Question

1. for the curves given below, please compute: i the unit tangent vector. ii the unit normal vector. iii the binormal vector. iv the curvature. v the tangential acceleration. vi the normal acceleration. you should use the “cheat sheet” and unless you want extra work, you should avoid “s” like the plague. (a) $vec{r}(t):=langlecos t,sin t,7

angle$. (b) $vec{r}(t):=langlecos t,sin t,2t
angle$. (c) $vec{r}(t):=langle1,t,t^{2}
angle$.

Explanation:

Step1: Recall relevant vector - valued function formulas

For a vector - valued function $\vec{r}(t)=\langle x(t),y(t),z(t)
angle$, the unit tangent vector $\vec{T}(t)=\frac{\vec{r}'(t)}{\vert\vec{r}'(t)\vert}$, the unit normal vector $\vec{N}(t)=\frac{\vec{T}'(t)}{\vert\vec{T}'(t)\vert}$, the binormal vector $\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)$, the curvature $\kappa(t)=\frac{\vert\vec{T}'(t)\vert}{\vert\vec{r}'(t)\vert}$, the tangential acceleration $a_T=\vec{r}''(t)\cdot\vec{T}(t)$ and the normal acceleration $a_N=\vert\vec{r}''(t)\times\vec{T}(t)\vert$.

Case (a): $\vec{r}(t)=\langle\cos t,\sin t,7

angle$

Step1: Find the first - derivative $\vec{r}'(t)$

$\vec{r}'(t)=\langle-\sin t,\cos t,0
angle$. Then $\vert\vec{r}'(t)\vert=\sqrt{(-\sin t)^2+\cos^2 t + 0^2}=1$.

Step2: Find the unit tangent vector $\vec{T}(t)$

$\vec{T}(t)=\frac{\vec{r}'(t)}{\vert\vec{r}'(t)\vert}=\langle-\sin t,\cos t,0
angle$.

Step3: Find the second - derivative $\vec{r}''(t)$

$\vec{r}''(t)=\langle-\cos t,-\sin t,0
angle$.

Step4: Find the unit normal vector $\vec{N}(t)$

$\vec{T}'(t)=\langle-\cos t,-\sin t,0
angle$, $\vert\vec{T}'(t)\vert = 1$, so $\vec{N}(t)=\langle-\cos t,-\sin t,0
angle$.

Step5: Find the binormal vector $\vec{B}(t)$

$\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-\sin t&\cos t&0\\-\cos t&-\sin t&0\end{vmatrix}=\langle0,0,1
angle$.

Step6: Find the curvature $\kappa(t)$

$\kappa(t)=\frac{\vert\vec{T}'(t)\vert}{\vert\vec{r}'(t)\vert}=1$.

Step7: Find the tangential acceleration $a_T$

$a_T=\vec{r}''(t)\cdot\vec{T}(t)=(-\cos t)(-\sin t)+(-\sin t)(\cos t)+0\times0 = 0$.

Step8: Find the normal acceleration $a_N$

$a_N=\vert\vec{r}''(t)\times\vec{T}(t)\vert=\vert\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-\cos t&-\sin t&0\\-\sin t&\cos t&0\end{vmatrix}\vert=\vert\langle0,0,-\cos^{2}t-\sin^{2}t
angle\vert = 1$.

Case (b): $\vec{r}(t)=\langle\cos t,\sin t,2t

angle$

Step1: Find $\vec{r}'(t)$

$\vec{r}'(t)=\langle-\sin t,\cos t,2
angle$, $\vert\vec{r}'(t)\vert=\sqrt{\sin^{2}t+\cos^{2}t + 4}=\sqrt{5}$.

Step2: Find $\vec{T}(t)$

$\vec{T}(t)=\frac{\vec{r}'(t)}{\vert\vec{r}'(t)\vert}=\langle-\frac{\sin t}{\sqrt{5}},\frac{\cos t}{\sqrt{5}},\frac{2}{\sqrt{5}}
angle$.

Step3: Find $\vec{r}''(t)$

$\vec{r}''(t)=\langle-\cos t,-\sin t,0
angle$.

Step4: Find $\vec{T}'(t)$

$\vec{T}'(t)=\langle-\frac{\cos t}{\sqrt{5}},-\frac{\sin t}{\sqrt{5}},0
angle$, $\vert\vec{T}'(t)\vert=\frac{1}{\sqrt{5}}$.

Step5: Find $\vec{N}(t)$

$\vec{N}(t)=\frac{\vec{T}'(t)}{\vert\vec{T}'(t)\vert}=\langle-\cos t,-\sin t,0
angle$.

Step6: Find $\vec{B}(t)$

$\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-\frac{\sin t}{\sqrt{5}}&\frac{\cos t}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\-\cos t&-\sin t&0\end{vmatrix}=\langle\frac{2\sin t}{\sqrt{5}},-\frac{2\cos t}{\sqrt{5}},-\frac{1}{\sqrt{5}}
angle$.

Step7: Find $\kappa(t)$

$\kappa(t)=\frac{\vert\vec{T}'(t)\vert}{\vert\vec{r}'(t)\vert}=\frac{1 / \sqrt{5}}{\sqrt{5}}=\frac{1}{5}$.

Step8: Find $a_T$

$a_T=\vec{r}''(t)\cdot\vec{T}(t)=(-\cos t)(-\frac{\sin t}{\sqrt{5}})+(-\sin t)(\frac{\cos t}{\sqrt{5}})+0\times\frac{2}{\sqrt{5}} = 0$.

Step9: Find $a_N$

$a_N=\vert\vec{r}''(t)\times\vec{T}(t)\vert=\vert\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-\cos t&-\sin t&0\\-\frac{\sin t}{\sqrt{5}}&\frac{\cos t}{\sqrt{5}}&\frac{2}{\sqrt{5}}\end{vmatrix}\vert=\sqrt{\frac{4\sin^{2}t}{5}+\frac{4\cos^{2}t}{5}+\frac{1}{5}} = 1$.

Case (c): $\vec{r}(t)=\langle1,t,t^{2}

angle$

Step1: Find $\vec{r}'(t)$

$\vec{r}'(t)=\langle0,1,2t
angle$, $\vert\vec{r}'(t)\vert=\sqrt{0 + 1+4t^{2}}$.

Step2: Find $\vec{T}(t)$

$\vec{T}(t)=\frac{\vec{r}'(t)}{\vert\vec{r}'(t)\vert}=\langle0,\frac{1}{\sqrt{1 + 4t^{2}}},\frac{2t}{\sqrt{1 + 4t^{2}}}
angle$.

Step3: Find $\vec{r}''(t)$

$\vec{r}''(t)=\langle0,0,2
angle$.

Step4: Find $\vec{T}'(t)$

$\vec{T}'(t)=\langle0,-\frac{4t}{(1 + 4t^{2})^{3/2}},\frac{2}{(1 + 4t^{2})^{3/2}}
angle$, $\vert\vec{T}'(t)\vert=\frac{2}{1 + 4t^{2}}$.

Step5: Find $\vec{N}(t)$

$\vec{N}(t)=\frac{\vec{T}'(t)}{\vert\vec{T}'(t)\vert}=\langle0,-\frac{2t}{\sqrt{1 + 4t^{2}}},\frac{1}{\sqrt{1 + 4t^{2}}}
angle$.

Step6: Find $\vec{B}(t)$

$\vec{B}(t)=\vec{T}(t)\times\vec{N}(t)=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0&\frac{1}{\sqrt{1 + 4t^{2}}}&\frac{2t}{\sqrt{1 + 4t^{2}}}\\0&-\frac{2t}{\sqrt{1 + 4t^{2}}}&\frac{1}{\sqrt{1 + 4t^{2}}}\end{vmatrix}=\langle1,0,0
angle$.

Step7: Find $\kappa(t)$

$\kappa(t)=\frac{\vert\vec{T}'(t)\vert}{\vert\vec{r}'(t)\vert}=\frac{2}{(1 + 4t^{2})^{3/2}}$.

Step8: Find $a_T$

$a_T=\vec{r}''(t)\cdot\vec{T}(t)=0\times0 + 0\times\frac{1}{\sqrt{1+4t^{2}}}+2\times\frac{2t}{\sqrt{1 + 4t^{2}}}=\frac{4t}{\sqrt{1 + 4t^{2}}}$.

Step9: Find $a_N$

$a_N=\vert\vec{r}''(t)\times\vec{T}(t)\vert=\vert\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0&0&2\\0&\frac{1}{\sqrt{1 + 4t^{2}}}&\frac{2t}{\sqrt{1 + 4t^{2}}}\end{vmatrix}\vert=\frac{2}{\sqrt{1 + 4t^{2}}}$.

Answer:

For (a):
i. $\vec{T}(t)=\langle-\sin t,\cos t,0
angle$
ii. $\vec{N}(t)=\langle-\cos t,-\sin t,0
angle$
iii. $\vec{B}(t)=\langle0,0,1
angle$
iv. $\kappa(t)=1$
v. $a_T = 0$
vi. $a_N = 1$

For (b):
i. $\vec{T}(t)=\langle-\frac{\sin t}{\sqrt{5}},\frac{\cos t}{\sqrt{5}},\frac{2}{\sqrt{5}}
angle$
ii. $\vec{N}(t)=\langle-\cos t,-\sin t,0
angle$
iii. $\vec{B}(t)=\langle\frac{2\sin t}{\sqrt{5}},-\frac{2\cos t}{\sqrt{5}},-\frac{1}{\sqrt{5}}
angle$
iv. $\kappa(t)=\frac{1}{5}$
v. $a_T = 0$
vi. $a_N = 1$

For (c):
i. $\vec{T}(t)=\langle0,\frac{1}{\sqrt{1 + 4t^{2}}},\frac{2t}{\sqrt{1 + 4t^{2}}}
angle$
ii. $\vec{N}(t)=\langle0,-\frac{2t}{\sqrt{1 + 4t^{2}}},\frac{1}{\sqrt{1 + 4t^{2}}}
angle$
iii. $\vec{B}(t)=\langle1,0,0
angle$
iv. $\kappa(t)=\frac{2}{(1 + 4t^{2})^{3/2}}$
v. $a_T=\frac{4t}{\sqrt{1 + 4t^{2}}}$
vi. $a_N=\frac{2}{\sqrt{1 + 4t^{2}}}$