Part (a)

Part (b)

$\mathbf{T}\=\frac{\mathbf{R}\prime }{\mathrm{\rho }}\=\frac{1}{2 p\+1\/2}\left[\begin{array}{c}2 p-1\/2\\ 2\sqrt{p}\end{array}\right]\=\frac{1}{1\+4 p}\left[\begin{array}{c}4 p-1\\ 4\sqrt{p}\end{array}\right]$

Part (c)

Initialize

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$\mathbf{R}\=⟨p^2-p\/2\,4\/3 p^{3\/2}⟩$$\stackrel{\text{assign}}{\to }$

Part (a)

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $\frac{\sqrt{{\left(1+4p\right)}^2}}{2}$$\stackrel{\text{assuming positive}}{\to }$$2p+\frac{1}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\rho }$$$

Part (b)

$\mathbf{T}\=\frac{1}{\mathrm{\ρ}} \left(\frac{\ⅆ}{\ⅆ p} \mathbf{R}\right)$$\stackrel{\text{assign}}{\to }$

$\mathbf{T}$ = $\left[\begin{array}{c}\frac{2p-\frac{1}{2}}{2p+\frac{1}{2}}\\ \frac{2\sqrt{p}}{2p+\frac{1}{2}}\end{array}\right]$$$

Part (c)

$\genfrac{}{}{0ex}{}{\mathbf{R}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{p\=\left(\sqrt{1\+16 s}-1\right)\/4}$

$\left[\begin{array}{c}{\left(\frac{\sqrt{1+16s}}{4}-\frac{1}{4}\right)}^2-\frac{\sqrt{1+16s}}{8}+\frac{1}{8}\\ \frac{4{\left(\frac{\sqrt{1+16s}}{4}-\frac{1}{4}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}\end{array}\right]$

$\stackrel{\text{assuming positive}}{\to }$

$\left[\begin{array}{c}\frac{1}{4}+s-\frac{\sqrt{1+16s}}{4}\\ \frac{{\left(\sqrt{1+16s}-1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{6}\end{array}\right]$

$\stackrel{\text{differentiate}}{\to }$

$\left[\begin{array}{c}1-\frac{2}{\sqrt{1+16s}}\\ \frac{2\sqrt{\sqrt{1+16s}-1}}{\sqrt{1+16s}}\end{array}\right]$

$\stackrel{\text{evaluate at point}}{\to }$

$\left[\begin{array}{c}1-\frac{2}{\sqrt{16p^2+8p+1}}\\ \frac{2\sqrt{\sqrt{16p^2+8p+1}-1}}{\sqrt{16p^2+8p+1}}\end{array}\right]$

$\stackrel{\text{assuming positive}}{\to }$

$\left[\begin{array}{c}\frac{4p-1}{1+4p}\\ \frac{4\sqrt{p}}{1+4p}\end{array}\right]$

Part (a)

$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \\ \mathrm{BasisFormat}\left(\mathrm{false}\right)\: \end{gathered}$$

$\mathbf{R}≔⟨p^2-p\/2\,4\/3 p^{3\/2}⟩\:$

$\mathrm{\ρ}≔\mathrm{simplify}\left(\mathrm{Norm}\left(\mathrm{diff}\left(R\,p\right)\right)\right) assuming p\>0$

$\mathrm{\rho }≔2p+\frac{1}{2}$

Part (b)

$\mathbf{T}≔\mathrm{diff}\left(\mathbf{R}\,p\right)\/\mathrm{\ρ}$

$T≔\left[\begin{array}{c}\frac{2p-\frac{1}{2}}{2p+\frac{1}{2}}\\ \frac{2\sqrt{p}}{2p+\frac{1}{2}}\end{array}\right]$

$$\begin{gathered} \mathbf{Temp}≔\mathrm{TangentVector}\left(\mathbf{R}\,p\,\mathrm{normalized}\right)\: \\ \mathrm{simplify}\left(\mathbf{Temp}\right) assuming p\>0 \end{gathered}$$

$\left[\begin{array}{c}\frac{4p-1}{1+4p}\\ \frac{4\sqrt{p}}{1+4p}\end{array}\right]$

Part (c)

Apply the diff and simplify commands.

$$\begin{gathered} \mathbf{Rs}≔\mathrm{eval}\left(\mathbf{R}\,p\=\left(\mathrm{sqrt}\left(1\+16 s\right)-1\right)\/4\right)\: \\ \mathbf{Q}≔\mathrm{simplify}\left(\mathrm{diff}\left(\mathbf{Rs}\,s\right)\right) \end{gathered}$$

$Q≔\left[\begin{array}{c}\frac{\sqrt{1+16s}-2}{\sqrt{1+16s}}\\ \frac{2\sqrt{\sqrt{1+16s}-1}}{\sqrt{1+16s}}\end{array}\right]$

Use the eval command to make the substitution $s\=p^2\+p\/2$ in $\mathbf{Q}\=\mathbf{R}\prime \left(s\right)$, then apply the simplify command to show equivalence with T obtained in Part (b).

$\mathrm{simplify}\left(\mathrm{eval}\left(\mathbf{Q}\,s\=p^2\+p\/2\right)\right) assuming p\>0$

$\left[\begin{array}{c}\frac{4p-1}{1+4p}\\ \frac{4\sqrt{p}}{1+4p}\end{array}\right]$