Find T(t) and N(t) at the given point. x=cosh t, y=sinh t, z=t, t=ln 2

In this exercise we will find the unit tangent vector $\bold T(t)$ and the the unit normal vector $\bold N(t)$ for the given vector-valued function (given to us as components, but we can assemble it into $\bold r$):

$$\bold r(t) = \cosh t \bold i + \sinh t \bold j + t \bold k$$

at $t= \ln 2$.

Is the derivative of $\cosh t$ negative?

$$\begin{gathered} x\left( t \right) = \cosh t,{\text{ }}y\left( t \right) = \sinh t,{\text{ }}z\left( t \right) = t \\ \textcolor{#4257b2}{ {\text{Therefore}}} \\ {\mathbf{r}}\left( t \right) = \cosh t{\mathbf{i}} + \sinh t{\mathbf{j}} + t{\mathbf{k}} \\ \textcolor{#4257b2}{{\text{Find }}\left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\|{\text{ and }}{\mathbf{T}}\left( t \right)} \\ {\mathbf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\cosh t} \right]{\mathbf{i}} + \frac{d}{{dt}}\left[ {\sinh t} \right]{\mathbf{j}} + \frac{d}{{dt}}\left[ t \right]{\mathbf{k}} \\ {\mathbf{r}}'\left( t \right) = \sinh t{\mathbf{i}} + \cosh t{\mathbf{j}} + {\mathbf{k}} \\ \left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\| = \sqrt {{{\left( {\sinh t} \right)}^2} + {{\left( {\cosh t} \right)}^2} + {{\left( 1 \right)}^2}} \\ \left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\| = \sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 1} \\ \textcolor{#4257b2}{{\text{Use the identity sin}}{{\text{h}}^2}t = {\cosh ^2}t - 1} \\ \left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\| = \sqrt {{{\cosh }^2}t - 1 + {{\cosh }^2}t + 1} \\ \left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\| = \sqrt 2 \cosh t \\ {\text{ }}{\mathbf{T}}\left( t \right) = \frac{{{\mathbf{r}}'\left( t \right)}}{{\left\| {{\mathbf{r}}'\left( t \right){\text{ }}} \right\|}} = \frac{{\sinh t{\mathbf{i}} + \cosh t{\mathbf{j}} + {\mathbf{k}}}}{{\sqrt 2 \cosh t}} \\ {\text{ }}{\mathbf{T}}\left( t \right) = \frac{1}{{\sqrt 2 }}\tanh t{\mathbf{i}} + \frac{1}{{\sqrt 2 }}{\mathbf{j}} + \frac{1}{{\sqrt 2 }}\operatorname{sech} t{\mathbf{k}} \\ \textcolor{#4257b2}{ {\text{Find the unit tanget vector }}{\mathbf{T}}\left( t \right){\text{ at }}t = \ln 2} \\ {\text{ }}{\mathbf{T}}\left( {\ln 2} \right) = \frac{1}{{\sqrt 2 }}\tanh \left( {\ln 2} \right){\mathbf{i}} + \frac{1}{{\sqrt 2 }}{\mathbf{j}} + \frac{1}{{\sqrt 2 }}\operatorname{sech} \left( {\ln 2} \right){\mathbf{k}} \\ \boxed{{\mathbf{T}}\left( {\ln 2} \right) = \frac{3}{{5\sqrt 2 }}{\mathbf{i}} + \frac{1}{{\sqrt 2 }}{\mathbf{j}}\frac{4}{{5\sqrt 2 }}{\mathbf{k}}} \\ \textcolor{#4257b2}{{\text{Find }}\left\| {{\text{ }}{\mathbf{T}}'\left( t \right)} \right\|{\text{ }}} \\ {\text{ }}{\mathbf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt 2 }}\tanh t} \right]{\mathbf{i}} + \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt 2 }}} \right]{\mathbf{j}} + \frac{d}{{dt}}\left[ {\frac{1}{{\sqrt 2 }}\operatorname{sech} t} \right]{\mathbf{k}} \\ {\text{ }}{\mathbf{T}}'\left( t \right) = \frac{1}{{\sqrt 2 }}{\operatorname{sech} ^2}t{\mathbf{i}} - \frac{1}{{\sqrt 2 }}\operatorname{sech} t\tanh t{\mathbf{k}} \\ \textcolor{#4257b2}{ {\text{Evaluate }}\left\| {{\text{ }}{\mathbf{T}}'\left( t \right)} \right\|{\text{ at }}t = \ln 2} \\ {\text{ }}{\mathbf{T}}'\left( {\ln 2} \right) = \frac{1}{{\sqrt 2 }}{\operatorname{sech} ^2}\left( {\ln 2} \right){\mathbf{i}} - \frac{1}{{\sqrt 2 }}\operatorname{sech} \left( {\ln 2} \right)\tanh \left( {\ln 2} \right){\mathbf{k}} \\ {\text{ }}{\mathbf{T}}'\left( 0 \right) = \frac{{16}}{{25\sqrt 2 }}{\mathbf{i}} - \frac{{12}}{{25\sqrt 2 }}{\mathbf{k}} \\ \left\| {{\text{ }}{\mathbf{T}}'\left( 0 \right)} \right\| = \sqrt {{{\left( {\frac{{16}}{{25\sqrt 2 }}} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( { - \frac{{12}}{{25\sqrt 2 }}} \right)}^2}} = \frac{4}{{5\sqrt 2 }} \\ \textcolor{#4257b2}{{\text{Find }}{\mathbf{N}}\left( t \right){\text{ at }}t = \ln 2} \\ {\text{ }}{\mathbf{N}}\left( t \right) = \frac{{{\mathbf{T}}'\left( t \right)}}{{\left\| {{\mathbf{T}}'\left( t \right)} \right\|}} \Rightarrow {\text{ }}{\mathbf{N}}\left( {\ln 2} \right) = \frac{{{\mathbf{T}}'\left( {\ln 2} \right)}}{{\left\| {{\mathbf{T}}'\left( {\ln 2} \right)} \right\|}} = \frac{{\frac{{16}}{{25\sqrt 2 }}{\mathbf{i}} - \frac{{12}}{{25\sqrt 2 }}{\mathbf{k}}}}{{\frac{4}{{5\sqrt 2 }}}} \\ \boxed{{\text{ }}{\mathbf{N}}\left( {\ln 2} \right) = \frac{4}{5}{\mathbf{i}} - \frac{3}{5}{\mathbf{k}}} \\ \end{gathered}$$