Relax a Little

Normally students ask me questions by e-mail and I post some responses here. Unfortunately for them, and fortunately for me, no one asked any questions. I would have been happy to reply. The instructor this semester has rules against explicitly posting answers to various material, but I still could have answered with a method. Anyways, here's a video I wanted to show my 3pm section, but never got around to doing. I just embedded the first in the series:



For multiple reasons, its too late to give any advice I didn't already give. I've already heavily emphasized knowing how to simplify $\sin(\theta)$, $\cos(\theta)$, etc. for common values of $\theta$. I also emphasized various formulas related to trigonometric functions, such as $\sin^2(\theta)+\cos^2(\theta)=1$ and $\cosh^2(\theta)-\sinh^2(\theta)=1$.

There's also
$\int \frac{1}{1+x^2}dx=\arctan(x)+C$
and
$\int \sec(x)dx=\ln \left| (\sec(x)+\tan(x)\right|+C$

Students should also be careful about using Euler's Method.
Suppose we are looking to approximate $y(x)$.
If the step size is $h$ and $x=k\cdot h+x_0$, then the value we seek is $y_k$.
Many students made the ERROR of determining $y_{k+1}$ (ERROR)
Recall some formulas:
$\frac{dy}{dx}=F(x,y)$
$x_n=x_{n-1}+h$
$y_n=y_{n-1}+h\cdot F(x_{n-1},y_{n-1})$

I like to remember the above by (note this is a way to remember the formula, it is not intended to justify or derive anything)
$\Delta y=\Delta x \cdot \frac{\Delta y}{\Delta x}$
$y_k - y_{k-1}=h \cdot \frac{dy}{dx}$
$y_k-y_{k-1}=h\cdot F(x_{k-1},y_{k-1})$
$y_k=y_{k-1}+h\cdot F(x_{k-1},y_{k-1})$

Finally for those hard to solve rate problems, just do your best to break down the problem and determine the rate in and the rate out. Suppose the quantity changing with respect to time is volume. Then let $V$ be the amount of volume. We set up a differential equation:
$\frac{dV}{dt}=$ {rate in} - {rate out}
There's a high chance either the rate in or the rate out will depend on $V$.
After you have completed the problem, remember to solve the initial value problem, if that is what the question is asking for.