the planes z = 0 and
y + z = 3.
R 2π R 2 R 3−r sin θ
R 2π R 2
R 2π R 2
sin θ
Solution: V =
rdzdrdθ = θ=0 r=0 r[z]3−r
drdθ = θ=0 r=0 (3r−r2 sin θ)drdθ =
0
z=0
R 2π 3 2 1 3 θ=02 r=0R 2π
[ r − 3 r sin θ]0 = 0 [6 − 38 sin θ]dθ = [6θ + 83 cos θ]02π = (12π + 83 ) − (0 + 83 ) = 12π.
θ=0 2
14. Find the mass of a plate that occupies the region D bounded by the parabola x = 1 − y 2 and the
coordinate axes in the first quadrant if the density function is δ(x, y) = y.
RR
R 1 R 1−y2
R1
R1
2
Solution: M = D δ(x, y)dA = y=0 x=0 ydxdy = y=0 y[x]1−y
dy = 0 (y − y 3 )dy = [ 21 y 2 −
0
1 4 1
1
4 y ]0 = 4 .
15. Evaluate the following expressions.
R2 R4
R2
R2
(a) −2 0 (4x3 + 3xy 2 )dydx = −2 [4x3 y + xy 3 ]40 dx = −2 [16x3 + 64x]dx = [4x4 + 32x2 ]2−2 =
(64 + 128) − (64 + 128) = 0
R 2 R x2 1
R2
R2
R2
2
(b) 1 0 x+y
dydx = 1 [ln |x + y|]x0 dx = 1 [ln(x + x2 ) − ln x]dx = 1 ln(1 + x)dx = [(x +
1) ln(1 + x) − x]21 = [3 ln 3 − 2] − [2 ln 2 − 1] = 3 ln 3 − 2 ln 2 − 1
Calculus III, Exam 2: Review Solutions
R 1 R √x
R1
y sin(x2 )dxdy = 0 0 y sin(x2 )dydx = 0
0 y2
1
1
1
1
2 1
2 [− 2 cos(x )]0 = 4 [− cos(1) + cos 0] = 4 [1 − cos 1]
(c)
R1R1
(d)
R 1 R x2 R y
(e)
(f)
4
√
x
sin(x2 )[ 12 y 2 ]0 dx =
R1
0
sin(x2 )[ 12 x]dx =
R 1 R x2
R 1 R x2
R 1 1 5 x2
R1 1 1
y 2 zdzdydx = 0 0 y 2 [ 12 z 2 ]y0 dydx = 0 0 12 y 4 dydx = 0 [ 10
y ]0 dx = 0 10
x 0dx
0 0
0
1
1
[ 110
x11 ]10 = 110
R 1 R 1 x/y
R1Ry
R1
R1
e dydx = 0 0 ex/y dxdy = 0 [yex/y ]y0 dy = 0 [ye − y]dy = [ 2e y 2 − 12 y 2 ]10 = 12 (e − 1)
0 x
R1R1 Ry
R1R1
R1R1
R1
R1
√
xydzdxdy = 0 √y xy[z]y0 dxdy = 0 √y xy 2 dxdy = 0 y 2 [ 12 x2 ]1√y dy = 0 [ 12 y 2 −
0
y 0
1 3
1 3
1 4 1
1
1
1
2 y ]dy = [ 6 y − 8 y ]0 = 6 − 8 = 24
16. Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height
12 cm if the tin is 0.04 cm thick.
Solution: The volume of tin is the volume of the exterior of the can minus the volume of the
interior of the can. The volume of a cylinder of radius r and height h is V = πr2 h, so ∆V ≈ dV =
Vr dr + Vh dh = [2πrh][dr] + [πr2 ][dh] = [2π(4)(12)][0.04] + [π(4)2 ][0.04] = 112π
25 .
p
17. Find the volume under the cone z = x2 + y 2 and above the ring 4 ≤ x2 + y 2 ≤ 25.
R 2π R 5 R r
R 2π R 5
R 2π R 5
R 2π
Solution: V = θ=0 r=2 z=0 rdzdrdθ = 0 2 r[z]r0 drdθ = 0 2 r2 drdθ = 0 [ 13 r3 ]52 dθ =
8
( 125
3 − 3 )[2π] = 78π.
18. Find the average height of the paraboloid z = x2 + y 2 over the rectangle R = {(x, y) : 0 ≤ x ≤
2, 0 ≤ y ≤ 1}.
RR
R2R1
R2
1
Solution: The average height is Area(R)
zdA = 12 0 0 (x2 + y 2 )dydx = 12 0 [x2 y + 13 y 3 ]10 dx =
R
R
1
1 1 3
1 2
1
5
1 2 2
2 0 [x + 3 ]dx = 2 [ 3 x + 3 x]0 = 6 [(8 + 2) − (0 + 0)] = 3 .
=