consider the following vector function. r(t) = <5√2t, e^5t, e^(-5t)>. (a) find the unit tangent and unit…

Explanation:

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

The derivative of $\mathbf{r}(t)=\langle5\sqrt{2}t,e^{5t},e^{- 5t}\rangle$ is $\mathbf{r}'(t)=\langle5\sqrt{2},5e^{5t},-5e^{-5t}\rangle$.

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

$|\mathbf{r}'(t)|=\sqrt{(5\sqrt{2})^2+(5e^{5t})^2+(-5e^{-5t})^2}=\sqrt{50 + 25e^{10t}+25e^{- 10t}}=\sqrt{25(e^{10t}+2 + e^{-10t})}=5\sqrt{(e^{5t}+e^{-5t})^2}=5(e^{5t}+e^{-5t})$.

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

$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\left\langle\frac{5\sqrt{2}}{5(e^{5t}+e^{-5t})},\frac{5e^{5t}}{5(e^{5t}+e^{-5t})},\frac{-5e^{-5t}}{5(e^{5t}+e^{-5t})}\right\rangle=\left\langle\frac{\sqrt{2}}{e^{5t}+e^{-5t}},\frac{e^{5t}}{e^{5t}+e^{-5t}},\frac{-e^{-5t}}{e^{5t}+e^{-5t}}\right\rangle$.

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

$\mathbf{T}'(t)=\left\langle\frac{-5\sqrt{2}(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2},\frac{5(e^{5t}+e^{-5t})^2-5e^{5t}(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2},\frac{5(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2}\right\rangle$.

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

$|\mathbf{T}'(t)|=\sqrt{\left(\frac{-5\sqrt{2}(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2}\right)^2+\left(\frac{5(e^{5t}+e^{-5t})^2-5e^{5t}(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2}\right)^2+\left(\frac{5(e^{5t}-e^{-5t})}{(e^{5t}+e^{-5t})^2}\right)^2}$. After simplification, $|\mathbf{T}'(t)|=\frac{5\sqrt{2}}{e^{5t}+e^{-5t}}$.

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

$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=\left\langle\frac{-(e^{5t}-e^{-5t})}{e^{5t}+e^{-5t}},\frac{(e^{5t}+e^{-5t})^2-e^{5t}(e^{5t}-e^{-5t})}{\sqrt{2}(e^{5t}+e^{-5t})},\frac{(e^{5t}-e^{-5t})}{\sqrt{2}(e^{5t}+e^{-5t})}\right\rangle$.

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

Since $|\mathbf{T}'(t)|=\frac{5\sqrt{2}}{e^{5t}+e^{-5t}}$ and $|\mathbf{r}'(t)| = 5(e^{5t}+e^{-5t})$, then $\kappa(t)=\frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}=\frac{\frac{5\sqrt{2}}{e^{5t}+e^{-5t}}}{5(e^{5t}+e^{-5t})}=\frac{\sqrt{2}}{(e^{5t}+e^{-5t})^2}$.

Answer:

$\mathbf{T}(t)=\left\langle\frac{\sqrt{2}}{e^{5t}+e^{-5t}},\frac{e^{5t}}{e^{5t}+e^{-5t}},\frac{-e^{-5t}}{e^{5t}+e^{-5t}}\right\rangle$ $\mathbf{N}(t)=\left\langle\frac{-(e^{5t}-e^{-5t})}{e^{5t}+e^{-5t}},\frac{(e^{5t}+e^{-5t})^2-e^{5t}(e^{5t}-e^{-5t})}{\sqrt{2}(e^{5t}+e^{-5t})},\frac{(e^{5t}-e^{-5t})}{\sqrt{2}(e^{5t}+e^{-5t})}\right\rangle$ $\kappa(t)=\frac{\sqrt{2}}{(e^{5t}+e^{-5t})^2}$