MAT397 M005
Calculus III
Spring 2011
The second midterm covers Sections 12.6–13.2 and Sections 13.4–14.6 of the
textbook, except pages 819–822 and pages 841–846.
The practice exam problems below are pretty representative of what you can
expect for the midterm in terms of the difficulty level, nature of problems,
and length of the exam, but not in terms of specific questions or topics
covered. The latter is due to the simple statistical fact that the topics that
can be covered on any given exam represent only a small random snapshot
of the entire exam material, and another such random snapshot is likely to
result in different questions and topics. Thus, simply studying the problems
below will not adequately prepare you. The only way to be fully prepared for
the exam is to work through the entire exam material, including all assigned
homework.
Practice Exam Problems:
1. Reduce the equation 4x2 + y 2 + 4z 2 − 4y − 24z + 36 = 0 to one of the
standard forms and classify the surface. No sketching is required.
Solution: Rewriting the equation as
4x2 + (y 2 − 4y + 4) + 4(z 2 − 6z + 9) = 4
and completing squares in y and z yields
4x2 + (y − 2)2 + 4(z − 3)2 = 4,
or x2 +
(y − 2)2
+ (z − 3)2 = 1, which is an ellipsoid with center (0, 2, 3)
4
2. Given the space curve r(t) = h3 sin t, −3 cos t, 4ti, compute the following quantities:
π
(a) The unit tangent vector T at t = .
2
Solution: r0 (t) = h3 cos t, 3 sin t, 4i,
p
√
|r0 (t)| = (3 cos t)2 + (3 sin t)2 + 42 = 32 + 42 = 5,
π 3 4
3
3
4
π
so T(t) =
cos t, sin t,
; at t = , T
= 0, ,
5
5
5
2
2
5 5
1
MAT397 M005
Calculus III
Spring 2011
π
(b) Parametric equations for the tangent line at t = .
2
π
passes through the point
Solution: The tangent line at t =
2
π π 4π
r
= 3, 0,
and has direction vector r0
= h0, 3, 4i,
2
2
2
so its vector equation is r(t) = h3, 0, 2πi + t h0, 3, 4i. Thus its
parametric equations are x = 3, y = 3t, z = 2π + 4t
3. The acceleration of a particle at any time t is given by a(t) = h− cos t, − sin t, 0i.
At time t = 0, it is at the point (1, 0, 0) and has velocity v(0) = h0, 1, 2i.
(a) Find the velocity of the particle at any time t.
Solution: We integrate a(t) = h− cos t, − sin t, 0i to get
Z
v(t) = a(t)dt = h− sin t, cos t, 0i + c,
where c is a constant vector. Substituting t = 0 and using the
given initial velocity v(0) = h0, 1, 2i, we get
h0, 1, 2i = v(0) = h0, 1, 0i + c,
so c = h0, 0, 2i. Thus the velocity at any given time t is
v(t) = h− sin t, cos t, 2i
(b) Find the position of the particle at any time t.
Solution: Integrating the velocity function v(t) = h− sin t, cos t, 2i,
we get
Z
r(t) = v(t)dt = hcos t, sin t, 2ti + d,
where d is a constant vector. Substituting t = 0 and using the
given initial position r(0) = h1, 0, 0i, we get
h1, 0, 0i = r(0) = h1, 0, 0i + d,
so d = h0, 0, 0i and the position at time t is
r(t) = hcos t, sin t, 2ti
2
MAT397 M005
Calculus III
Spring 2011
(c) When is the particle closest to the origin?
Solution: The distance to the origin at time t is just the magnitude of this vector
p
√
√
|r(t)| = (cos t)2 + (sin t)2 + (2t)2 = 1 + 4t2 ≥ 1 + 0 = 1.
This minimum value of 1 is achieved when 4t2 = 0, or t = 0 .
Alternatively, one may also use either the 1st or the 2nd Derivative
Tests to find the minimum.
Comments: This minimization problem is similar to Problem
19 of Section 13.4. The main point of such problems is that the
magnitude of a vector function is a scalar (or real-valued) function,
which could have a maximum or a minimum. On the other hand, it
does not make sense to talk about the maximum or the minimum
of a vector function.
4. The ideal gas law states that
P V = cT
(1)
where P (pressure), V (volume) and T (temperature) are variables and
c is a constant. Compute the product
T
∂P ∂V
,
∂T ∂T
simplifying as much as possible. (A complete, step-by-step argument is required here; an answer alone will not earn credit.)
∂P
∂V
and
, we solve (1) for
Solution: To compute T and the partials
∂T
∂T
each of the variables, and then compute T and the appropriate partial
derivatives.
PV
,
c
cT
P =
=⇒
V
cT
V =
=⇒
P
T =
3
∂P
c
= ,
∂T
V
∂V
c
= .
∂T
P
MAT397 M005
Calculus III
Spring 2011
Multiplying these expressions gives
T
PV c c
∂P ∂V
=
= c.
∂T ∂T
c V P
5. Find the tangent plane to the surface given by the equation
z = x2 + y 2 − 2
at the point (1, 0, −1).
Solution: The surface is of the form z = f (x, y), so we can use the
formula z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ). We have
(x0 , y0 , z0 ) = (1, 0, −1), fx (x, y) = 2x, fy (x, y) = 2y, fx (1, 0) = 2,
fy (1, 0) = 0, so the equation becomes z + 1 = 2(x − 1) + 0(y − 0), or
z − 2x + 3 = 0
6. Find the linear approximation of the function
p
f (x, y) = ln x2 + y 2
at the point (2, 2) and use it to approximate f (2.01, 2.01).
p
1
Solution: Using ln x2 + y 2 = ln(x2 + y 2 ) and then differentiating
2
x
1 2x
=
and simiwith the chain rule, we obtain fx =
2 x2 + y 2
x2 + y 2
y
larly, fy = 2
. At the given point, the value of f is f (2, 2) =
x + y2
√
√
3
3
ln 22 + 22 = ln 8 = ln 2 2 = ln 2, and all three partial derivatives
2
2
1
are equal to 2
= . Hence the linearization for f at the given
2 + 22
4
point is
L(x, y) =
3
1
1
3
1
1
ln 2 + (x − 2) + (y − 2) = ln 2 − 1 + x + y
2
4
4
2
4
4
and the linear approximation for f is given by
f (x, y) ≈
3
1
1
ln 2 − 1 + x + y .
2
4
4
4
MAT397 M005
Calculus III
Spring 2011
Substituting x = 2.01, y = 2.01 into the formula for L(x, y), we obtain
1
1
1
3
3
f (x, y) ≈ ln 2+ (0.01)+ (0.01) = ln 2 +
as the approximate
2
4
4
2
200
value sought.
7. Let
z = f (x, y),
x = r 2 + s2 ,
y = r 2 − s2 .
Use the Chain Rule to compute
1 ∂z 1 ∂z
+
,
r ∂r s ∂s
simplifying as much as possible.
Solution: By the Chain Rule,
∂z
∂z ∂x ∂z ∂y
∂z
=
+
=
2r +
∂r
∂x ∂r ∂y ∂r
∂x
∂z
∂z ∂x ∂z ∂y
∂z
=
+
=
2s −
∂s
∂x ∂s ∂y ∂s
∂x
Hence
∂z
2r,
∂y
∂z
2s.
∂y
1 ∂z 1 ∂z
∂z
+
=4
r ∂r s ∂s
∂x
8. Suppose the temperature at a point (x, y, z) in space is given by the
function
1
f (x, y, z) =
.
2
1 + x + y2 + z2
A bug is located at a point P = (1, 1, 1) in space.
(a) Determine the direction (specified as a 3-dimensional vector, e.g.,
h2, 4, 2i) in which the bug needs to fly in order to encounter the
maximal rate of increase in temperature.
Solution: The direction of maximal increase is in the direction
of the gradient: ∇f (x, y, z) =
−2x
−2y
−2z
,
,
.
(1 + x2 + y 2 + z 2 )2 (1 + x2 + y 2 + z 2 )2 (1 + x2 + y 2 + z 2 )2
5
MAT397 M005
Calculus III
Spring 2011
Evaluating the gradient at the given point (1, 1, 1) gives
1 1 1
∇f (1, 1, 1) = − , − , −
8 8 8
(b) Find the rate of increase in temperature if it flies in the direction
found in (a).
Solution: The rate of change in this direction is the magnitude
of the gradient at this point, i.e.
r
√
3
1 2
1 2
1 2
|∇f (1, 1, 1)| = (− ) + (− ) + (− ) =
8
8
8
8
(c) Suppose the bug dies and fall straight down from the point (1, 1, 1),
i.e., moves in the direction of the vector h0, 0, −1i. Determine the
rate of change of temperature.
Solution: The rate of change is given by the directional derivative in the direction u = h0, 0, −1i (Note that this is already a
unit vector, so no normalization is necessary):
1
1 1 1
· h0, 0, −1i =
Du f (1, 1, 1) = − , − , −
8 8 8
8
9. Consider the surface
xyz = cos(x + y + z)
π π
and the point P 0, ,
, which lies on this surface.
4 4
(a) Find an equation of the tangent plane to this surface at the point
P.
Solution: The given surface is of the form F (x, y, z) = 0 with
F (x, y, z) = xyz − cos(x + y + z). Now ∇F (x, y, z) =
hyz + sin(x + y + z), xz + sin(x + y + z), xy + sin(x + y + z)i ,
π π π2
∇F 0, ,
=
+ 1, 1, 1 , and the latter is a normal vector
4 4
16
for the tangent plane. Thus, the vector equation for the tangent
6
MAT397 M005
Calculus III
Spring 2011
D π π E
π2
plane is
+ 1, 1, 1 · hx, y, zi − 0, ,
= 0. Multiplying
16
4 4
π2
π
π
out, we get
+ 1 (x − 0) + 1 y −
+1 z −
= 0, which
16
4
4
2
π
π
+1 x+y+z =
simplifies to
16
2
(b) If z is regarded as a function of x and y defined implicitly by
the above equation (i.e., xyz = cos(x + y + z)), find the partial
∂z
at the point P . (Your answer should not involve
derivative
∂x
any variables!)
Solution: With F (x, y, z) = xyz −cos(x+y +z) = 0, the implicit
differentiation formula gives
∂z
Fx
yz + sin(x + y + z)
=−
=−
.
∂x
Fz
xy + sin(x + y + z)
Substituting x = 0, y = z =
7
π2
π
∂z
, we get
=− −1
4
∂x
16