Exam 1 (Sample: Solutions)
Chapter 9 and 10
Math 240 (Uricchio) - Summer 2016
(1) Given that the equation of a sphere is −62 = x2 − 6x + y 2 − 12y + z 2 − 18z find the
coordinates of the point on the sphere that is furthest from the origin (0, 0, 0).
−62 + 9 + 36 + 81 = x2 − 6x + 9 + y 2 − 12y + 36 + z 2 − 18z + 81
64 = (x − 3)2 + (y − 6)2 + (z − 9)2
So then the center of the sphere is (3, 6, 9) and is of radius 8. The furthest point
on the sphere will be on the line from the origin to the center of √
the circle. The
vector from the √
origin to √
the center is h3, 6, 9i which has a length of 32 + 62 + 92 =
√
9 + 36 +√81 = 126 = 3 14. So then the distance√from the origin to the furthest
point is 3 14 plus the radius, 8 units, which is 8 + 3 14. Multiplying this against a
unit vector that points towards the furthest point will give us the coordinates of the
point.
h3, 6, 9i
1
2
3
√
= √ ,√ ,√
is our unit vector
3 14
14 14 14
√ 1
2
3
8
16
24
√ ,√ ,√
8 + 3 14 = √ + 3,√ + 6,√ + 9
14 14 14
14
14
14
Hence our furthest point is
√8
14
√24 + 9 ≈ (5.14, 10.28, 15.42)
+ 3, √16
+
6,
14
14
(2) Several points are given: A(−3, 4, 1), B(−4, −10, 1), C(9, −7, −3), D(−1, 7, 8),
X(−1, −2, 7), Y (4, −2, 3), Z(−9, 4, 6). What point W needs to be added so that
the volume of the parallelepiped determined by X, Y , Z, and W is equal to the
volume of the parallelepiped determined by A, B, C, and D?
−→
AB = h−4 − 3, −10 − 4, 1 − 1i
= h−1, −14, 0i
−→
AC = h9 − 3, −7 − 4, −3 − 1i
= h12, −11, −4i
−→ −→
AB × AC = h(−14)(−4) − (−11)(0), (0)(12) − (−4)(−1), (−1)(−11) − (−14)(12)i
= h, −4, 179i
−→ −→ p
AB × AC = (56)2 + (−4)2 + (179)2
√
= 35193
≈ 187.60
−−→
AD = h−1 − 3, 7 − 4, 8 − 1i
= h2, 3, 7i
−−→ −→ −→
AD · AB × AC = |(2)(56) + (3)(−4) + (7)(179)|
= 1353
−−→
XY = h4 − 1, −2 − 2, 3 − 7i
= h5, 0, −4i
−−→
XZ = h−9 − 1, 4 − 2, 6 − 7i
= h−8, 6, −1i
−−→ −−→
XY × XZ = h(0)(−1) − (−4)(6), (−4)(−8) − (5)(−1), (5)(6) − (0)(−8)i
= h24, 37, 30i
−−→ −−→ p
XY × XZ = (24)2 + (37)2 + (30)2
√
= 2845
≈ 53.34
−−→ −−→
XY × XZ
unit normal = −−→ −−→
XY × XZ ≈ h0.45, 0.69, 0.56i
volume
height =
area
1353
=
53.34
≈ 25.37
−−→
XW = unit normal · height
≈ h0.45, 0.69, 0.56i · 25.37
≈ h11.42, 17.6, 14.27i
2
√
√ (3) Two forces (measured in Newtons), F1 = 53 3i + 43 j and F2 = 13 3, 32 , are applied to
a box to move it up a slope which has angle 15o up from horizontal. Find the work
done on the box by the two forces if the box is moved a distance of 10 meters.
F~ = F~1 + F~2 =
5√ 4
3,
3
3
+
1√ 2
3,
3
3
D√ E
= 2 3, 2
~ = h10 · cos(15o ), 10 · sin(15o )i ≈ h9.66, 2.59i
D
√
~ = (2 3)(10 cos(15o )) + (2)(10 sin(15o )) ≈ 38.64
W = F~ · D
(4) Find the angle of intersection between the planes 5x+10y +7z = 7 and 8x−6y −4z =
2. Then calculate the distance from the point P (5, 5, 3) to each plane and determine
which plane is closer to the point.
n1 = h5, 10, 7i
n2 = h8, −6, −4i
p
|n1 | = (5)2 + (10)2 + (7)2
≈ 13.19
p
|n2 | = (8)2 + (−6)2 + (−4)2
≈ 10.77
n1 · n2 = (5)(8) + (10)(−6) + (7)(−4)
= −48
n1 · n2
cos(θ) =
|n1 ||n2 |
−48
θ = arccos
(13.19)(10.77)
o
≈ 70.25
|ax1 + by1 + cz1 + d|
√
a2 + b2 + c2
|5(5) + 10(5) + 7(3) − 7|
D1 = p
(5)2 + (10)2 + (7)2
≈ 6.75
|8(5) + (−6)(5) + (−4)(3) − 2|
p
D2 =
(8)2 + (−6)2 + (−4)2
≈ 0.37
D=
3
Plane 2 is much closer.
√
√
(5) Convert the following point P ( 2, π3 , 2) from cylindrical coordinates to spherical
coordinates.
π √2
√
=
x = r cos θ = 2 cos
3
2
π √6
√
=
x = r sin θ = 2 sin
2
√ 3
z=z= 2
2
ρ = x2 + y 2 + z 2 = 4
ρ=2
√ !
z
2
π
φ = arccos
= arccos
=
ρ
2
4
!
√
x
2/2
π
θ = arccos
sin φ = arccos
sin(π/4) =
ρ
2
3
!
√ √
√ π √ π π
2 6√
Cylin 2, , 2
⇒ Rect
, , 2
⇒ Sphere 2, ,
3
2 2
3 4
(6) Identify the surface whose equation is given and show how you determined your
answer. (If it is a sphere give its center and radius. If it is a cylinder, give its radius
and axis of revolution. Etc.)
(a) z = 4 − r2
z = 4 − (x2 + y 2 )
This is an upside down (pointed downward) elliptical paraboloid, whose vertex
is shifted upwards along the z-axis 4 units from the origin.
(b) ρ = 8 sin(θ) sin(φ) + 6 cos(θ) sin(φ)
ρ2
x2 + y 2 + z 2
x2 − 6x + y 2 − 8y + z 2
x2 − 6x + 9 + y 2 − 8y + 16 + z 2
(x − 3)2 + (y − 4)2 + z 2
= 8ρ sin(θ) sin(φ) + 6ρ cos(θ) sin(φ)
= 8y + 6x
=0
= 25
= 25
4
This is a sphere of radius 5 centered at the point (3, 4, 0).
(c) r = 12 cos(θ)
r2
x2 + y 2
x2 − 12x + y 2
x2 − 12x + 36 + y 2
(x − 6)2 + y 2
= 12r cos(θ)
= 12x
=0
= 36
= 36
This is a elliptical cylinder of radius 6 centered at the point (6, 0, 0) and runs
parallel to the z-axis.
(d) ρ2 (sin2 (φ) sin2 (θ) + cos2 (φ)) = 9
9 = ρ2 sin2 (φ) sin2 (θ) + ρ2 cos2 (φ)
9 = y2 + z2
This is a elliptical cylinder of radius 3 centered at the point (0, 0, 0) and runs
parallel to the x-axis.
(7) Find the point which has maximum curvature for the function f (x) = ex and give
the value of the curvature at that point.
κ(x) =
|f 00 (x)|
[1 +
f 0 (x) = ex
(f 0 (x))2 ]3/2
κ(x) =
κ0 (x) =
f 00 (x) = ex
|ex |
[1 + e2x ]3/2
ex − 2e3x
(1 + e2x )5/2
We set κ0 (x) = 0 and solve for x. Notice that the denominator can never be zero, so
we only worry about the numerator being 0.
√
1
ln(1/2)
0 = ex − 2e3x ⇒ 0 = 1 − 2e2x ⇒ e2x = ⇒ x =
= ln( 2/2)
2
2
5
√
√
2
f (ln( 2/2)) =
2
√
√
√
The point with maximum curvature is (ln( 2/2), 2/2) and its curvature is (2 3)/9 ≈
0.39.
(8) Reparametrize the vector function r(t) = het cos(t), et sin(t), et i with respect to the
arc length measured from (1, 0, 1) in the direction of t increasing. Find the point on
the space curve a particle is at if it has moved 16 units along the curve. (Round to
two decimals.)
r0 (u) = heu cos(u) − eu sin(u), eu cos(u) + eu sin(u), eu i
p
|r0 (u)| = (eu cos(u) − eu sin(u))2 + (eu cos(u) + eu sin(u))2 + (eu )2
√
= eu 3
Z t
s=
|r0 (u)| du
Z0 t √
=
eu 3 du
0
√
√
t
=e 3− 3 !
√
s+ 3
√
t = ln
3
*
r(t(s)) =
ln
e
√ s+
√ 3
3
cos ln
√ !!
√ s+ 3
√ 3
ln s+
3
√
,e
sin ln
3
+
√ !!
√ s+
3
s+ 3
√
ln
3
√
,e
3
r(t(16)) = h−7.02, 7.45, 10.24i
The point is (−7.02, 7.45, 10.24).
(9) A particle moves with position function r(t) = ht, 2t2 , 3t3 i. Find the tangential and
normal components of acceleration in general and then when t = 6.
6
r0 (t) = 1, 4t, 9t2
p
|r0 (t)| = (1)2 + (4t)2 + (9t2 )2
√
= 1 + 16t2 + 81t4
r00 (t) = h0, 4, 18ti
r0 (t) × r00 (t) = (4t)(18t) − (9t2 )(4), (9t2 )(0) − (1)(18t), (1)(4) − (4t)(0)
= 36t2 , −18t, 4
p
|r0 (t) × r00 (t)| = (36t2 )2 + (−18t)2 + (4)2
√
= 2 324t4 + 81t2 + 4
r0 (t) · r00 (t) = (1)(0) + (4t)(4) + (9t2 )(18t)
= 16t + 162t3
|r0 (t) × r00 (t)|
|r0 (t)|
√
2 324t4 + 81t2 + 4
= √
1 + 16t2 + 81t4
aN (t) =
r0 (t) · r00 (t)
|r0 (t)|
16t + 162t3
=√
1 + 16t2 + 81t4
aN (6) ≈ 4
aT (6) ≈ 108
aT (t) =
(10) Find a parametric representation in cylindrical coordinates for the upper half of the
hyperboloid of one sheet x2 + y 2 − z 2 = 1.
Relations:
r 2 = x2 + y 2
x = r cos θ
y = r sin θ
r2 − z 2 = 1
z 2 = r2 + 1
√
z = r2 + 1
D
E
√
~r(r, θ) = r cos θ, r sin θ, r2 − 1
7