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In this course, you will be asked to plot functions using either Mathematica, Matlab, or Python. The university has a license that allows undergraduate students to download and use these computer algebra systems on their personal computers, but you have to send an e-mail to receive an access code:
for Mathematica: https://softwarelist.oregonstate.edu/software/mathematica (Link on Canvas Too)
for Matlab: https://softwarelist.oregonstate.edu/software/matlab (Link on Canvas Too)
for Python: https://www.anaconda.com/products/individual (Link on Canvas Too)
(There's nothing to turn in for this problem. This is just to prepare you for future homework.)
Needing help getting any or all of the above software working is a perfectly good reason to come to office hours.
Rules for Plots in this Course:
For help with plotting in Mathematica, see:
(modified from Taylor 2.33 & 2.34)
Hyperbolic trigonometry is going to be useful for us both for describing the motion of objects experiencing quadratic drag force and also for geometric approaches to special relativity. You should be familiar with the properties of these hyperbolic trig functions for later parts of this course.
The hyperbolic functions \(\cosh z\) and \(\sinh z\) are defined as follows: \begin{align} \cosh z &= \frac{e^z+e^{-z}}{2} \\ \sinh z &= \frac{e^z-e^{-z}}{2} \\ \end{align}
for any \(z\), real or complex.
First, I'll plot \(\cosh(z)\) vs. \(z\)
At \(z=0\), I get \(\cosh(0) = 1\) and
at \(z=\pm\infty\), I get \(\cosh(\pm\infty) = +\infty\).
\(\cosh z\) is even: \(\cosh(-z)=\cosh)z)\).
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Now, I plot \(\sinh(z)\) vs. \(z\)
At \(z=0\), I get \(\sinh(0) = 0\) and
at \(z=\pm\infty\), I get \(\sinh(\pm\infty) = \pm\infty\).
\(\sinh z\) is odd: \(\sinh(-z)=-\sinh(z)\)
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Show that \(\cosh^2z-\sinh^2z = 1\)
First, I'll square my expressions for sinh and cosh. Then I'll subtract and simplify: \begin{eqnarray*} \cosh^2z - \sinh^2z &=& \left(\frac{e^z+e^{-z}}{2}\right)^2-\left(\frac{e^z-e^{-z}}{2}\right)^2\\[6pt] &=&\frac{1}{4}\left(e^z+e^{-z}\right)^2 - \frac{1}{4}\left(e^z-e^{-z}\right)^2 \\[6pt] &=& \frac{1}{4}\left[\left(\cancel{e^{2z}}+\cancel{e^{-2z}}+2\right)- \left(\cancel{e^{2z}}+\cancel{e^{-2z}}-2\right)\right]\\[6pt] &=& \frac{4}{4}\\ &=& 1\checkmark \end{eqnarray*}
\begin{eqnarray*} \frac{d}{dz}\left(\sinh{z}\right) &=& \frac{d}{dz}\left(\frac{e^z-e^{-z}}{2}\right) \\[6pt] &=& \frac{e^z - (-e^{-z})}{2} \\[6pt] &=& \cosh{z}\\[6pt] \end{eqnarray*}
and \begin{eqnarray*} \frac{d}{dz}\left(\cosh{z}\right) &=& \frac{d}{dz}\left(\frac{e^z+e^{-z}}{2}\right) \\[6pt] &=& \frac{e^z+ (-e^{-z})}{2} \\[6pt] &=& \sinh{z}\\[6pt] \end{eqnarray*} These derivatives have zeroes at \(z=0\) (the local extrema of the original functions). The signs (\(\pm\)) of the derivative functions corresponds to the sign of the slope of the original functions: The slope of \(\cosh(z)\) is negative with \(z<0\) and positive when \(z>0\) (matching the sign of \(\sinh\)). The slope of \(\sinh(z)\) is always positive (matching the sign of \(\cosh\)).
Using the hint given:
\(x=\sinh(z)\) and \(dx=\cosh(z)dz\) \begin{eqnarray*} \int \frac{dx}{\sqrt{1+x^2}}&=& \int \frac{\cosh(z)dz}{\sqrt{1+\sinh^2(z)}} \end{eqnarray*} Now use the trig identity: \(1+\sinh^2(z)=\cosh^2(z)\) \begin{eqnarray*} \int \frac{dx}{\sqrt{1+x^2}}&=&\int \frac{\cosh(z)dz}{\sqrt{\cosh^2(z)}}\\ &=& \int dz\\ &=& z\\ &=&\mbox{arcsinh}\;x \end{eqnarray*} Depending on the range of integration, you may need to be careful about signs when using this identity.