Wave Mechanics Examples: Partial Derivatives for Waves
This companion article provides exercises for WM13, wave mechanics: Partial Derivatives
for Waves. All exercises are presented first. Complete worked solutions follow in Part
II.
WM13 introduced the calculus of a one-dimensional wave field
The first partial derivatives answer two different local questions:
and
For transverse string displacement, ut is the transverse velocity of the material point labeled by x,
while
measures spatial curvature and
is local transverse acceleration [1, 2, 3, 4].
For a right-moving translating disturbance
WM13 derived
and
The exercises below reinforce those results without treating this as the physical Newton’s-law
derivation of the string wave equation. That derivation comes later in the series.
How to use this problem set
Attempt all exercises in Part I before consulting Part II. For each derivative, state explicitly which
independent variable is being varied and which is being held fixed. When a derivative has physical
units, include them. In chain-rule problems, identify the intermediate phase or translating
coordinate before differentiating.
Part I: Exercises
Exercise 1: What is being held fixed?
For a field u(x,t), describe in words what each derivative means and state which independent
variable is held fixed.
- ux
- ut
- uxx
- utt
For transverse displacement of a string, give the physical interpretation of each derivative where
appropriate.
Exercise 2: Units of wave derivatives
A transverse string displacement u is measured in meters, position x is measured in meters, and
time t is measured in seconds.
Determine the units of
- ux,
- ut,
- uxx,
- utt,
- c2u
xx if c has units of meters per second.
Explain why the units in part (e) are consistent with the relation utt = c2u
xx.
Exercise 3: Estimate derivatives from finite differences
At time t0, measurements near x0 give
and
At the fixed position x0, measurements at nearby times give
Use forward difference quotients to estimate
- ux(x0,t0),
- ut(x0,t0).
State the sign and physical meaning of each result.
Exercise 4: Read derivative signs from a spatial profile
The following spatial snapshot marks four points on a sinusoidal profile.
Figure. A spatial snapshot used to reason about the signs of ux and uxx.
Without calculating exact numerical values, determine the sign of ux and the sign of uxx at points
P, Q, R, and S.
Use the local slope for ux and the local concavity for uxx.
Exercise 5: Differentiate a sinusoidal traveling wave
Consider
Starting from the phase
derive expressions for
- ux,
- uxx,
- ut,
- utt.
Then show that
and
Exercise 6: Numerical derivatives at one event
A traveling wave is
where x is in meters and t is in seconds.
At the event
find
- u,
- ux,
- ut,
- uxx,
- utt.
Interpret the signs of ux and ut physically.
Exercise 7: Verify the wave-equation relation for a sinusoid
For the wave in Exercise 6,
- determine the propagation speed from c = ω∕k,
- compute c2u
xx symbolically,
- show that c2u
xx = utt,
- explain why this verification is not yet the physical derivation of the string wave
equation.
Exercise 8: Chain rule for an arbitrary right-moving profile
Let
Define
Use the chain rule to derive
- ux,
- ut,
- uxx,
- utt.
Then prove
and
Exercise 9: Left-moving profile and the sign change
Now let
- Derive the relationship between ut and ux.
- Derive the relationship between utt and uxx.
- Explain why the first-order relation changes sign but the second-order wave-equation
relation does not.
Exercise 10: Interpret the right-moving slope-rate relation
The following figure compares the spatial slope of a translating profile with its local time
rate.
Figure. For a right-moving profile F(x−ct), the graphs of ux and ut∕c have opposite signs.
Answer the following.
- If ux > 0 at some event and c > 0, what is the sign of ut?
- If ux < 0, what is the sign of ut?
- Give a geometric explanation using a right-moving pulse.
- What relation would replace this one for a left-moving profile?
Exercise 11: Partial derivative versus total derivative
The figure below shows three directions through the (x,t) domain.
Figure. Partial derivatives move along coordinate directions. The total derivative of
u(x(t),t) follows a path through the (x,t) domain.
Suppose an observer moves according to x = x(t) through a field u(x,t).
- State the multivariable chain-rule formula for du∕dt.
- For a right-moving profile u = F(x−ct), substitute ut = −cux into the total derivative.
- What value of dx∕dt makes du∕dt = 0?
- Interpret that result physically.
Exercise 12: Mixed partial derivatives
Consider
Calculate
- ux,
- ut,
- uxt,
- utx.
Verify that uxt = utx for this smooth function.
Exercise 13: Diagnose conceptual statements
For each statement, decide whether it is correct. If incorrect, rewrite it accurately.
- “ut is the speed at which the wave pattern propagates.”
- “When calculating ux, time must be held fixed.”
- “If uxx = 0 at a point, the displacement must also be zero there.”
- “For a right-moving profile F(x − ct), ut and ux always have the same sign.”
- “Showing that a trial waveform satisfies utt = c2u
xx proves that every physical medium
obeys that wave equation.”
Exercise 14: Synthesis problem
A right-moving sinusoidal wave is
At the event
complete the following.
- Identify A, k, ω, and ϕ.
- Find λ, T, f, and c.
- Find ux and ut at the event.
- Verify numerically that ut = −cux at that event.
- Find uxx and utt at the event.
- Verify numerically that utt = c2u
xx.
- In one sentence, distinguish the local material velocity ut from the propagation speed
c.
Part II: Complete Worked Solutions
Solution 1: What is being held fixed?
- ux is the spatial rate of change of the field. Position varies while time is held fixed. For
string displacement it is the local slope of a spatial snapshot.
- ut is the local time rate of change. Time varies while position is held fixed. For string
displacement it is the transverse velocity of the material point labeled by x.
- uxx is the rate at which spatial slope changes with position while time remains fixed.
In one-dimensional wave problems it measures local curvature.
- utt is the rate at which ut changes with time while position remains fixed. For string
displacement it is local transverse acceleration.
Solution 2: Units of wave derivatives
The field u has units of meters.
-
For displacement versus position, the slope is dimensionless.
-
-
-
- Since
we obtain
This is exactly the unit of utt, so the relation utt = c2u
xx is dimensionally consistent.
Solution 3: Estimate derivatives from finite differences
For the spatial derivative,
| ux(x0,t0) | ≈ | (32)
|
| =  | (33)
|
| = . | (34) |
Converting 0.6 mm = 6.0 × 10−4 m,
The negative sign means the profile slopes downward as x increases at that instant.
For the time derivative,
| ut(x0,t0) | ≈ | (36)
|
| =  | (37)
|
| =  | (38)
|
| = 0.080 m/s . | (39) |
The positive sign means the local string point is moving in the positive transverse direction at that
event.
Solution 4: Read derivative signs from a spatial profile
For a cosine-shaped spatial snapshot:
- At P, the curve is descending, so ux < 0. The profile is concave downward there, so
uxx < 0.
- At Q, the profile crosses zero while descending most steeply, so ux < 0. At the zero
crossing of a pure cosine, the curvature is zero, so uxx = 0.
- At R, the profile is at a trough. The slope is zero, so ux = 0, while the curve is concave
upward, so uxx > 0.
- At S, the profile is rising, so ux > 0. Since the displacement there is positive for a
cosine and uxx = −k2u, the curvature is negative: u
xx < 0.
Solution 5: Differentiate a sinusoidal traveling wave
Let
For the spatial derivative, hold t fixed:
| ux | = (A cos 𝜃) | (41)
|
| = −A sin 𝜃 | (42)
|
| = −Ak sin 𝜃. | (43) |
Differentiate again:
| uxx | = −Ak cos 𝜃 | (44)
|
| = −Ak2 cos 𝜃 | (45)
|
| = −k2u. | (46) |
Thus
For the time derivative, hold x fixed:
| ut | = (A cos 𝜃) | (48)
|
| = −A sin 𝜃 | (49)
|
| = Aω sin 𝜃. | (50) |
Differentiate again:
| utt | = Aω cos 𝜃 | (51)
|
| = −Aω2 cos 𝜃 | (52)
|
| = −ω2u. | (53) |
Therefore
Solution 6: Numerical derivatives at one event
The wave is
At
the phase is
| 𝜃 | = 5(0.20) − 20(0.050) | (57)
|
| = 1.0 − 1.0 | (58)
|
| = 0. | (59) |
Therefore
-
-
so
-
so
-
| uxx | = −k2u | (66)
|
| = −25(0.030) | (67)
|
| = −0.750 m−1 . | (68) |
-
| utt | = −ω2u | (69)
|
| = −400(0.030) | (70)
|
| = −12.0 m/s2 . | (71) |
At this event the string is at a crest of the spatial profile and also at an instantaneous turning
point of the local motion. Thus both slope and transverse velocity are zero, while the curvature
and acceleration are negative.
Solution 7: Verify the wave-equation relation for a sinusoid
The wave has
-
- Since
and
we obtain
- But
Therefore
- This calculation verifies that the proposed sinusoidal traveling wave satisfies the differential
relation. It does not explain why a real stretched string must obey that equation or why its
speed is set by the Tension and linear mass density. That physical derivation requires force
balance and Newton’s second law.
Solution 8: Chain rule for an arbitrary right-moving profile
Let
The needed derivatives of ξ are
Then
| ux | = F′(ξ) | (81)
|
| = F′(ξ), | (82) |
and
| ut | = F′(ξ) | (83)
|
| = −cF′(ξ). | (84) |
Hence
Differentiate again:
and
Therefore
The derivation does not require F to be sinusoidal. It only requires the translating form and
sufficient differentiability.
Solution 9: Left-moving profile and the sign change
Let
Then
Therefore
and
Thus
for a left-moving profile.
The second derivatives are
and
so once again
The first derivative contains one factor of the sign of the propagation term. Differentiating
twice produces the square of that sign, so the second-order relation is the same for both
directions.
Solution 10: Interpret the right-moving slope-rate relation
For a right-moving profile,
with c > 0.
- If ux > 0, then
- If ux < 0, then
- At a fixed position, a right-moving profile brings material from slightly to the left into the
observation point as time advances. If the spatial profile rises with increasing x, the values
just to the left are smaller, so the local field decreases with time. This produces the opposite
sign between ut and ux.
- For a left-moving profile G(x + ct),
Solution 11: Partial derivative versus total derivative
For a moving observer x = x(t), the measured quantity is u(x(t),t). The multivariable chain rule
gives
For a right-moving profile,
Therefore
 | = −cux + ux | (103)
|
| = ux. | (104) |
Thus
when
An observer moving to the right at exactly the wave speed follows a fixed feature of the translating
profile. Along that path, the value of F(x − ct) remains constant.
Solution 12: Mixed partial derivatives
Given
first differentiate with respect to x:
Differentiate with respect to t:
Now
| uxt | = (2xt3 + 4t) | (110)
|
| = 6xt2 + 4 . | (111) |
And
| utx | = (3x2t2 + 4x) | (112)
|
| = 6xt2 + 4 . | (113) |
Therefore
This agrees with equality of mixed partial derivatives for sufficiently smooth functions.
Solution 13: Diagnose conceptual statements
- Incorrect. ut is the local time rate of change at fixed position. For string displacement
it is local transverse material velocity. The propagation speed of the wave pattern is a
separate quantity such as c = ω∕k.
- Correct. In ux, position changes while time is held fixed.
- Incorrect. uxx = 0 means the local curvature measure is zero. The displacement itself
can be nonzero.
- Incorrect. For F(x − ct),
so for c > 0 the two derivatives have opposite signs whenever they are nonzero.
- Incorrect. Showing that one waveform satisfies utt = c2u
xx only verifies compatibility with
that PDE. A physical derivation is required to show why a particular medium obeys that
equation and what determines c.
Solution 14: Synthesis problem
The wave is
- By comparison with A cos(kx − ωt + ϕ),
-
| λ | = = = m ≈ 1.047 m , | (118)
|
| T | = = = s ≈ 0.262 s , | (119)
|
| f | = = = 3.82 Hz , | (120)
|
| c | = = = 4.0 m/s . | (121) |
- At the event x = 0.10 m and t = 0.025 s, the phase is
| 𝜃 | = 6(0.10) − 24(0.025) +  | (122)
|
| = 0.6 − 0.6 +  | (123)
|
| = . | (124) |
Therefore
The spatial derivative is
| ux | = −Ak sin 𝜃 | (126)
|
| = −(0.015)(6) | (127)
|
| ≈−0.0779 . | (128) |
The time derivative is
| ut | = Aω sin 𝜃 | (129)
|
| = (0.015)(24) | (130)
|
| ≈ 0.312 m/s . | (131) |
- Since c = 4.0 m/s,
This equals ut to rounding:
- First find the displacement at the event:
| u | = 0.015 cos  | (134)
|
| = 0.015 | (135)
|
| = 0.0075 m. | (136) |
Then
| uxx | = −k2u | (137)
|
| = −36(0.0075) | (138)
|
| = −0.270 m−1 , | (139) |
and
| utt | = −ω2u | (140)
|
| = −576(0.0075) | (141)
|
| = −4.32 m/s2 . | (142) |
-
| c2u
xx | = (4.0)2(−0.270) | (143)
|
| = 16(−0.270) | (144)
|
| = −4.32 m/s2. | (145) |
Therefore
- The quantity ut is the transverse velocity of the material point at a fixed spatial label, while
c is the speed at which the wave pattern propagates through the medium.
Common mistakes
- Mistake: differentiating both x and t when taking a partial derivative. Change one
independent variable at a time.
- Mistake: confusing ut with wave propagation speed. They describe different motions.
- Mistake: dropping the chain-rule factors k or ω when differentiating a sinusoidal
phase.
- Mistake: forgetting the second minus sign when differentiating twice, leading to an
incorrect sign in uxx or utt.
- Mistake: using ut = −cux for a left-moving profile. The left-moving relation is
ut = +cux.
- Mistake: treating a PDE verification as a physical derivation of the governing wave
equation.
What WM13E1 reinforces
The exercises reinforce the interpretation of partial derivatives as directional rates of change
through the (x,t) domain:
and, for transverse displacement,
For translating profiles,
and
while both satisfy
This completes the calculus preparation needed for the later physical derivation of the
one-dimensional string wave equation.
References
References
[1] Massachusetts Institute of Technology, 18.02SC Multivariable Calculus, Unit 2,
“Partial Derivatives,” MIT OpenCourseWare.
[2] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1,
OpenStax, 2016, Section 16.2, “Mathematics of Waves,” especially partial derivatives,
traveling-wave functions, and the linear wave equation.
[3] A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W.
Norton & Company, 1971.
[4] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill,
1968.
[5] Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves
- The Physics of Waves, Fall 2016, MIT OpenCourseWare, sections introducing
traveling-wave solutions and the one-dimensional wave equation.