Theoretical Mechanics: Fall-2026
HW 2 (SOLUTION): Due F Day 8 MathBits

1. Zapping with d S1 5343S Zap the following equations with “d”:

   1. \(f= 3x^3-e^{2x}\)

      \(df=(9x^2-2e^{2x})\, dx\)

   2. \(g = \sin(y^2)\)

      \(dg = \cos{(y^2)}\, 2y\,dy\)

   3. \(5z^4 = \ln(p)\)

      \(20\, z^3\, dz = \frac{1}{p} \, dp\)

2. Separable ODEs S1 5343S

   Solve the following differential equations by separation. \(b\) and \(c\) are constants with the appropriate dimensions.

   1. \(\frac{dv}{dt}=ct\) where \(v(t=0)=v_0\)

      \begin{eqnarray*} dv = ct\;dt\\ \int_{v_0}^v dv' = \int_0^t ct'\;dt'\\ v-v_0 = c\frac{t^2}{2}\\ v(t) = \frac{c}{2}t^2+v_0 \end{eqnarray*}

   2. \(\frac{dp}{dm}=-cp\) where \(p(m=0)=p_0\)

      \begin{eqnarray*} \frac{dp}{p} &=& -c\;dm\\ \int_{p_0}^p \frac{dp'}{p'} &=& -\int_0^m c\;m'\\ \ln{p'}\bigg|_{p_0}^{p} &=& -cm\\ \ln{p}-\ln{p_0} &=& -cm \\ \ln{\left(\frac{p}{p_0}\right)} &=& -cm \\ p(m) &=& p_0\;e^{-cm} \end{eqnarray*}

   3. \(\frac{dz}{dk}=-b-cz\) where \(z(k=0)=z_0\)

      \begin{eqnarray*} \frac{dz}{b+cz} &=& -dk\\ \int_{z_0}^z \frac{dz'}{b+cz'} &=& -\int_0^k dk'\\ \mbox{let }u &=& b+cz\\ du &=& cdz\\ \int_{b+cz_0}^{b+cz} \frac{du}{cu} &=& -\int_0^k dk'\\ \frac{1}{c}\ln{u}\bigg|_{b+cz_0}^{b+cz} &=& -k\\ \ln{b+cz}-\ln{b+cz_0} &=& -ck \\ \ln{\left(\frac{b+cz}{b+cz_0}\right)} &=& -ck \\ z(k) &=&\left(\frac{b}{c}+z_0\right)\;e^{-ck}-\frac{b}{c} \end{eqnarray*}

3. Speed Squared in Curvilinear Coordinates S1 5343S

   Show that the speed squared written in different standard coordinates systems is:

   1. in rectangular coordinates: \(v^2 = \dot{x}^2 + \dot{y}^2+\dot{z}^2\)

      \begin{align*} v^2 &= \vec{v} \cdot \vec{v}\\ &= (\dot{x}\hat{x}+\dot{y}\hat{y}+\dot{z}\hat{z}) \cdot (\dot{x}\hat{x}+\dot{y}\hat{y}+\dot{z}\hat{z})\\ &= \dot{x}^2 + \dot{y}^2+\dot{z}^2 \checkmark \end{align*}

   2. in cylindrical coordinates: \(v^2 = \dot{s}^2 + s^2\dot{\phi}^2+\dot{z}^2\)

      First approach:

      \begin{align*} x &= s \cos \phi\\ \dot{x} & = \dot{s}\cos\phi - s\sin\phi\dot\phi \\[6pt] y &= s \sin \phi \\ \dot{y} &= \dot{s}\sin\phi + s\cos\phi\dot\phi \\[6pt] z &= z\\ \dot{z} &= \dot{z} \\[6pt] v^2 &= \dot{x}^2 + \dot{y}^2+\dot{z}^2 \\ &= (\dot{s}\cos\phi - s\sin\phi\dot{\phi})^2\\ &+ (\dot{s}\sin\phi + s\cos\phi\dot{\phi})^2\\ &+\dot{z}^2 \\[6pt] v^2 &= \dot{s}^2\color{red}{\cos^2\phi} + s^2\color{blue}{\sin^2\phi}\dot{\phi}^2 \cancel{- 2 s\dot{s}\dot{\phi}\sin\phi\cos\phi} \\ & + \dot{s}^2\color{red}{\sin^2\phi} + s^2\color{blue}{\cos^2\phi}\dot{\phi} + \cancel{2 s\dot{s}\dot{\phi}\sin\phi\cos\phi}\\ & + \dot{z}^2 \\[6pt] &= \dot{s}^2 \cancelto{1}{\color{red}{(\cos^2\phi + \sin^2\phi)}} + s^2 \dot{\phi}^2 \cancelto{1}{\color{blue}{(\sin^2\phi + \cos^2\phi)}} + \dot{z}^2 \\[6pt] &= \dot{s}^2 + s^2 \dot{\phi}^2 + \dot{z}^2 \checkmark \end{align*}

   3. in spherical coordinates: \(v^2 = \dot{r}^2 + r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2\)

      First approach:

      \begin{align*} x &= r \cos\phi \sin\theta\\ \dot{x} & = \dot{r} \cos\phi \sin\theta - r \sin\phi \dot\phi \sin\theta + r \cos\phi \cos\theta \dot{\theta} \\[6pt] y &= s \sin \phi \sin\theta \\ \dot{y} &= \dot{r} \sin\phi \sin\theta + r \cos\phi \dot\phi \sin\theta + r \sin\phi \cos\theta \dot{\theta} \\[6pt] z &= r \cos\theta \\ \dot{z} &= \dot{r} \cos\theta - r \sin\theta \dot{\theta} \\[6pt] v^2 &= \dot{x}^2 + \dot{y}^2+\dot{z}^2 \\ &= (\dot{r} \cos\phi \sin\theta - r \sin\phi \dot{\phi} \sin\theta + r \cos\phi \cos\theta \dot{\theta})^2\\ &+ (\dot{r} \sin\phi \sin\theta + r \cos\phi \dot{\phi} \sin\theta + r \sin\phi \cos\theta \dot{\theta})^2\\ &+ (\dot{r} \cos\theta - r \sin\theta \dot{\theta})^2 \\[6pt] v^2 &= \dot{r}^2 \color{red}{\cos^2\phi} \sin^2\theta + r^2 \color{blue}{\sin^2\phi} \dot{\phi}^2 \sin^2\theta + r^2 \color{green}{\cos^2\phi} \cos^2\theta \dot{\theta}^2 \\ & \cancel{- 2r\dot{r} \sin\phi \cos\phi \sin^2\theta} - \bcancel{2r \dot{r}\sin\phi \cos\phi \dot{\phi} \dot{\theta} \sin\theta \cos\theta} + 2r \dot{r} \color{purple}{\cos^2\phi} \dot{\theta} \sin\theta \cos\theta \\ & + \dot{r}^2 \color{red}{\sin^2\phi} \sin^2\theta + r^2 \color{blue}{\cos^2\phi} \dot{\phi}^2 \sin^2\theta + r^2 \color{green}{\sin^2\phi} \cos^2\theta \dot{\theta}^2 \\ & \cancel{+ 2r\dot{r} \sin\phi \cos\phi \sin^2\theta} + \bcancel{2r^2 \sin\phi \cos\phi \dot{\phi} \dot{\theta} \sin\theta \cos\theta} + 2r \dot{r} \color{purple}{\sin^2\phi} \dot{\theta} \sin\theta \cos\theta \\ & + \dot{r}^2 \cos^2\theta +r^2 \sin^2\theta \dot{\theta}^2 -2r \dot{r} \sin\theta \cos\theta \dot{\theta} \\[6pt] &= \dot{r}^2 \color{magenta}{\sin^2\theta} \cancelto{1}{\color{red}{(\cos^2\phi + \sin^2\phi)}} + r^2 \dot{\phi}^2 \sin^2\theta \cancelto{1}{\color{blue}{(\sin^2\phi + \cos^2\phi)}} + r^2 \color{orange}{\cos^2\theta} \dot{\theta}^2 \cancelto{1}{\color{green}{(\sin^2\phi + \cos^2\phi)}} \\ & \bcancel{2r\dot{r}\dot{\theta}\sin\theta\cos\theta} \cancelto{1}{\color{purple}{(\cos^2\phi + \sin^2\phi)}} \\ & + \dot{r}^2 \color{magenta}{\cos^2\theta} +r^2 \color{orange}{\sin^2\theta} \dot{\theta}^2 \bcancel{-2r \dot{r} \sin\theta \cos\theta \dot{\theta}} \\[6pt] &= \dot{r}^2 \cancelto{1}{\color{magenta}{(\sin^2\theta + \cos^2\theta)}} + r^2 \dot{\phi}^2 \sin^2\theta + r^2 \cancelto{1}{\color{orange}{(\cos^2\theta + \sin^2\theta)}}\dot{\theta}^2\\ &= \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta \dot{\phi}^2 \checkmark \end{align*}

   where:

   
   

   The algebra for these calculations is a little tedius, but it's good to do them once in your career to convince yourself that these expressions are true!

   See: GMM: Curvilinear Coordinates for more discussion of curvilinear coordinates. Note that the labels for the aximuthal and polar coordinates in different for math and physics (\(\phi\) and \(\theta\) are swapped.)