for the curve given by (r(t)=langle7sin(t),3t, - 7cos(t)\rangle), find the unit tangent (t(t)=langle\rangle)…

Explanation:

Step1: Find the derivative of $\mathbf{r}(t)$

Given $\mathbf{r}(t)=\langle7\sin(t),3t, - 7\cos(t)\rangle$. Then $\mathbf{r}'(t)=\langle7\cos(t),3,7\sin(t)\rangle$.

Step2: Calculate the magnitude of $\mathbf{r}'(t)$

$|\mathbf{r}'(t)|=\sqrt{(7\cos(t))^{2}+3^{2}+(7\sin(t))^{2}}=\sqrt{49\cos^{2}(t)+9 + 49\sin^{2}(t)}=\sqrt{49(\cos^{2}(t)+\sin^{2}(t))+9}=\sqrt{49 + 9}=\sqrt{58}$.

Step3: Find the unit - tangent vector $\mathbf{T}(t)$

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\left\langle\frac{7\cos(t)}{\sqrt{58}},\frac{3}{\sqrt{58}},\frac{7\sin(t)}{\sqrt{58}}\right\rangle$.

Step4: Find the derivative of $\mathbf{T}(t)$

$\mathbf{T}'(t)=\left\langle-\frac{7\sin(t)}{\sqrt{58}},0,\frac{7\cos(t)}{\sqrt{58}}\right\rangle$.

Step5: Calculate the magnitude of $\mathbf{T}'(t)$

$|\mathbf{T}'(t)|=\sqrt{\left(-\frac{7\sin(t)}{\sqrt{58}}\right)^{2}+0^{2}+\left(\frac{7\cos(t)}{\sqrt{58}}\right)^{2}}=\sqrt{\frac{49\sin^{2}(t)}{58}+\frac{49\cos^{2}(t)}{58}}=\sqrt{\frac{49(\sin^{2}(t)+\cos^{2}(t))}{58}}=\frac{7}{\sqrt{58}}$.

Step6: Find the unit - normal vector $\mathbf{N}(t)$

$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=\langle-\sin(t),0,\cos(t)\rangle$.

Step7: Calculate the curvature $\kappa(t)$

$\kappa(t)=\frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}=\frac{\frac{7}{\sqrt{58}}}{\sqrt{58}}=\frac{7}{58}$.

Answer:

$\mathbf{T}(t)=\left\langle\frac{7\cos(t)}{\sqrt{58}},\frac{3}{\sqrt{58}},\frac{7\sin(t)}{\sqrt{58}}\right\rangle$; $\mathbf{N}(t)=\langle-\sin(t),0,\cos(t)\rangle$; $\kappa(t)=\frac{7}{58}$