# Find the curvature of the space curve. (1) r(t) = (t + 10)i + 6j + (ln(\sec t) + 4)k (2) r(t) =...

## Question:

Find the curvature of the space curve.

(1) {eq}\displaystyle r(t) = (t + 10)i + 6j + (ln(\sec t) + 4)k {/eq}

(2) {eq}\displaystyle r(t) = \frac{28}{9}(1 + t)^{3/2}i + \frac{28}{9}(1 - t)^{3/2}j + \frac{7}{3}tk {/eq}

## Finding Curvature:

The curvature of the curve formula is {eq}\displaystyle \kappa = \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} {/eq}. The higher order derivatives are evaluated with respect to the variable. The norm of the derivative has to be evaluated using the square root of the sum of the squares .

(1) Consider the space curve {eq}\displaystyle r(t) = (t+10) \vec{i} + 6 \vec{j} + (ln(\sec t)+4) \vec{k} {/eq}.

Finding the curvature of the space curve:

{eq}\begin{align*} \displaystyle r'(t) &= 1 \vec{i} + 0 \vec{j} + \tan \left(t\right) \vec{k} \\ \displaystyle \left \| r'(t) \right \| &=\sqrt{(1)^{2}+(\tan \left(t\right))^{2}} \\ \displaystyle \left \| r'(t) \right \| &=\sec \left(t\right) \\ \displaystyle {r}''(t) &=0 \vec{i} + 0 \vec{j} + \sec ^2\left(t\right) \vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=\begin{vmatrix} i & j & k\\ 1 & 0 & \tan \left(t\right)\\ 0 & 0 & \sec ^2\left(t\right) \end{vmatrix} \\ \displaystyle &=((\sec ^2\left(t\right))(0)-(0)(\tan \left(t\right))) \vec{i} - ((\sec ^2\left(t\right))(1)-(0)(\tan \left(t\right)))\vec{j}+((0)(1)-(0)(0))\vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=0 \vec{i} -\sec ^2\left(t\right) \vec{j}+ 0 \vec{k} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{(-\sec ^2\left(t\right))^{2}} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sec ^2\left(t\right) \\ \displaystyle \kappa &= \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} \\ \displaystyle \kappa &= \frac{\sec ^2\left(t\right)}{(\sec \left(t\right))^{3}} \\ \displaystyle \kappa &=\frac{1}{\sec \left(t\right)} \\ \displaystyle \kappa &=\cos \left(t\right) \end{align*} {/eq}

The curvature of the space curve is {eq}\ \displaystyle \mathbf{\color{blue}{ \kappa =\cos \left(t\right) }} {/eq}.

(2) Consider the space curve {eq}\displaystyle r(t) = \frac{28}{9}(1+t)^{\frac{3}{2}} \vec{i} + \frac{28}{9}(1-t)^{\frac{3}{2}} \vec{j} + \frac{7}{3}t \vec{k} {/eq}.

Finding the curvature of the space curve:

{eq}\begin{align*} \displaystyle {r}'(t) &=\left( \frac{14\sqrt{t+1}}{3} \right) \vec{i} + \left( -\frac{14\sqrt{-t+1}}{3} \right) \vec{j} + \left( \frac{7}{3} \right) \vec{k} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{\left(\:\frac{14\sqrt{t+1}}{3}\:\right)^2+\left(\:-\frac{14\sqrt{-t+1}}{3}\:\right)^2+\left(\frac{7}{3}\right)^2} \\ \displaystyle \left \| {r}'(t) \right \| &=7 \\ \displaystyle {r}''(t) &=\left( \frac{7}{3\left(t+1\right)^{\frac{1}{2}}} \right) \vec{i} + \left( \frac{7}{3\left(-t+1\right)^{\frac{1}{2}}} \right) \vec{j} + 0 \vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=\begin{vmatrix} i & j & k\\ \frac{14\sqrt{t+1}}{3} & -\frac{14\sqrt{-t+1}}{3} & \frac{7}{3}\\ \frac{7}{3\left(t+1\right)^{\frac{1}{2}}} & \frac{7}{3\left(-t+1\right)^{\frac{1}{2}}} & 0 \end{vmatrix} \\ \displaystyle &=\left( \left( 0 \right)\left( -\frac{14\sqrt{-t+1}}{3} \right)-\left( \frac{7}{3\left(-t+1\right)^{\frac{1}{2}}} \right)\left( \frac{7}{3} \right) \right) \vec{i} - \left( \left( 0 \right)\left( \frac{14\sqrt{t+1}}{3} \right)-\left( \frac{7}{3\left(t+1\right)^{\frac{1}{2}}} \right)\left( \frac{7}{3} \right) \right) \vec{j} + \left( \left( \frac{7}{3\left(-t+1\right)^{\frac{1}{2}}} \right)\left( \frac{14\sqrt{t+1}}{3} \right)-\left( \frac{7}{3\left(t+1\right)^{\frac{1}{2}}} \right)\left( -\frac{14\sqrt{-t+1}}{3} \right) \right) \vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=\left( -\frac{49}{9\left(-t+1\right)^{\frac{1}{2}}} \right) \vec{i} +\left( \frac{49}{9\left(t+1\right)^{\frac{1}{2}}} \right) \vec{j} + \left( \frac{196}{9\left(t+1\right)^{\frac{1}{2}}\left(-t+1\right)^{\frac{1}{2}}} \right) \vec{k} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{\left(-\frac{49}{9\left(-t+1\right)^{\frac{1}{2}}}\right)^2+\left(\frac{49}{9\left(t+1\right)^{\frac{1}{2}}}\right)^2+\left(\frac{196}{9\left(t+1\right)^{\frac{1}{2}}\left(-t+1\right)^{\frac{1}{2}}}\right)^2} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\frac{49\cdot \:2^{\frac{1}{2}}}{3\left(\left(t+1\right)\left(-t+1\right)\right)^{\frac{1}{2}}} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\frac{49\cdot \:2^{\frac{1}{2}}}{3\left(-t^2+1\right)^{\frac{1}{2}}} \\ \displaystyle \kappa &= \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} \\ \displaystyle \kappa &= \frac{\frac{49\cdot \:2^{\frac{1}{2}}}{3\left(-t^2+1\right)^{\frac{1}{2}}}}{(7)^{3}} \\ \displaystyle \kappa &= \frac{\sqrt{2}}{21\sqrt{-t^2+1}} \end{align*} {/eq}

The curvature of the plane curve is {eq}\ \displaystyle \mathbf{\color{blue}{ \kappa= \frac{\sqrt{2}}{21\sqrt{-t^2+1}} }} {/eq}.