Solved problems from math review worksheet

By popular demand, I am going to solve a few problems from the basic math review worksheet here.

 5. a. The amount an object's speed v changes with time is proportional to its current speed.


$$ \frac{dv}{dt}\propto v(t).$$

Note this is a differential equation that can be solved by rearranging: $$\frac{dv}{v}=dt.$$ Integrating both sides we find $$\ln v=t+C,$$ C a constant of integration, meaning that $$v=v_0\exp(t),$$ where we set the value of C by saying the initial velocity is $$v_0.$$ In more detail, $$v=e^{t+C}=e^te^C=v_0e^t$$ with $$e^C=v_0$$ (we can choose C such that this is true because C was just an arbitrary integration constant!)

b.  A is inversely proportional to the square root of B.

$$A\propto \frac{1}{\sqrt{B}}.$$

c.  An object's speed v varies sinusoidally with a period P and a phase $$\phi.$$

$$v(t)=\sin \bigg(\frac{2\pi}{P}t+\phi \bigg).$$

d.  The flux F of an astronomical source decreases with distance according to the inverse square law.

This law comes from considering the radiation from the source to be emitted isotropically into a sphere of radius d, with surface area $$4\pi d^2.$$  So we have

$$F=\frac{L}{4\pi d^2},$$  where L is the luminosity, which is the energy output per unit time of the source.

e.  The number of stars, N, per unit mass, M, is given by a powerlaw in mass with an index $$-\alpha.$$

This problem was not perfectly worded!  What we meant was the amount of stars dN with masses in the range M, M+dM, where dM is small (really, infinitesimally small!), is given by a powerlaw in mass.  This translates as
$$\frac{dN}{dM}\propto M^{-\alpha}.$$

This is a typical way of specifying the distribution of initial masses with which stars are born.  That is, not all stars are like our Sun, just like babies are born with different weights---some heavier, some lighter.  There is a pattern to the distribution of stars' initial masses, called the IMF, for Initial Mass Function, and it is a subject of intense debate and study.  It turns out that it is set partially by turbulence in the gas from which stars form and partially by the scale on which pressure from the gas is able to balance gravity (called the Jeans length).

f.  The pressure in the atmosphere decreases exponentially from an initial value $$P_0$$ to a value $$P(z)$$ with an e-folding length H.

e-folding means the time a quantity takes to change by one factor of e = 2.818 (note e is also given by the infinite series
$$\sum_{n=0}^{\infty}\frac{x^n}{n!},$$ which I mention to justify the seemingly totally random value of e as having some kind of symmetrical, elegant origin).

We have
$$P(z)=P_0 e^{-z/H},$$ with z one's height above ground and H more conventionally called the "scale height."  You might wonder why this is true?  Fundamentally, it is because the atmosphere is made up of billions of air molecules that are bouncing around. All of these particles are bouncing in a gravitational field that wants them to all end up on the Earth's surface.  The faster ones get to go up higher.  So the density will decrease as you go up from the Earth's surface, because there are fewer particles with high speeds than with low.  That is because for an ideal gas in equilibrium (a reasonable model here), the velocities are "Maxwellian":  the probability of finding a particle with velocity v is proportional to $$v^2e^{-v^2}$$---you'll learn more about that in a statistical mechanics course!  Now, pressure scales with density, and as we've seen, the density is determined by the velocity probability distribution.  Exponential in, exponential out.

7.  Write the ideal gas law in terms of gas pressure P, mass density $$\rho,$$ and temperature T.

It is $$P=\frac{\rho}{\mu m_p}k_{\rm B}T,$$  where $$\mu$$ is the mean particle mass in units of the proton mass $$m_p$$ and $$k_{\rm B}$$ is Boltzmann's constant.

8.  Calculate the derivative of:
a. 1/(1-x)

Rewrite as $$(1-x)^{-1}$$ and then use the Chain Rule: $$-(1-x)^{-2}(-1)=(1-x)^{-2}$$ is the derivative.

b. $$e^{-x}$$.  Derivative: $$-e^{-x}.$$

c.  $$\sin x.$$ Derivative: $$\cos x.$$

d.  $$\cos x.$$ Derivative: $$-\sin x.$$

e.  $$\ln(1-x).$$  Again use Chain Rule.  Derivative: $$\frac{1}{1-x}(-1)=-\frac{1}{1-x}.$$

f.  $$\bigg(e^{-x}-1\bigg)^{-1}.$$  Again use Chain Rule.  $$-\bigg(e^{-x}-1\bigg)^{-2}\bigg(-e^{-x}\bigg),$$ which simplifies to
$$e^{-x}\bigg(e^{-x}-1\bigg)^{-2}.$$

9.  Taylor expansions represent functions as sums of powers of their arguments.  They do this by taking successive approximations to the function: first-order Taylor series models the function as a straight line, second-order as a straight line plus a parabola, etc.  For the second-order series, we need the first and second derivatives.  We have the first derivatives from problem 8.; we differentiate again to get the second derivatives.  Let's do $$e^{-x.}$$  The second derivative is
$$e^{-x}.$$  For a Taylor expansion around x=0, we evaluate the derivatives at zero and then multiply them by respectively, x and $$x^2.$$  We also weight the contributions by 1/n!, where n is the order of the derivative.  Finally, we need a constant term: the value of the function itself at x=0.  Think of that as a vertical offset to the whole curve we are trying to model. Putting it together, we have
$$e^{-x}\approx 1-x + \frac{x^2}{2}.$$  We could easily have gotten this from the series representation of $$e^{x}$$ given earlier by letting x go to -x.

Let's do one more: $$\ln(1-x)$$ about x=0.  We have
$$\ln(1-x)\approx -x+\frac{x^2}{2}.$$

10.  Integrals of the functions in 8. (note: these should be evaluated at $$x_{min}$$ and subtracted from their value at $$x_{max}$$ for the definite integral.)

a.  Use the substitution u=1-x, du=-dx to see that
$$\int \frac{1}{1-x}dx=-\ln (1-x).$$

b.$$\int e^{-x}dx=-e^{-x}.$$

c.  $$\int \sin x dx =-\cos x.$$

d.$$\int \cos x dx=\sin x.$$

e.  $$\int \ln(1-x)dx$$ is tricky.  We should integrate by parts (a very similar example is done in the Basic Math Review page on integration!) to find
$$x\ln(1-x)+\int\frac{x}{1-x}dx,$$ and can then set $$u=1-x$$ to do the remaining integral.

f.  $$\int \frac{1}{e^{-x}-1}dx.$$  This one is also quite hard!  I actually didn't know how to approach it immediately, so still working on a good intuitive explanation!