18.311 — MIT (Spring 2015)
Turn it in before 3:00 PM, in the boxes provided in Room E18-366.
IMPORTANT:
— Turn in the regular and the special problems stapled in two SEPARATE packages.
— Print your name in EACH PAGE of your answers.
1
— In page one of each package print the names of the other members of your group.
1 Regular Problems.
1
1.1
ExID04 statement: Single variable implicit differentiation . . . . . . . . . . . . . . .
1
1.2
ExID13 statement: Two variable implicit differentiation . . . . . . . . . . . . . . . .
1
1.3
ExID41 statement: Differentiation within integrals . . . . . . . . . . . . . . . . . . .
2
1.4
ExID60 statement: Direct Taylor Expansions . . . . . . . . . . . . . . . . . . . . . .
2
1.5
ExID70 statement: Change of variables for an ode . . . . . . . . . . . . . . . . . . .
2
2 Special Problems.
2
2.1
CoLa07 statement: Small vibrations, 2D string under tension . . . . . . . . . . . .
2
In each case compute y
0
= dy dp as a function of y and p , given that y = y ( p ) satisfies:
1.
p
3 y + 2 p + 3 y
3
= 0.
2.
y = cos( p + y
2
).
3. ln(1 + y ) = e p
.
4. cos
2
( y
2
) = e
− p
2
.
5.
y = f (1 + y p ).
6. 1 = y f ( p + y ).
Note: In (5) and (6) f is an arbitrary function.
In each case compute u x
=
∂u
∂x satisfies: and u p
=
∂u
∂p
(as functions of u , x , and p ), given that u = u ( x, p )
1.
x
3 p + 2 x u + p u
3
= 0.
2.
p = f ( p + u x ).
3. ln(1 + p ) = u e x u
Note: In (2) f is an arbitrary function of a single variable, f = f ( ζ ). Assume that f
0
= 0.
.
1 Pages, quite often, become separated from the rest of the package.
1

∂u
In each case compute u x
= and u y
=
∂x satisfies:
1.
2.
u =
3.
y y
=
=
Z u exp y sin( s ) + x s
2
0
Z x ds sin y u ( s
2
, s ) + x s ds .
.
0
Z u cos y sin( s ) + x s
2 ds .
x
∂u
∂y
(as functions of u , x , and y ), given that u = u ( x, y )
For the examples below, calculate the Taylor expansion up to the order indicated — for example:
1 cos( x ) = 1 − or 2 /
√
2 = x
2
+ O ( x
4
√
2 /π or cos(3). On the other hand, do simplify when possible — e.g., tan( π/ 4) = 1
√
2.
). Do not use a calculator to evaluate constants that appear in the expan-
1.
Expand, up to O ( x
4
), f ( x ) = sin( x ) cos(
√ x ).
2.
Expand, up to O ( x
5
), f ( x ) = sin(1 + x ).
3.
Expand, up to O ( x
5
), f ( x ) = sin(1 + x + x
3
).
4.
Let G = G ( x, y ) be some smooth 2 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 .
Consider the second order, nonlinear, ode x
2 cos w d
2 w dx 2
− x
2 sin w dw
!
2 dx
+ x cos w dw dx
+ sin w = 0 for w = w ( x ), where x > 0. Rewrite it in terms of u = u ( y ), where u = sin w and y = ln x .
(1.1)
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 2.1
, #1–3.
2 The partial derivatives of f , to any order, exist and are continuous
2

3.
The string has no bending strength.
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 such that the equation for the curve describing the string is 3
— The tension remains constant throughout — see remark 2.1
, #4.
u y
=
= u u
(
( x, t x, t
),
).
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 u x 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.
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,
4 and it is directed along the direction tangent to the string.
5
#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 .
THE END
3 The x -coordinate of any mass point on the string does not change in time.
4
If you were to cut the string, this would be the force needed to keep the lips of the cut from separating.
5 There are no normal forces, see item 3.
3