Question

for each vector function r(t) below, compute the unit tangent vector t(t) at the given value of t. a) let r(t)=⟨cos 5t,sin 5t⟩. then t(π/4)=⟨ ⟩. b) let r(t)=⟨t²,t³⟩. then t(4)=⟨ ⟩. c) let r(t)=e^(5t)i + e^(-4t)j+tk. then t(-1)= i + j + k.

Explanation:

Step1: Recall the formula for the unit - tangent vector

The unit - tangent vector $\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$, where $\mathbf{r}'(t)$ is the derivative of the vector function $\mathbf{r}(t)$ and $\|\mathbf{r}'(t)\|$ is the magnitude of $\mathbf{r}'(t)$.

Part A

Let $\mathbf{r}(t)=\langle\cos(5t),\sin(5t)
angle$.

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

Differentiate each component: $\mathbf{r}'(t)=\langle - 5\sin(5t),5\cos(5t)
angle$.

Step2: Find $\|\mathbf{r}'(t)\|$

$\|\mathbf{r}'(t)\|=\sqrt{(-5\sin(5t))^{2}+(5\cos(5t))^{2}}=\sqrt{25\sin^{2}(5t) + 25\cos^{2}(5t)}=\sqrt{25(\sin^{2}(5t)+\cos^{2}(5t))}=5$.

Step3: Find $\mathbf{T}(t)$

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}=\langle - \sin(5t),\cos(5t)
angle$.
When $t = \frac{\pi}{4}$, $\mathbf{T}(\frac{\pi}{4})=\langle-\sin(\frac{5\pi}{4}),\cos(\frac{5\pi}{4})
angle=\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}
angle$.

Part B

Let $\mathbf{r}(t)=\langle t^{2},t^{3}
angle$.

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

Differentiate each component: $\mathbf{r}'(t)=\langle 2t,3t^{2}
angle$.

Step2: Find $\|\mathbf{r}'(t)\|$

$\|\mathbf{r}'(t)\|=\sqrt{(2t)^{2}+(3t^{2})^{2}}=\sqrt{4t^{2}+9t^{4}}=|t|\sqrt{4 + 9t^{2}}$.

Step3: Find $\mathbf{T}(t)$

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}=\frac{\langle 2t,3t^{2}
angle}{|t|\sqrt{4 + 9t^{2}}}$.
When $t = 4$, $\mathbf{r}'(4)=\langle 8,48
angle$, $\|\mathbf{r}'(4)\|=\sqrt{8^{2}+48^{2}}=\sqrt{64 + 2304}=\sqrt{2368}=8\sqrt{37}$.
$\mathbf{T}(4)=\frac{\langle 8,48
angle}{8\sqrt{37}}=\langle\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}
angle$.

Part C

Let $\mathbf{r}(t)=e^{5t}\mathbf{i}+e^{-4t}\mathbf{j}+t\mathbf{k}$.

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

Differentiate each component: $\mathbf{r}'(t)=5e^{5t}\mathbf{i}-4e^{-4t}\mathbf{j}+\mathbf{k}$.

Step2: Find $\|\mathbf{r}'(t)\|$

$\|\mathbf{r}'(t)\|=\sqrt{(5e^{5t})^{2}+(-4e^{-4t})^{2}+1^{2}}=\sqrt{25e^{10t}+16e^{-8t}+1}$.

Step3: Find $\mathbf{T}(t)$

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}=\frac{5e^{5t}\mathbf{i}-4e^{-4t}\mathbf{j}+\mathbf{k}}{\sqrt{25e^{10t}+16e^{-8t}+1}}$.
When $t=-1$, $\mathbf{r}'(-1)=5e^{-5}\mathbf{i}-4e^{4}\mathbf{j}+\mathbf{k}$, $\|\mathbf{r}'(-1)\|=\sqrt{25e^{-10}+16e^{8}+1}$.
$\mathbf{T}(-1)=\frac{5e^{-5}\mathbf{i}-4e^{4}\mathbf{j}+\mathbf{k}}{\sqrt{25e^{-10}+16e^{8}+1}}$.

Answer:

A. $\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}
angle$
B. $\langle\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}
angle$
C. $\frac{5e^{-5}\mathbf{i}-4e^{4}\mathbf{j}+\mathbf{k}}{\sqrt{25e^{-10}+16e^{8}+1}}$