Course Announcements

« PHYS290F Fall 2010

Final Examination

Sat 11Dec2010 9:56PM

Our Final Examination will be on:

Wed 12/15, 3:20p - 5:20p, in HORT 117

Same deal with the previous exams; no crib sheets, no electronics.

Covers all topics covered after the 2nd exam plus one problem

that involves contour integrals.

See you there. Good luck, all!

Some help for Homework 11

Mon 06Dec2010 10:45AM

Problem 1:

As you learned in PHYS 272 (or equivalent), the electrostatic potential is V(r) is related to the (static) electric field E by E=-∇V . Thus, if V(r) = f(x) (i.e., a function only of x and independent of y or z), then E=-∇V will have only the x-component. Its sign and the trend in its magnitude can be read off from the graph given for f(x). Similarly, we derived the local (differential) version of the Gauss' Law in class. According to this, the charge density ρ(r) = ∇•E/ε₀. Just apply this to the E above and read off whatever you can on the charge density from the graph of f(x). In particular, estimate the sign and trend of f'(x) and f''(x) to find out how E and ρ behave as a function of x.

Problem 2:

(i) Use the parametrization: x=t, y=t².

(ii) Use the parametrization: x=cos(t), y=sin(t), where t goes from 0 to 2π to do the actual line integral of F. As directed, also calculate curl F and its surface integral over the unit disk.

[Note] The integrand F is not conservative as you can show it cannot be expressed as a gradient of another function (by e.g. noting ∇×F ≠ 0).

Problem 3:

First, do the surface integral directly. Here, note that you can divide the 6 faces of the unit cube into 3 parallel pairs, and that for each pair the surface normal unit vector n points in the opposite directions. That should tell you what the sum of the integrals for each pair is.

Second, calculate the divergence of V. Use the divergence theorem and equate the volume integral of this divergence with the required surface integral. The result should agree with that of the first part, of course.

Problem 4:

Make sure to draw a picture. The force F you use must be one that is just enough to keep the pendulum in equilibrium (and not accelerate it). However, this still leaves F undetermined (think about why). The trick is to write F = F_θ + F_r where F_θ is tangential to the arc traversed by the pendulum bob while F_r is in the radial direction. Then, only F_θ does work because the displacement of the pendulum is purely in the tangential direction. The line integral of F_θ•dr does not involve the tension at all and can be done without knowing it.

Problem 5:

To see if a vector field is conservative or not is equivalent to see if it is a gradient of some scalar function. So you can try to see if it is possible to construct a scalar function whose gradient is the given vector field. Another, simpler way is to calculate the curl of the given vector field because this is in turn equivalent to see if its curl is zero. [This is due to the Stokes Theorem.]

Problem 6:

There is a misprint in this problem. The radius shown as 2 in some editions of this text should be 1. Also, for specificity, let's traverse the semicircle above the x-axis counterclockwise.

Problem 7:

Essentially the same as Problem 3 except that, here, the integral results in a non-zero value.

Some help for Homework 10

Mon 22Nov2010 10:46AM

Problem 1:

(a) and (b) In both these parts, you can work either with the Cartesian coordinates or polar coordinates as your choice. If you are to obtain answers in the Cartesian coordinates of v(t), a(t), and L, these values will be expressed in terms of the time derivatives of r and θ as well as trigonometric functions of θ. Use known trigonometric identities whereever possible to simplify. If you choose to work with the polar coordinates, then you should make use of (7.2.10) and (7.2.11); i.e., the unit vectors e_r and e_θ at the location of the particle are NOT constant in time since the particle is moving. Then in (b), use the fact that e_r×e_θ=e_z. That is you would be using the Cylindrical coordinates in effect.

(c) Since the unit vectors in polar coordinates (in xy-plane) can be expressed in terms of their (x,y) coordinates as e_r = ( cosθ, sinθ ) and e_θ = ( -sinθ, cosθ ), if you answered (a) and (b) with Cartesian coordinates, the acceleration in part (a) should first be converted to a simple answer in the polar coordinates. Central force means that the acceleration is purely in the radial direction, i.e., the coefficient of e_θ is zero. Consider what that condition translates to in terms of the derivative dL/dt using L you calculated in (b).

Problem 2:

Here, I am asking you to do the problem without invoking the Stokes Theorem (which I have not yet covered in class). We will see soon that that theorem can be used to solve this problem simply as well but that comes later.

You can do this problem by brute force using the polar coordinates, e.g., but it is also possbile to considerably reduce the complications if you follow the following approach:

First, write the integral in the canonical form of a line integral, that is, as the integral of a vector field f(x,y) in the form of ∫_Γf•dr. Now, write the vector field f as f = a (x,y) + b (-y,x). Note that the vector (x,y) is proportional to e_r and (-y,x) is proportional to e_θ. Since the path Γ is totally in e_θ direction, only b contributes to the integral. Solve for a and b, and you only need to use b. Then, the integral will be reduced something of the form ∫₀^2π b(-y,x)•e_θ r dθ. At this point, you can use x = r cosθ, y = r sinθ (r=1 for the unit circle), and do the integral over θ (r=1 on the unit circle).

Problem 3:

Pick any path from O to A=(1,1). For example, you can pick a parabolic path through y=x². This can be parametrized by x=t, y=t², where t goes from 0 to 1. Then, note that dr=(dx,dy)=(dt,2t dt)=(1, 2t) dt and carry on the integral over t. Actually, this particular integral is independent of path - try another path and see for yourself. Can you see why? (But don't use this fact to do this problem; actually do the integration over a chosen path instead.)

Problem 4:

Since the sphere is described by x²+y²+z² = R², you can write the height z as h(x,y). Then just take the gradient of h(x,y). Remember that the gradient of h is in the direction where h(x,y) increases most rapidly.

Problem 5:

Just look at each component of ∇(φχ) separately.

Problem 6:

∇T is the direction of most rapid increase in T, while -∇T is that of most rapid decrease in T. If a unit vector in the direction of -∇T = e, then a step of size a in that direction is a vector ae. You can plug this into the Taylor series expansion (cf. p.52 ) and keep only up to the 1st order term.

Resubmitted Exam II has been graded and available for pick-up

Sat 20Nov2010 10:04PM

Resubmitted Exam II has been graded and the average score, after taking the mean with the original score, is 33.5 with the standard diviation of 11.5. 41 students took advantage of the resubmission opportunity while 4 did not. The resulting average is virtually identical to the similarly rescored average of 33.6 from Exam I.

The original and resubmitted Exam II are available for your pick-up in PHYS, Room 144.

Another typo in Homework 9! Sorry...

Fri 19Nov2010 10:32AM

In Homework 9, Problem 4(b), on the right-hand side of Problem 4b, the coefficient "2/|a|" should have been "1/(2|a|)". Another careless error on my part. I am sorry about that...

With the two typos (both related to a careless factor of 2!), I think I need to extend the deadline to Monday. So, if you wish to turn it in by 12:30 pm on Monday Nov. 22 (without penalty) instead of today in class, that is fine. If you do that, please either put it in my department mail box or under my door (rm.264).

Thanksgiving Week and the Final Examination

Thu 18Nov2010 9:47AM

1. As we announced previously, this course will not have a lecture on Monday, November 22 (in lieu of the evening exam we had earlier).

2. Final Exam will be given on Wed., Dec. 15, 3:20 - 5:20 pm in HORT 117. The Horticulture building is to the south of W. State St. on the Agricultural Mall.

The Final Exam will NOT be cumulative, but will cover only the materials covered after the Exam II plus one problem that involves a contour integral in the complex plane. It will be in the same format as in the first two exams, and carries the same weight as each of the first two exams toward the final grade. Closed book, no electronics, etc. the same way as for those.

Homework 9 Correction

Thu 18Nov2010 9:25AM

In Homework 9, Problem 3, you are asked to show a contour integral to be equal to "g(2i/n)". This should have been "g(i/n)", i.e., no factor of 2 in the argument of the function g. I apologize for this typo.

Some help for Homework 9

Mon 15Nov2010 10:01AM

Problem 1:

First write the integrand on the right-hand side |f^~(ω)|² as the product f^~(ω) f^~(ω)* and then substitute (1) into those. Then, rearrange the integration on the right-hand side of (2) to pull out an integral that is a form of the Dirac delta function. Recall that the integral of exp(±ikx) over x (or k) is 2π δ(k) (or 2π δ(x)) as shown in class.

Problem 2:

(b) Use integration by parts. The Heaviside step function is often used to write a function that is truncated at certain value.

Problems 3 and 4:

In both problems, you show something is a delta function by showing that (1) it is zero unless its argument is zero, and (2) the integral of its product with a function g(x) over all x is equal to g(0) just as in Problem 2. For Problem 3, this will require you to use the contour integral in question. In Problem 4(b), you would need to show that the integral of δ(x²-a²) g(x) is equal to (2/|a|)(g(-a)+g(a)). Since the argument of δ(x²-a²) is zero at two values of x, you will need to split the integral into two parts (around the two zeros).

Problem 5:

(a) Note that e_r = (cosθ,sinθ), e_θ = (-sinθ,cosθ) in Cartesian coordinates, and invert the relations.

(b) You can use the fact that e_r = (sinθcosφ,sinθsinφ,cosθ), e_θ = (cosθcosφ,cosθsinφ,-sinθ), e_φ = (-sinφ,cosφ,0), in Cartesian system, and invert the relations.

Problem 6:

Note that |axb| = |a||b||sinθ| is twice the area of the triangle formed by the three sides a, b, and a-b.

Exam II results

Mon 08Nov2010 1:53PM

As discussed in class today, the Exam II average was about the same as the Exam I (before reworking) at about 26/50. I am offering the same bonus opportunity to you for raising your scores by reworking problems. The details have already been communicated to you in class and in private Emails to those who were not present in class today. There are requirements and conditions. So please make sure to observe them.

Some help for Homework 8

Fri 29Oct2010 10:03AM

Problem 1:

(a) Find z² first, and then take the 1/2-power of them.

(b) We always have z^a = e^{a ln z}. So write our number as e^{i ln (3-4i)} first. The logarithm in the exponent can be written in the Cartesian form (x+iy) by first expressing (3-4i) in the polar form and then taking its log.

Problem 2:

(a) I had forgotten to specify the direction in which the contour is traversed in the integration; please take it to be counterclockwise. To remind you, in order to find the residue at, say, z₁, factor out 1/(z-z₁) from the integrand and substitute z₁ into z in the remaining parts of the integrand. (For example, the residue of f(z)=1/(z²+1) at z=i is 1/(2i).) Once you know the poles within the contour and the residues at those simple poles, you have the integrals.

(b)(c) For non-closed contours, you need to explicitly evaluate the integral by parametrizing the contour and performing the integral. If the integrand is analytic throughout the contour and you know the indefinite integral, then you can do this more easily, of course.

Problem 3:

I added a slide on the integration discussed in Friday class to the file "Topics for Chapter 6 #2" under Topics. It is the last page of that file. If you understand what it says (which is exactly the same as what I did in Friday class), then it should be easy to modify it by changing the numerator of the integrand of the complex contour integrand to e^-iz and the contour to a rectangle in the LHP (lower half plane) as this problem askes for.

Problem 5 (6.2.9):

We discussed this problem in class. z^N = 1 means z=1^1/N and you can express the N distinct values of the right-hand side by first writing 1 = e^i(2nπ). Generally, these roots can be written as ωⁿ where n=0,1,2,...,N-1 where ω= e^i(2π/N). So the sum of the N distinct roots is the sum of a geometric sequence.

Problem 6 (6.3.1):

Show that the Taylor series of ln(1+z) about z=0, differentiated term by term, becomes the Taylor series expansion of 1/(1+z), and the same with the series for sin(z) which becomes the one for cos(z) upon term-by-term differentiation. Show that the radii of convergence of the two series are the same. In fact, you can show in general that the radii of convergence of a Taylor series and the series obtained by differentiating it term-by-term are identical.

Problem 7 (6.3.3):

You need to do these integrals explicitly parametrizing the paths as directed (no "cheating" using your knowledge of the open contour integrals of analytic functions). Integrate f(z)=z² from the origin to 1+i by the two different paths and show that the results are the same. The first path is composed of two parts, 0 to i on the imaginary axis and then from i to 1+i parallel to the real axis. The second path is a straight line from the origin to 1+i (at 45 degrees to the real axis).

Problem 8 (6.3.5):

The last part of this problem which tells you to get the result "treating z as if it were x" means that you are to evaluate the contour integral (of these open contours) as the defference between the indefinite integral evaluated at the final point (1+i) and the same evaluated at the intial point (origin).

Exam II coming up next week

Mon 25Oct2010 4:37PM

Evening Exam II will be given on Tuesday, Nov. 2, 8 - 10 pm in Physics Building, Rm. 112. This is again a closed book exam with no crib sheets or equation sheets; the format will be the same as in Exam I. Only pens, pencils, and erasers are allowed. No calculators of any kind, and no electronic equipment are allowed (no cell phones, etc.) please.

Exam II coming up next week

Mon 25Oct2010 4:37PM

Some help for Homework 7

Wed 20Oct2010 2:19PM

Problem 1:

You just need to calculate the various partial derivatives of the real and imaginary parts of the given functions and find out whether CRE are satisfied or not. If they are, then you know the function should be expressible as f(z). In that case, you could take a guess at what this form should be and check to make sure that your guess is indeed correct. (It shouldn't be hard to make a good guess.) Or, note that when you set y = Im{z} = 0, we get f(z)=f(x).

Problem 2:

(a) This is a problem that shows a geometric implication of analyticity of a function of a complex variable. In (a), consider an equation u(x,y) = C (and v(x,y) = D) where C (and D) are some chosen constant(s). Say, u(x,y) = 1. This equation then relates x and y, or defines y=y(x) implicitly. So, if you now rewrite the equation as u(x,y(x)) = 1, then you can take the (total) derivative of u with respect to x. (This corresponds to differentiating u with respect to x when y is constrained to vary along the curve u(x,y) = 1 if C is chosen to be 1 as in the example case I am talking about now.) This results in u_x + u_yy' = 0. You can then solve this to obtain the slope y' = dy/dx of the curve y=y(x). You can do the same for v(x,y).

(b) Two lines are perpendicular to each other if the product of their slopes is -1. E.g, a line with a slope of 2 (such as y=2x) and one with a slope of -1/2 (such as y=-(1/2)x) intersect at 90 degrees. Use the CRE for analyticity and the results from (a) to show that this condition is satisfied.

(c) For f(z) = z², u(x,y) = x²-y², v(x,y)=2xy. So the curves u=constant and v=constant are hyperbolas (of different axis orientations).

Prolem 3:

In this problem, you are asked to show where each singularity is and what type of singularity it is (i.e., a pole (of what order?), essential singularity, or branch point). If it is a simple pole (pole of first order), then also calculate the residue there. Simple poles are singularities of the form C/(z-z₀) where C is either a constant or something that approach a non-zero, finite constant as z approaches z₀. In this case that limit of C is the residue.

(a) Factor the denominator into the product of 4 terms of the form (z-z_i), where z_i is the i-th root of z⁴=-1.

(b) tan(z) = sin(z)/cos(z). Since neither sin(z) nor cos(z) has singularities, the only singularities of tan(z) are where the denominator cos(z) = 0. For what values of z is cos(z)=0? This is one of the questions in Problem 8 (6.2.4). So you may do that problem first and use the result here. In order to obtain the residues at each simple pole, you can replace cos(z) near a pole at, say, π/2, by the first non-vanishing term of its Taylor series expansion about z=π/2.

(c) There are some poles and also one other type of singularity that comes from the numerator. Because of this last singularity, the residue from a simple pole is triply-valued.

Problem 4:

(a) You can always use z^a = e^{a ln z}. When you write z in the polar form, allowing adding any integer multiple of 2π to the polar angle θ, you will get all possible values of z^a. Sketching is easy when using the polar form. Convert to Cartesian form later.

(b) Again, start in polar form and write 1 = e^{2nπ i} and raise this expression to the 1/4 power. There are 4 distinct roots (which can be written as 1, ω, ω², ω³), and you can easily plot them on the complex plane. Then you can convert them to the Cartesian form.

Problem 5 (6.1.1):

The problem has 3 parts. First, (a) investigate the limits when you approach the origin along x and y axes. Second, (b) write x = r cos θ, y = r sin θ, and take the limit of r → 0 at fixed θ. Lastly, (c) do the same for approaching any point other than the origin.

Problem 6 (6.1.8):

Work this problem in 3 parts. First, (a) apply the known fact that, for any f(r,θ), df = (∂f/∂r)dr + (∂f/∂θ)dθ to the particular function f(z) = z = r e^iθ. Second, (b) "interpret" the two factors of dz, that is, (dr + ir dθ) and e^iθ. Use geometry to show that the first factor is the variation dz when r is changed by dr and θ =0 is changed to dθ. Interpret the second factor in terms of rotation. Next, (c) apply the same approach as when we derived the Cauchy-Riemann Equations in cartesian form to the polar form. That is, calculate (∂f/∂r) and (∂f/∂θ) in terms of (df/dz). Then relate the two partial derivatives to each other using the fact that (df/dz) is common in the two equations. Finally, write the relationship between the two partial derivatives in terms of the real part u(r,θ) and imaginary part v(r,θ) of f(r,θ).

Problem 7 (6.1.12):

This is a very important problem. Analytic functions of a complex variable are so special that knowing just its real (or imaginary) part allows you to reconstruct the whole function (up to a constant). For example, if u(x,y) is given, then CRE tells you what v_x and v_y. Each of these can be integrated to get (parts of) v(x,y). For example, integrating v_x with respect to x (considering y to be a constant) leads to v(x,y) up to an integratino 'constant', which in this case is an unknown function of y (because y is held fixed like a constant). Analogous procedure on v_y will also lead to v(x,y) but up to an unknown function of x. Compare the two expressions and use the fact that they have to be the same because they are both expressions of the same function v(x,y).

Problem 8 (6.2.4):

For sin z and cos z, one way to do this is to first derive the results shown in (6.2.5) and use them, but that's like doing two problems for the price of one. So, my suggestion is to follow the suggestion given in the problem and use the exponential formulations of these functions. That is, e.g., sin z = (e^iz - e^-iz)/(2i). Substitute z=x+iy, and rewrite sin z in Cartesian form (u+iv). For sin z = 0, then both the real part u and imaginary part v must be zero. This tells you that the zeros must satisfy sin x = 0 and also sinh y = 0 . Then you can write down all zeroes of sin z and sketch them on the complex plane. Similarly for the rest of the functions. You should be able to see some relations emerge between sin, cos, sinh, and cosh.

Some help for Homework 6

Mon 11Oct2010 11:46AM

(This will be added to later.)

Problem 7 (5.3.5):

Note that for a geometric sequence of n terms:
S = a₁ + a₂ + ... + a_n
where a_n = a₁r^n-1, the sum S_n is given by
S_n = a₁ ( 1 - rⁿ)/(1 - r). This is true independent of whether any of the numbers a₁ and r are real or complex. Make sure that you have the correct r, a₁, and n.

When you calculate something like |(1-e^2iθ)|², it may be useful to do it by proceeding as:

|(1-e^2iθ)|² = (1-e^2iθ)(1-e^-2iθ) = 2 ( 1 - Re {e^2iθ} )

rather than immediately applying Euler theorem and writing 1-e^2iθ = (1 - cos 2θ) - i sin 2θ and getting into lots of sines and cosines immediately.

Also, when you run into sums/differences of trigonometric functions like cos (2n-1)θ - cos (2n+1)θ , you will need to convert them into the products like 2 sin 2nθ sin θ

Problem 8 (5.3.6):

We actually already did the second part of this problem in class. For the first part, the idea is to write the de Moivre's theorem for n=4, and then expanding the 4th power product into the form of x+iy (where x, y are real). After that, just compare the real and imaginary parts of this (x, y) with the corresponding part of the RHS (i.e., cos 4θ and sin 4θ).

Then, calculate the real and imaginary parts of the sum from S_n and equate the corresponding part of the sequence (i.e., cos θ + cos 3θ + ... + cos (2n-1)θ for the real part and sin θ + sin 3θ + ... + sin (2n-1)θ for the imaginary part). You will have to use the trigonometric identities for addition and subtraction of sines and cosines to simplify the results.

Problem 10 (5.4.6):

This problem is for the parallel LRC circuit whereas we did the series LRC circuit in class. What is different for a parallel circuit as opposed to the series circuit is that the potential differences between the terminals of each element (L, R, and C) are the same (common) while, for the series circuit, the current through each element is the same (common). Otherwise, the relationships between the current and voltage for L, R, and C (individual components) are common to both parallel and series circuits.

For example, V(t) = L (dI_L(t)/dt). Substituting V(t) = V₀ e^iωt and I_L(t) = I₁ e^iωt, we get I₁ = V₀/(iωL). Do the same for I₂ and I₃. This allows you to write the total current I₀ = I₁ + I₂ + I₃ in terms of V₀, ω, L, and R. This in turn makes it possible to calculate Z ≡ V₀/I₀. Once this is done, you can calculate its modulus r and polar angle (phase) φ in the same way as we did in class for the series LRC circuit.

Class cancelled for Fri., Oct. 8

Thu 07Oct2010 11:36AM

Our class for Friday, Oct. 8 is cancelled in lieu of the evening exam we just had. Another class on Monday, Nov. 22 will also be cancelled in lieu of the second evening exam to come on Tuesday, Nov. 2.

Exam I results

Wed 06Oct2010 5:04PM

Exam I results are out. The average was 26.9/50, the highest score was 49.5 and the lowest was 4. This average is lower than I would like, but more than that, I am very concerned about those scores clustered at the low end. So as I have outlined in class today, I am offering an optional opportunity to you to recover up to half of the points lost in the exam. If you weren't in class today and did not receive your scored exam back, then you would need to come to see me and talk to me to get your exam and to hear about the bonus opportunity. Good luck.

Evening Exam I coming up on Sept. 28

Wed 22Sep2010 3:29PM

Exam I will be held in PHYS Room 203 on Tuesday, Sept. 28, from 8 - 10 pm. This is a closed book exam and you are allowed to use only pens, pencils, and erasers during the exam - no calculators of any kind, please. There will be no crib sheets or formula sheets either.

I have posted a couple of past exams under "Exam Archive" on this site. The Monday class on Sept. 27 will be a review session, which will be an excellent opportunity for you to ask questions on past homework, the posted past exams, or any class material.

Some help for Homework 5

Mon 20Sep2010 2:05PM

Problems 1 and 2: Since the ellipse and ellipsoid are so close to circle and sphere, the first impulse should be transforming the domains to those more symmetric ones. You can do this by introducing variable changes such as x'=x/a, y' = y/b, and z' = z/c. Once that is done, you should then be able to change (x',y') or (x',y',z') system into the polar or spherical coordinate system with no difficulty. Just don't forget that there are two stages of Jacobians (i.e., the factors that need to be included to accompany the changes of variables) to take into account here. First, the introductinon of the primed variables yields a set of Jacobian(s). Then, changing that Cartesian system into polar or spherical system requires another set.

Problem 3: In this problem, you integrate f(x,y) = x²|y| instead of 1 over the disk because you are calculating the total charge on the disk rather than the area of the disk. Since the domain of integration has circular symmetry, your first impulse should be try to take advantage of this by using the polar coordinates.

Problem 4: (a) Just remember that the Taylor series for (1+x)^p = 1 + px + p(p-1) x²/2 + ... and apply this to the problem at hand where there are two terms with p=-2 and x in the above expansion will be replaced by (+/-)d/(2x) or (+/-)(2x/d) depending on the region of x. While p=-1 for the potential, we have p=-2 for the electric field. For |x|>d/2, the expansion is in (positive) powers of (d/2x), but for |x| < d/2, the expansion is in (positive) powers of (2x/d). Be careful that the expansions are rather different for the two cases.

Problem 5 (4.2.4):

(i) You can just compare with known series.

(ii) Ratio test will work.

(iii) Integral test

(iv) Telescopic!

(v) Look at a_n as n → ∞.

(vi) a_n = 1/(n n^1/n). Note that n^1/n → 1 as n → ∞. Therefore, if you pick any value, say, 2, n^1/n < 2 for any n that is large enough (i.e., for n > n₀). Use this fact and find a lower bound to the original series that is itself divergent. [n^1/n → 1 because n^1/n = e^{(1/n)ln n} and the exponent → 0 as n → ∞.]

Problem 6 (4.3.4): First, show that E²=P²+m² exactly. Then, use expansions like you used in Problem 4 (but with p= - 1/2).

Problem 7 (4.3.5): Though you can consider the equation for the integral as given, the reason why this is true should be obvious by using a change of variable u = tan(v). Now, just follow the instructions in the problem and expand (1+u²)^-1 around u=0 as in some of the other problems in this assignment. Then integrate each term of the resulting series one by one.

Problems removed from Homework 4

Thu 16Sep2010 3:19PM

Apologies! Since I have not discussed in class as much material as I hoped to so far this week, I decided to move Problems 7 and 8 on Homework #4 (due Monday, Sept. 20) to the next set (Homework #5) instead. So you need only to turn in your solutions to Problems 1 - 6 of the previously distributed Homework Set #4.

Office Hours and graded Homework #2

Tue 14Sep2010 4:12PM

Updated office hours:

Nakanishi: Fridays, 10:30 am - 12:20 pm. Rm.264

Shen: Thursdays, 3 - 4 pm, Rm. 105

Graded Homework #2 is available for pick-up in Rm. 144. Ask Pam for yours at the counter. There are separate stacks (in alphabetical order) for each homework set.

Some help for Homework 4

Tue 14Sep2010 3:25PM

Problem 1 (2.2.9): You can start from I₀(a) = (1/2) (π/a)^1/2. As we discussed in class, this should lead to I_2n(a). To do I₃(a), you can do the same starting from I₁(a), where the integrand in the latter integral is an exact differential (as shown in class).

Problem 2 (2.2.10): Do this in 4 steps as guided by the problem statement.

(a) First, write
I(a) = ∫₀¹ (t^a-1)/ln(t) dt

and differentiate both sides with restpect to a. This leads to an easy integral on the right-hand side. Perform that integral and evaluate I'(a). Then (b) use this result to write I(a) = ∫ I'(a) da and perform this integral. Third (c) evaluate the constant of integration that appears as a result of (b) by using the knowledge of I(0). Finally, (d) substitute a=1 to obtain I(1) that is equal to the original integral which you wanted to evaluate.

Problem 4: Take the coordinates of the upper right corner of the rectangle as (u,v). Then, since you want it to be on the ellipse, you have (u²/a²)+(v²/b²)=1. That is your constraint is g(u,v) = (u²/a²)+(v²/b²) - 1 = 0. The area of the rectangle is 4 uv. So you can maximize 4 uv (or, equivalently, just uv). So you can take f(u,v) = uv, and apply the method of Lagrange multipliers to F(u,v,λ) = f(u,v) - λ g(u,v). Don't forget that the length of the horizontal edge is 2u and that of the vertical edge is 2v when you give the side lengths of the rectangle desired.

Problem 5 (3.1.4): "The temperature problem done earlier" refers to the problem of finding the maximum temperature where the temperature f(x,y) is given by (3.1.17) and the region to consider is limited to the circle defined by (3.1.18). This problem was solved without invoking the Lagrange multipliers on pp.55-56, but now you are asked to do it using the Lagrange multipliers as done in class and also, for a different problem, on pp.57-58.

Problem 6 (3.1.5): The first part is to use the method of Lagrange mulitipliers to minimize the distance betweent he origin and the line x+2y=4. The second part is to confirm the result by directly finding the distance between the origin and the point of intersection between the above line and the line through the origin that intersects it perpendicularly (because from plane geometry, that is the shortest distance desired).

To remind you, the equation of the line through the origin that intersects a line defined by ax + by= c is given by bx - ay = 0. [Why?]

Some help for Homework 3

Wed 08Sep2010 5:15PM

Please note that Homework #3 is due on MONDAY, Sep. 13.

Problem 1:

For (a), all you need is the fact that E(t=0) = E(t) [conservation of energy]. Just express E(t=0) in terms of the initial angular velocity (ω=0 because the pendulum starts from rest) and initial angle (θ_m). For (b), note that, since θ is a function of time t, you can also consider time t as a function of the angle θ and find out what the derivative dt/dθ must be equal to. For (c), just use the substitution as suggested and carry through the calculation carefully. The final form is a standard form of one of the so-called elliptic integrals (which is known not to be reducible to any combination of elementary functions).

Problems 2:

Hyperbolic identities like below come in handy in these problems: cosh²x - sinsh²x = 1, (cosh x)' = sinsh x, etc.

Problem 3:

For (a), the above hyperbolic identities may be useful here too, but you will need more than that. It may be useful to first write the integrand as sinh x/(sinh x)², before trying to use a hyperbolic identity. Then, use a change of variable and break up the resulting integrand into a difference of two terms.

For (b), you can use integration by parts a couple of times to come up with the same expression as part of the right-hand side as you started from on the left-hand side. Then solve for that.

Problem 4:

We actually did an indefinite integral version of the first one in class. The second integral has the Gamma function evaluated at 5/4 on the righthand side. This will require the obvious change of variable first. Then you use the definition of the Gamma function as given in problem (2.1.3) on p.43 of Shankar:

Γ (n) = ∫₀^∞ x^n-1 e^-x dx

One more ingredient you need is the identity Γ (n) = (n-1) Γ(n-1) [that you can derive from the definition of Gamma function by integration by parts].

Some help for Homework 2

Mon 06Sep2010 11:48AM

Some hopefully helpful remarks about Homework #2:

Problem 1: Since this is a continuation from Problem 1 of Homework #1, you would need to have the correct answer for the earlier problem to work on it. The correct answer to part (b) of that previous problem was:

y' = [ v sin ωt + v t ω cos ωt ]/[ v cos ωt + v t ω sin ωt ]

For part (a) of the current problem, you use the addition formulas for the sin and cos of the same argument as discussed in class to convert this into tan ( ωt + φ ). the formulas are:

a sin x + b cos x = (a²+b²)^1/2 sin (x+y)
a cos x - b sin x = (a²+b²)^1/2 cos (x+y)

where the angle y is given implicitly by cos y = a/(a²+b²)^1/2 and sin y = b/(a²+b²)^1/2

For part (b), just consider the ranges of cos φ and sin φ as t ranges from 0 to ∞. Since they are both between 0 and 1, you can take 0 < φ < π/2. For the limit t → 0, note the limits for cos φ(t) and sin φ(t) and find what value φ must go to in order to achieve that. For part (d), compare the motion with the constant speed circular motion to make your argument. In particular, a • v ≠ 0 unlike in the latter motion.

Problem 2: We discussed this problem in some detail in class. In fact we already worked on part (a) in class previously. For part (c), try an "intelligent guess" and see if a solution in the form of M(t) = A exp(-t/λ) + B exp(-t/τ) would work for a suitable choice of A and B. [Challenge: see what the solution for M(t) looks like. Can you interpret it? Can you see what happens if λ = τ ? What if M(0) ≠ 0 ?]

Problem 3: Always consider things like the limits x → ±∞ and x → 0±, and any other aymptotes. Also, always evaluate the derivative and see where they are zeroes (maxima, minma, or saddle point) or infinity. Anything else of note for each function enhances the knowledge of how they behave as a function of x. To see what the limit of x → 0 is, use L'Hospital's Rule.

Problem 4: (Shankar 1.6.5) This problem is in the Ising class, that is the magnetic moment μ can only point parallel (up) or anti-parallel (down) to the external magnetic field h. It has 4 parts:

(a) You are given the ratio of the probabilities P_up and P_down and the fact that they add up to 1. Then you are to calculate the P_up and P_down individually. Do just like you would any problem where there are two unknowns and you know their ratio and sum.

(b) The average magnetic moment m is given by
m = μ P_up + (-μ) P_down
just like the average (expected) value when you know values of all possible cases (outcomes) and their probabilities of occurrences.

Problem 5: (Shankar 1.6.10) This problem, if presented as one of a light ray refraction at an interface between two different media, amounts to that of Fermat's Principle. You can set up the problem as I explained in class and differentiate the total time required for the path with respect to x (location where the man or light enters water) to find the equation satisfied when the total time is at the minimum. You need not solve this equation explicitly for x, but just note what it means in terms of the angles involved. You will then immediately get something that is the analog of Snell's Law of refraction.

Some help for Homework 1

Fri 27Aug2010 3:15PM

Problem 5 (1.4.1): If you don't know where to start on this problem, read p.8-11, particularly p.10. The main part asks you to derive (1.4.18) and (1.4.20) from the general result (1.3.16), using your knowledge of the derivatives of cos x and sin x evaluated at x=0. The second part is asking for the radius of convergence of the Taylor series for cos x and sin x. Use the ratio test just as they do on p.10-11 for the exp(x) function.

Problem 6 (1.5.1): We didn't cover L'Hospital's rule in class, but this is something you all did (I am sure) in your calculus classes. Write xⁿe^-x as xⁿ/e^x and apply the L'Hospital's rule (p.24 about the middle).

Problem 7 (1.6.4): This problem has 4 parts (though they are not explicitly marked as separate parts). In the third part (starting from the bottom of p.26 to the top 4 lines of p.27), they define θ" and v" as those that relate (x",t") directly to (x,t), while θ' and v' relate (x",t") to (x',t') and θ and v relate (x',t') to (x,t). In this part, it is useful to realize that:

cosh(θ+θ') = cosh θ cosh θ' + sinh θ sinh θ'

sinh(θ+θ') = sinh θ cosh θ' + cosh θ sinh θ'

Don't forget the fourth part where you will consider the limit of small v and v'.

Office Hours

Thu 26Aug2010 8:41PM

Mr. Shen (grader) will hold office hours on Thursdays, 3 - 4 pm in Physics Room. 105.

Prof. Nakanishi plans to hold office hours on Fridays, 11:30 am - 12:20 pm (just before class) in Physics Room 264, starting this week. Also, just for next week, he will hold an extra office hour on Monday, 11:30 am - 12:20 pm.

Textbook situation

Thu 26Aug2010 2:01PM

Some of you mentioned that the text by Shankar is not available in bookstores. Upon some investigation, it seems that the University Bookstore on W. State St. has two copies on the shelf now and 8 on order. One of the Follet bookstores has 4 on order. Borders on E. State St. can take orders, and once ordered, it will come in a week. You can also get the book (new or used) through the Internet. Places like Amazon.com, Buy.com, and Alibris.com have copies on sale now.

In the mean time, I will post the scanned images of the homework problems from the book on this site under "Homework". One copy is also on reserve at the Physics Library. You cannot take the reserve copy out of the Library but can use it in the Library for a couple of hours (such as copying necessary parts - or just to read and take notes).

I hope this helps.

Welcome to PHYS 290F in Fall 2010

Mon 23Aug2010 4:47PM

Welcome all to Math Methods of Physics I. This is a course created

by physicists primarily to address the needs of physics students,

using a text written by a physicist with a physicist's perspective

on the balance of rigor and practicality. I hope you will enjoy it,

but then I am sure you will be working pretty hard, too. This is

definitely not an easy ride - but should prove beneficial for further

studies of physics. Bon voyage!

Last Updated: May 9, 2023 8:31 AM