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Homework & Assignments

HW# 7 | 6 | 5 | 4 | 3 | 2 | 1

Homework #1

chapter 1- 3*, 4, 5, 6*, 10*, 12*, (from sections 1.1,2)
24, 27, 30 (from sections 1.3-5)

Notes

* - include a sketch for these.

Sometimes Paul uses primes (') to distinguish between two reference frames, e.g. $r'(t), \phi'(t)$ and $r(t), \phi(t)$: this does not mean "the derivative of...".

1.4

$\theta=\arccos\left( \frac{12}{21}\right)=$$55.15{}^o$

1.5

$\theta=\arccos\left( \frac{2}{\sqrt 6}\right)=$ $0.615 \text{ Radians}=35.3{}^o$.

1.6

1.10

Show that $$\myv r(t)=R\cos(\omega t)\uv x + R\sin(\omega t)\uv y$$ Then taking the first and second derivative... $$\myv v(t)=\dot{\myv r}(t)=-R\omega\sin(\omega t)\uv x+R\omega\cos(\omega t)\uv y$$ $\myv v(t)$ is a tangent vector to the circle, always pointing in the direction of motion. And the acceleration is: $$\myv a(t)=\dot{\myv v}(t)=-R\omega^2\cos(\omega t)\uv x-R\omega^2\sin(\omega t)\uv y=-R\omega^2 \myv r(t).$$ So, the acceleration is pointing back towards the origin at all times, and its magnitude is $$a=R\omega^2\sqrt{\cos^2(\omega t)+\sin^2(\omega t)}=R\omega^2$$ And since $\omega=v/R$, the acceleration is $a=v^2/R$ which we recognize as the equation for centripetal acceleration.

1.12
You can pick particular values for $b$, $c$, $\omega$, and $v_0$, and make a "sketch" in geogebra. Look up how to use geogebra's Curve(...) command to make a parametric plot in Cartesian coordinates.

If you do it in geogebra, paste the URL of your plot into our moodle assignment for HW 01, and describe in words what the trajectory looks like.

For example: www.geogebra.org/3d/srtf9vxg: The projection in the $x-y$ plane is an ellipse, crossing the $x$ axis at $x=b$ and crossing the $y$ axis at $y=c$. It's moving with a constant velocity, $v_0$ in the $z$ direction. So its trajectory looks like an "elliptical spring", or "elliptical spiral" around the $z$ axis.

1.24

Use the method of Separation of variables (Calculus I notes) to re-arrange the differential equation, integrate, and solve for $f(t)$.

The highest-order derivative in this equation is a 1st order derivative ($df/dt$) so there should be 1 arbitrary constant in the solution.

Another example: The differential equation for 1-d vertical motion near Earth's surface is: $$\text{acceleration}=\frac{d^2z}{dt}=-g.$$ This equation has a 2nd derivative of position, $z$. It is a 2nd order diff. eq. You might remember the solution to this equation from GPhys: $$ z= -\frac12gt^2+v_0t+z_0.$$ There are 2 integration constants in the solution: $v_0$ and $z_0$.

In general, the number of integration constants is equal to the order of the highest-order differential in a differential equation.

Starting with $df/dt=f$, we separate variables and integrate: $$\begineq \frac{1}{f}df&=&dt\\ \int\frac{1}{f}\,df&=&\int\,dt\\ \ln f &=& t + C\\ e^{\ln f} &=& e^{t + C}\\ f &=& e^{t} e^{C}=Ae^t \endeq$$ where we were left with one free constant, $A\equiv e^C$.