Use the results of Problem 1to find the velocity and acceleration components in spherical coordinates. Find the velocity in two ways: starting with ds and starting with$\mathit{s}\mathbf{=}\mathit{r}{\mathit{e}}_{\mathbf{r}}$.

Spherical coordinates are given below.

$$\begin{gathered} x=r\cos \varphi \sin \theta \\ y=r\sin \varphi \sin \theta \\ z=r\cos \theta \end{gathered}$$

$$\begin{gathered} x=r\cos \varphi \sin \theta \\ y=r\sin \varphi \sin \theta \\ z=r\cos \theta \end{gathered}$$

The coordinate system primarily utilized in three-dimensional systems is the spherical coordinates of the system. The spherical coordinate system is used to find the surface area in three-dimensional space.

Spherical coordinates are given below.

$$\begin{gathered} x=r\cos \varphi \sin \theta \\ y=r\sin \varphi \sin \theta \\ z=r\cos \theta \end{gathered}$$

The formula states that $ds=idx\times jdy+kdz$.

$$\begin{gathered} dx=\cos \varphi \sin \theta dr+r\cos \varphi \cos \theta d\theta -r\sin \varphi \sin \theta d\varphi \\ dx=\sin \varphi \sin \theta dr+r\sin \varphi \cos \theta d\theta +r\cos \varphi \sin \theta d\varphi \\ dz=\cos \theta dr-r\sin \theta d\theta \end{gathered}$$

Substitute the above value in the formula.

$$\begin{gathered} ds=idx+jdy+kdz \\ ds=\left(i\cos \varphi \sin \theta +j\sin \varphi +k\cos \theta \right)dr+\left(ir\cos \varphi \cos \theta +jr\sin \varphi \cos \theta -kr\sin \theta \right)d\theta \\ +\left(-ir\sin \varphi \sin \theta +jr\sin \varphi \sin \theta d\theta \right)d\varphi \end{gathered}$$

Find the other values.

localid="1659347919099" $$\begin{gathered} h_r{e}_r=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +k\cos \theta \\ {h}_{\theta }{e}_{\theta }=ir\cos \varphi \cos \theta +jr\sin \varphi \cos \theta -kr\sin \theta \\ {h}_{\varphi }{e}_{\varphi }=-ir\sin \varphi \sin \theta +jr\cos \varphi \sin \theta \end{gathered}$$

Find the other values.

localid="1659348180974" $$\begin{gathered} e_r=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +k\cos \theta \\ {e}_{\theta }=i\cos \varphi \cos \theta +j\sin \varphi \cos \theta -k\sin \theta \\ {e}_{\varphi }=-i\sin \varphi +j\cos \varphi \end{gathered}$$

$$\begin{gathered} e_r=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +k\cos \theta \\ {e}_{\theta }=i\cos \varphi \cos \theta +j\sin \varphi \cos \theta -kr\sin \theta \\ {e}_{\varphi }=-i\sin \varphi +j\cos \varphi \end{gathered}$$

Derivate the values mentioned above.

$$\begin{gathered} \stackrel{\mathbf{.}}{e_r}=\stackrel{}{\stackrel{\mathbf{.}}{\varphi }\sin \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\theta }{e}_{\theta } \\ \stackrel{\mathbf{.}}{e_{\theta }}=\stackrel{}{\stackrel{\mathbf{.}}{\varphi }\cos \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\theta }{e}_r \\ \stackrel{\mathbf{.}}{e_{\varphi }}=\stackrel{\mathbf{.}}{\varphi }\left(\stackrel{}{\sin \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\cos \theta }{e}_{\theta }\right) \end{gathered}$$

$$\begin{gathered} \stackrel{\mathbf{.}}{e_r}=\stackrel{}{\stackrel{\mathbf{.}}{\varphi }\sin \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\theta }{e}_{\theta } \\ \stackrel{\mathbf{.}}{e_{\theta }}=\stackrel{}{\stackrel{\mathbf{.}}{\varphi }\cos \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\theta }{e}_r \\ \stackrel{\mathbf{.}}{e_{\varphi }}=\stackrel{\mathbf{.}}{\varphi }\left(\stackrel{}{\sin \theta e_{\varphi }}+\stackrel{\mathbf{.}}{\cos \theta }{e}_{\theta }\right) \end{gathered}$$

Let the velocity be$\frac{ds}{dt}$.

The value of ds is mentioned below.

$$\begin{gathered} ds=e_r{h}_{r}dr+e_{\theta }{h}_{\theta }d\theta +e_{\varphi }{h}_{\varphi }d\varphi \\ ds=e^{r}dr+e_{\theta }rd\theta +e_{\varphi }r\sin \theta d\varphi \end{gathered}$$

The value of velocity becomes as follows.

$$\begin{gathered} \frac{ds}{dt}=\frac{d}{dt}\left(i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +kr\sin \theta \right) \\ \frac{ds}{dt}=\frac{d}{dt}\left(r{e}_r\right) \\ \frac{ds}{dt}=e_r\stackrel{\mathbf{.}}{r+e_{\theta }}+r\stackrel{\mathbf{.}}{\theta }+e_{\varphi }r\sin \theta \stackrel{\mathbf{.}}{\varphi } \end{gathered}$$

Let the acceleration be $\frac{d^{2}s}{d{t}^2}$.

$$\begin{gathered} \frac{d^{2}s}{d{t}^2}=e_r\stackrel{••}{r}+\stackrel{•}{e_r+}\stackrel{•}{r}\stackrel{•}{e_{\theta }}\stackrel{••}{\theta }+r\stackrel{•}{e_{\theta }}\stackrel{•}{\theta }+\stackrel{•}{r}\stackrel{•}{\varphi }{e}_{\varphi }{\sin }_{\theta }+\stackrel{•}{e_{\varphi }}r\stackrel{•}{\varphi \sin \theta } \\ \frac{d^{2}s}{d{t}^2}=\left(\stackrel{••}{r}-r{\stackrel{•}{\theta }}^2-r{\stackrel{•}{\varphi }}^2{\sin }^2\theta \right)e_2+\left(\stackrel{••}{r\theta }-2\stackrel{•}{r}{\stackrel{•}{\theta }}^2-r{\stackrel{•}{\varphi }}^2\sin \theta \cos \theta \right)+\left(\stackrel{••}{2r\varphi }\sin \theta +2\stackrel{•}{r}\stackrel{•}{\stackrel{•}{\varphi }\cos \theta }+r{\stackrel{•}{\varphi }}^2\sin \theta \right)e_{\varphi } \end{gathered}$$

The velocity is$\frac{ds}{dt}=e_r\stackrel{•}{r}+e_{\theta }r\stackrel{•}{\theta +e_{\varphi }}r\sin \theta \stackrel{•}{\varphi }$ $\frac{ds}{dt}=e_r\stackrel{•}{r}+e_{\theta }r\stackrel{•}{\theta +e_{\varphi }}r\sin \theta \stackrel{•}{\varphi }$.

The acceleration is mentioned belowlocalid="1659350430667" $$\begin{gathered} (\stackrel{••}{r}-\stackrel{•}{r}{\theta }^2-r{\stackrel{•}{\varphi }}^{2}si{n}^2\theta )e_r+(r\stackrel{••}{\theta }-2\stackrel{•}{r}\stackrel{•}{\theta }-r{\stackrel{•}{\varphi }}^{2}sin\theta cos\theta ) \\ {e}_{\theta }+(2\stackrel{•}{r}\varphi sin\theta +2r\stackrel{•}{\theta }\stackrel{•}{\varphi }┴•cos\theta +r\stackrel{••}{\varphi }sin\theta )e_{\varphi }. \end{gathered}$$