# The three-dimensional parametric curve C is defined by r(t)= \left \langle \cos (t), \sin (t),...

## Question:

The three-dimensional parametric curve C is defined by {eq}r(t)= \left \langle \cos (t), \sin (t), t^2 \right \rangle. {/eq}

Verify that the curvature is given by the formula {eq}k(t)= \sqrt{\frac{5 + 4t^2}{(4t^2 + 1)^3}} {/eq}

## Verifying the Curvature Value:

By using the given three-dimensional parametric curve, we have to verify that the given curvature value is true or false. So, we are going to use the curvature formula {eq}\displaystyle \kappa = \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} {/eq}. Derivatives and magnitude of the derivatives are the most important to produce the curvature.

## Answer and Explanation:

Given parametric curve is {eq}\displaystyle r(t)= \left \langle \cos (t), \sin (t), t^2 \right \rangle. {/eq}.

Verifying the curvature is given by the formula {eq}\displaystyle k(t)= \sqrt{\frac{5 + 4t^2}{(4t^2 + 1)^3}} {/eq}:

{eq}\begin{align*} \displaystyle {r}'(t) &= \left \langle -\sin \left(t\right), \cos \left(t\right), 2t \right \rangle \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(-\sin \left(t\right))^{2}+(\cos \left(t\right))^{2}+(2t)^{2}} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{4t^2+1} \\ \displaystyle {r}''(t) &=\left \langle -\cos \left(t\right), -\sin \left(t\right), 2 \right \rangle \\ \displaystyle {r}'(t) \times {r}''(t) &=\begin{vmatrix} i & j & k\\ -\sin \left(t\right) & \cos \left(t\right) & 2t\\ -\cos \left(t\right) & -\sin \left(t\right) & 2 \end{vmatrix} \\ \displaystyle &=((2)(\cos \left(t\right))-(-\sin \left(t\right))(2t))\vec{i}-((2)(-\sin \left(t\right))-(-\cos \left(t\right))(2t))\vec{j}+((-\sin \left(t\right))(-\sin \left(t\right))-(-\cos \left(t\right))(\cos \left(t\right)))\vec{k} \\ \displaystyle {r}'(t) \times {r}''(t) &=(2\cos \left(t\right)+2t\sin \left(t\right)) \vec{i}+ (2\sin \left(t\right)-2t\cos \left(t\right))\vec{j}+(1)\vec{k} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{(2\cos \left(t\right)+2t\sin \left(t\right))^{2}+(2\sin \left(t\right)-2t\cos \left(t\right))^{2}+(1)^{2}} \\ \displaystyle \left \| {r}'(t) \times {r}''(t) \right \| &=\sqrt{4t^2+5} \\ \displaystyle \kappa &= \frac{ \left \| {\vec{r}}'(t)\times {\vec{r}}''(t) \right \|}{ \left \| {\vec{r}}'(t) \right \|^{3}} \\ \displaystyle \kappa &= \frac{\sqrt{4t^2+5}}{\left( \sqrt{4t^2+1} \right)^{3}} \\ \displaystyle \kappa &=\sqrt{\frac{4t^2+5}{\left(4t^2+1\right)^3}} \end{align*} {/eq}

Hence, it is verified that the given curvature value is true.

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from UExcel Physics: Study Guide & Test Prep

Chapter 2 / Lesson 8