18.311 — MIT (Spring 2015)
Rodolfo R. Rosales (MIT, Math. Dept., E17-410, Cambridge, MA 02139).
February 8, 2015.
Answers to Problem Set # 01.
Contents
1 ExID04. Single variable implicit differentiation
1.1 ExID04 statement: Single variable implicit differentiation . . . . . . . . . . . . . . .
1.2 ExID04 answer: Single variable implicit differentiation . . . . . . . . . . . . . . . .
2
2
2
2 ExID13. Two variable implicit differentiation
2.1 ExID13 statement: Two variable implicit differentiation . . . . . . . . . . . . . . . .
2.2 ExID13 answer: Two variable implicit differentiation . . . . . . . . . . . . . . . . .
3
3
3
3 ExID41. Differentiation within integrals
3.1 ExID41 statement: Differentiation within integrals . . . . . . . . . . . . . . . . . . .
3.2 ExID41 answer: Differentiation within integrals . . . . . . . . . . . . . . . . . . . .
4
4
4
4 ExID60. Direct Taylor Expansions
4.1 ExID60 statement: Direct Taylor Expansions . . . . . . . . . . . . . . . . . . . . . .
4.2 ExID60 answer: Direct Taylor Expansions . . . . . . . . . . . . . . . . . . . . . . .
5
5
5
5 ExID70. Change of variables for an ode
5.1 ExID70 statement: Change of variables for an ode . . . . . . . . . . . . . . . . . . .
5.2 ExID70 answer: Change of variables for an ode . . . . . . . . . . . . . . . . . . . .
6
6
6
6 CoLa07. Small vibrations of a 2D string under tension
6.1 CoLa07 statement: Small vibrations, 2D string under tension . . . . . . . . . . . .
6.2 CoLa07 Answer: Small vibrations of a 2D string under tension . . . . . . . . . . .
6.2.1 Conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
8
8
1
18.311 MIT, (Rosales)
1
Single variable implicit differentiation.
ExID04.
2
ExID04. Single variable implicit differentiation
1.1
ExID04 statement: Single variable implicit differentiation
In each case compute y 0 =
dy
as a function of y and p, given that y = y(p) satisfies:
dp
1. p3 y + 2 p + 3 y 3 = 0.
2
4. cos2 (y 2 ) = e−p .
2. y = cos(p + y 2 ).
3. ln(1 + y) = ep .
5. y = f (1 + y p).
6. 1 = y f (p + y).
Note: In (5) and (6) f is an arbitrary function.
1.2
ExID04 answer: Single variable implicit differentiation
1. p3 y + 2 p + 3 y 3 = 0 yields 3 p2 y + 2 + p3 y 0 + 9 y 2 y 0 .
Thus y 0 = −
2. y = cos(p + y 2 ) yields y 0 = −(1 + 2 y y 0 ) sin(p + y 2 ).
3. ln(1 + y) = e
p
Thus y 0 = −
y0
yields
= ep .
1+y
3 p2 + 2
.
p3 + 9 y 2
sin(p + y 2 )
.
1 + 2 y sin(p + y 2 )
Thus y 0 = (1 + y) ep = exp(p + ep ).
2
2
2
4. cos (y ) = e
−p2
0
2
2
−p2
yields 4 y y cos(y ) sin(y ) = 2 p e
p e−p
.
Thus y =
y sin(2 y 2 )
0
.
5. y = f (1 + y p) yields y 0 = (y + p y 0 ) f 0 (1 + y p).
Thus y 0 =
y f 0 (1 + y p)
.
1 − p f 0 (1 + y p)
6. 1 = y f (p + y) yields 0 = y 0 f (p + y) + y (1 + y 0 ) f 0 (p + y).
Thus y 0 = −
y f 0 (p + y)
.
f (p + y) + y f 0 (p + y)
18.311 MIT, (Rosales)
2
Two variable implicit differentiation.
ExID13.
3
Two variable implicit differentiation
2.1
ExID13 statement: Two variable implicit differentiation
In each case compute ux =
∂u
∂u
and up =
(as functions of u, x, and p), given that u = u(x, p)
∂x
∂p
satisfies:
1. x3 p + 2 x u + p u3 = 0.
2. p = f (p + u x).
3. ln(1 + p) = u ex u .
Note: In (2) f is an arbitrary function of a single variable, f = f (ζ). Assume that f 0 6= 0.
2.2
ExID13 answer: Two variable implicit differentiation
1. Upon taking partial derivatives with respect to x and p, x3 p + 2 x u + p u3 = 0 yields:
3 x2 p + 2 u + 2 x ux + 3 p u2 ux = 0
Thus:
ux = −
3 x2 p + 2 u
2 x + 3 p u2
and
and
x3 + 2 x up + u3 + 3 p u2 up = 0.
up = −
x3 + u3
.
2 x + 3 p u2
2. Upon taking partial derivatives with respect to x and p, p = f (p + u x) yields:
0 = (u + x ux ) f 0 (p + u x)
Thus:
ux = −
1
x
and
and
up =
1 = (1 + x up ) f 0 (p + u x).
1 − f 0 (p + u x)
.
x f 0 (p + u x)
3. Upon taking partial derivatives with respect to x and p, ln(1 + p) = u ex u yields:
1
= up e x u + x u u p e x u .
0 = ux ex u + u2 ex u + x u ux ex u and
1+p
u2
e−x u
Thus: ux = −
and up =
.
1 + xu
(1 + p) (1 + x u)
18.311 MIT, (Rosales)
3
Differentiation within integrals.
ExID41.
4
ExID41. Differentiation within integrals
3.1
ExID41 statement: Differentiation within integrals
In each case compute ux =
∂u
∂u
and uy =
(as functions of u, x, and y), given that u = u(x, y)
∂x
∂y
satisfies: Z
u
1. y =
exp y sin(s) + x s2 ds.
0
2. u =
Z x
sin y u(s2 , s) + x s ds.
0
3. y =
Z u
cos y sin(s) + x s2 ds.
x
3.2
ExID41 answer: Differentiation within integrals
In each case take partial derivatives of the expressions satisfied by u, and then solve to obtain
formulas for ux and uy .
1. 1 = exp y sin(u) + x u2 uy +
Z u
exp y sin(s) + x s2 sin(s) ds,
0
1−
Z u
Z u
exp y sin(s) + x s2 s2 ds,
0
2. Clearly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . uy =
Z x
Z u
2
Z x
+
s cos y u(s2 , s) + x s ds
0
sin(s) sin y sin(s) + x s2 ds,
x
0 = cos y sin(u) + x u2 ux − cos y sin(x) + x3 −
Z u
sin(s) sin y sin(s) + x s2 ds
x
so that . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . uy =
.
exp (y sin(u) + x u2 )
u(s2 , s) cos y u(s2 , s) + x s ds,
2
1+
0
and . . . . . . . . . . . . . . . . . . . . . ux = sin y u(x , x) + x
exp y sin(s) + x s2 s2 ds
0
so that . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ux = −
.
exp (y sin(u) + x u2 )
Z u
3. 1 = cos y sin(u) + x u2 uy −
0
so that . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . uy =
0 = exp y sin(u) + x u2 ux +
exp y sin(s) + x s2 sin(s) ds
cos (y sin(u) + x u2 )
Z u
x
sin y sin(s) + x s2 s2 ds,
.
18.311 MIT, (Rosales)
Direct Taylor Expansions.
cos y sin(x) + x3 +
so that . . . . . . . . . . . . . . . . . . . ux =
4
Z u
ExID60.
5
sin y sin(s) + x s2 s2 ds
x
cos (y sin(u) + x u2 )
.
ExID60. Direct Taylor Expansions
4.1
ExID60 statement: Direct Taylor Expansions
For the examples below, calculate the Taylor expansion up to the order indicated — for example:
1
cos(x) = 1 − x2 + O(x4 ). Do not use a calculator to evaluate constants that appear in the expan2√
sions √
— e.g.,
√ 2/π or cos(3). On the other hand, do simplify when possible — e.g., tan(π/4) = 1
or 2/ 2 = 2.
√
1. Expand, up to O(x4 ), f (x) = sin(x) cos( x).
2. Expand, up to O(x5 ), f (x) = sin(1 + x).
3. Expand, up to O(x5 ), f (x) = sin(1 + x + x3 ).
4. Let G = G(x, y) be some smooth1 function of two variables. For z ≥ 0, expand up to O(z 3 ),
f (z) = G(z, z 1.5 ). Express the expansion coefficients in terms of partial derivatives of G.
4.2
ExID60 answer: Direct Taylor Expansions
We have
1
1 2
1
1
1− x+
x + O(x3 ) = x − x2 − x3 + O(x4 ).
2
24
2
8
1
1
1
2. f (x) = sin(1) + cos(1) x − sin(1) x2 − cos(1) x3 +
sin(1) x4 + O(x5 ).
2
6
24
3. Using the answer to part 2, we get
f (x) = sin(1) + cos(1) (x + x3 ) − 12 sin(1) (x2 + 2 x4 + O(x6 ))−
1
1
cos(1) (x3 + O(x5 )) + 24
sin(1) (x4 + O(x6 )) + O(x5 )
6
23
= sin(1) + cos(1) x − 21 sin(1) x2 + 76 cos(1) x3 − 24
sin(1) x4 + O(x5 ).
1
1. f (x) = x − x3 + O(x5 )
6
4. We have G(x, y) = G0 + G0x x + G0y y + 21 G0x x x2 + G0x y x y + 12 G0y y y 2 + O ((x2 + y 2 )1.5 ),
where: The superscript 0 denotes evaluation at (0, 0) — e.g., G0 = G(0, 0), etc.
The subscripts denote partial derivatives — e.g., Gx = ∂G
, etc.
∂x
1
Thus f (z) = G0 + G0x z + G0y z 1.5 + G0x x z 2 + G0x y z 2.5 + O(z 3 ).
2
1
The partial derivatives of f , to any order, exist and are continuous
18.311 MIT, (Rosales)
5
Change of variables for an ode.
ExID70.
6
Change of variables for an ode
5.1
ExID70 statement: Change of variables for an ode
Consider the second order, nonlinear, ode
d2 w
x cos w 2 − x2 sin w
dx
2
dw
dx
!2
+ x cos w
dw
+ sin w = 0
dx
(5.1)
for w = w(x), where x > 0. Rewrite it in terms of u = u(y), where u = sin w and y = ln x.
5.2
ExID70 answer: Change of variables for an ode
d
From the relationship between y and x, and the chain rule, it follows that
dy
=x
d
dx
.
Hence
du
dw
= x cos w
,
dy
dx
(5.2)
dw
d2 u
= x cos w
− x2 sin w
2
dy
dx
dw
dx
!2
+ x2 cos w
d2 w
,
dx2
(5.3)
d2 u
d
dw
where the second equation follows from expanding 2 = x
x cos w
. Equation (5.2) could
dy
dx
dx
2
2
be used to write dw
in terms of du
. Then (5.3) would yield ddxw2 in terms of du
and ddyu2 . Finally,
dx
dy
dy
these expressions, substituted into (5.1), would produce the final answer. However, here it is easy
to see directly from (5.3) that the transformed equation is
!
d2 u
+ u = 0.
dy 2
6
6.1
(5.4)
CoLa07. Small vibrations of a 2D string under tension
CoLa07 statement: Small vibrations of a 2D string under tension
Here you are asked to derive the equation for the transversal vibrations of a thin elastic string under
tension, under the following hypotheses
1. The string is homogeneous, with a mass density (mass per unit length) ρ — ρ is a constant.
2. The motion is restricted to the x-y plane. At equilibrium the string is described by y = 0 and
0 ≤ x ≤ L = string length. The tension T > 0 is then constant. See remark 6.1, #1–3.
3. The string has no bending strength.
18.311 MIT, (Rosales)
Small vibrations of a string under tension in 2-D.
CoLa07.
7
4. The amplitude of the vibrations is very small compared with their wavelength, and any longitudinal motion can be neglected. Thus:
— The string can be described in terms of a deformation function
u = u(x, t),
2
such that the equation for the curve describing the string is
y = u(x, t).
— The tension remains constant throughout — see remark 6.1, #4.
Use the conservation of the transversal momentum to derive an equation for u. Thus:
i) Obtain a formula for the transversal momentum density along the string (momentum per unit length).
ii) Obtain a formula for the transversal momentum flux along the string (momentum per unit time).
iii) Use the differential form of the conservation of transversal momentum to write a pde for u.
Hint: when doing (ii) remember that forces are momentum flux! Since there is no longitudinal motion,
momentum can flow only via forces. Remember also that ux is very small. Be careful with the sign here!
The pde for u derived above applies for 0 < x < L. For a solution to this equation to be determined,
extra conditions on the solution u at each end are needed. These are called boundary conditions,
and must be derived/modeled as well. Later on we will see that exactly one boundary condition (a
restriction on u and its derivatives) is needed at each end, x = 0 and x = L.
Answer the following two extra questions:
iv) What boundary condition applies at an end where the string is tied? (e.g., as in a guitar).
v) What boundary condition applies at an end where the string is free?
For example, imagine a no-mass cart that can slide up and down (without friction) a vertical rod at
x = 0, and tie the left end of the string to this cart. This keeps the left end of the string at x = 0,
with the rod canceling the horizontal component of the tension force, but u(0, t) can change.
A no-mass-no-friction cart is an idealization, approximating a setup where the mass and friction are small.
Remark 6.1 Some details:
#1 We idealize the string as a curve. This is justified as long as λ d, where λ = scale over which
motion occurs, and d = string diameter. This condition justifies item 3 as well.
#2 The tension is generated by the elastic forces (assume that the string is stretched). At a point along
the string, the tension is the force with which each side (to the right or left of the point) pulls on the
other side,3 and it is directed along the direction tangent to the string.4
#3 At equilibrium the tension must be constant. For imagine that the tension is different at two points
a < b along the string. Then the segment a ≤ x ≤ b would receive a net horizontal force (the
difference in the tension values), and thus could not be at equilibrium.
#4 The hypothesis imply that any changes in the string length can be neglected. Thus the amount of
stretching, which generates the tension, does not change significantly anywhere. Hence the tension
remains, essentially, equal to the equilibrium tension T .
2
The x-coordinate of any mass point on the string does not change in time.
If you were to cut the string, this would be the force needed to keep the lips of the cut from separating.
4
There are no normal forces, see item 3.
3
18.311 MIT, (Rosales)
6.2
Small vibrations of a string under tension in 2-D.
CoLa07.
8
CoLa07 Answer: Small vibrations of a 2D string under tension
We have
i) The transversal momentum per unit length along the string is given by
ρmom = ρ ut .
ii) Momentum flux along the string happens because of transversal forces.
That is: the force in the y-direction that, at every point, one side of the string applies on the
other side. This is just the y-component of the tension, which is given by T sin θ, where θ is
the angle that the string tangent makes with the x-axis. However, since
ux is small, θ ≈ ux , so and the transversal momentum flux is given by
qmom = −T ux .
Note about the sign here: when ux < 0, the left side of the string pulls the
right side up, thus generating a positive momentum flux. Vice-versa, when ux > 0, momentum flows
from right to left (negative). Thus the sign is as above.
Note also that the x-component of the tension (horizontal force along the string) is T cos θ = T , since
ux is small. Thus it is constant, consistent with the approximation of no motion in the x direction.
iii) From conservation (ρmom )t + (qmom )x = 0. That is
Alternatively
ρ utt − T uxx = 0.
(6.1)
utt − c2 uxx = 0,
(6.2)
where c2 = T /ρ is a speed (the wave speed).
iv) Tied end. Suppose the left end of the string is tied. Then its
position is prescribed there (say: y = a at x = 0), which leads to
where a is a constant. More generally, we can have u(0, t) = σ(t).
v) Free end. Suppose the string’s left end is free, so there is no force
there to balance any transversal component of the tension. Hence
u(0, t) = a,
ux (0, t) = 0.
In the problem statement a no-mass, no-friction, car on a rod is mentioned as a way to achieve this.
But the car has some mass, say M , and there is friction (say, a force proportional to the cart velocity,
opposing its motion). Then, since the cart position is given by y = u(0, t), we can write
M utt (0, t) = −ν ut (0, t) + T ux (0, t).
(6.3)
If τ is a typical time scale for the vibrations of the string, and λ is a typical wavelength, then
ux (0, t) = 0 is a good approximation to this provided that Tλ M
and Tλ ντ . That is, provided
τ2
that the two a-dimensional numbers
νλ
Mλ
and
2
Tτ
Tτ
are small.
6.2.1
Conservation of energy
In the derivation above we obtained an equation for the string using, solely, the conservation of the
transversal momentum. What happens with the other quantities that are conserved? They should
be conserved. Let us check that this is so.
Conservation of mass. This is guaranteed by the parameterization we use. That is, each point
(x, u) follows a specific bit of the spring, with mass ρ dx. Thus mass is automatically conserved.
This is typical of solid mechanics problems, where the parameterization tracks mass points in the
object. In fluids, where individual particles are not tracked, and instead the flow velocity at each
18.311 MIT, (Rosales)
Small vibrations of a string under tension in 2-D.
CoLa07.
9
point in space is prescribed, mass conservation is not automatic, and must be enforced. On the
other hand, in fluids the equations involve only first derivatives in time of the densities — while
solids second order equations occur. This is because the velocities are variables in fluid problems,
while in solids they result from the time derivatives of the displacements.
Conservation of longitudinal momentum. Since there is no motion in the x-direction, and the
horizontal component of the tension forces is constant (see the end of item (ii) of the answer), the
longitudinal momentum is conserved as well (trivially so). In fact, the statement that the derivation
above does not use the conservation of the longitudinal momentum is incorrect! This is used,
effectively, when concluding that the tension on the string is constant.
Conservation of energy. To show that energy is conserved, we first need to calculate its density and
its flux. The energy density has two components: kinetic energy per unit length = 12 ρ u2t , and the
potential energy stored in the stretching of the string, which we need to compute. In fact we only
need to compute the difference between the potential energy at any time, and some constant energy
— specifically: the potential energy of the string at equilibrium. But this is easy, for then the tension
is constant through the whole stretching process, so that the energy is q
the product of T times the
change in length. This yields potential energy per unit length = T
1 + u2x − 1 = 21 T u2x ,
q
since the arc-length is ds = 1 + u2x dx, and ux is small. Finally, the flux of energy is given by
the work (per unit time) done by the transversal force (which we already calculated in (ii) of the
answer). That is: energy flux = −T ux ut . It follows that the equation for the conservation of
energy is
1 2 1
ρ ut + T u2x − (T ux ut )x = 0.
(6.4)
2
2
t
It is easy to see that (6.1) guarantees this.
This calculation does not involve an internal energy, because in this model there is no mechanism for energy
exchanges between mechanical and internal. This is the reason why energy conservation is “automatic”.
THE END.