for each vector function r(t) below, compute the unit tangent vector t(t) at the given value of t. a) let…

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.

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.

Answer

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

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

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

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 $\mathbf{T}(t)$

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

Part B

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

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

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

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 $\mathbf{T}(t)$

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

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