4. find the arclength of the curve r(t) = <9√(2t), e^(9t), e^(-9t)>, 0 ≤ t ≤ 1
4. find the arclength of the curve r(t) = <9√(2t), e^(9t), e^(-9t)>, 0 ≤ t ≤ 1
Answer
Explanation:
Step1: Find the derivative of each component
If $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle=\langle9\sqrt{2t},e^{9t},e^{- 9t}\rangle$, then $x(t)=9\sqrt{2t}=9\sqrt{2}t^{\frac{1}{2}}$, $y(t)=e^{9t}$, $z(t)=e^{-9t}$. $x^\prime(t)=\frac{9\sqrt{2}}{2\sqrt{t}}$, $y^\prime(t)=9e^{9t}$, $z^\prime(t)=-9e^{-9t}$.
Step2: Calculate the magnitude of $\mathbf{r}^\prime(t)$
The magnitude of $\mathbf{r}^\prime(t)$ is given by $\vert\mathbf{r}^\prime(t)\vert=\sqrt{(x^\prime(t))^{2}+(y^\prime(t))^{2}+(z^\prime(t))^{2}}$. $(x^\prime(t))^{2}=\frac{162}{4t}=\frac{81}{2t}$, $(y^\prime(t))^{2}=81e^{18t}$, $(z^\prime(t))^{2}=81e^{-18t}$. $\vert\mathbf{r}^\prime(t)\vert=\sqrt{\frac{81}{2t}+81e^{18t}+81e^{-18t}}=\sqrt{81\left(\frac{1}{2t}+e^{18t}+e^{-18t}\right)} = 9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}$.
Step3: Use the arc - length formula
The arc - length formula for a vector - valued function $\mathbf{r}(t)$ from $t = a$ to $t = b$ is $L=\int_{a}^{b}\vert\mathbf{r}^\prime(t)\vert dt$. Here, $a = 0$, $b = 1$. $L=\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt$. We know that $e^{18t}+e^{-18t}\geq2$ (by the AM - GM inequality $a + b\geq2\sqrt{ab}$ for $a = e^{18t}$ and $b=e^{-18t}$). $L=\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt=\int_{0}^{1}9\sqrt{\frac{1}{2t}+2\cosh(18t)}dt$. $\int_{0}^{1}9\sqrt{\frac{1}{2t}+2\cosh(18t)}dt=9\int_{0}^{1}\sqrt{\frac{1}{2t}+2\cosh(18t)}dt$. First, consider the integral $\int\sqrt{\frac{1}{2t}+2\cosh(18t)}dt$. We know that $\cosh(18t)=\frac{e^{18t}+e^{-18t}}{2}$. $\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt=9\int_{0}^{1}\sqrt{\frac{1}{2t}+\frac{e^{18t}+e^{-18t}}{2}+\frac{e^{18t}+e^{-18t}}{2}}dt$. $\int_{0}^{1}\vert\mathbf{r}^\prime(t)\vert dt=\int_{0}^{1}\sqrt{(x^\prime(t))^{2}+(y^\prime(t))^{2}+(z^\prime(t))^{2}}dt$ $=\int_{0}^{1}\sqrt{\frac{81}{2t}+81e^{18t}+81e^{-18t}}dt=\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt$. We calculate the integral: [ \begin{align*} L&=\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt\ &=9\int_{0}^{1}\sqrt{\frac{1}{2t}+2\cosh(18t)}dt \end{align*} ] [ \begin{align*} \int_{0}^{1}\sqrt{\frac{1}{2t}+2\cosh(18t)}dt&=\int_{0}^{1}\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt\ \end{align*} ] [ \begin{align*} \int_{0}^{1}\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt&=\int_{0}^{1}\sqrt{\left(\frac{9}{\sqrt{2t}}\right)^{2}+(9e^{9t})^{2}+(- 9e^{-9t})^{2}}dt \end{align*} ] [ \begin{align*} L&=\int_{0}^{1}\sqrt{(x^\prime(t))^{2}+(y^\prime(t))^{2}+(z^\prime(t))^{2}}dt\ &=\int_{0}^{1}9\sqrt{\frac{1}{2t}+e^{18t}+e^{-18t}}dt\ &=9\int_{0}^{1}\sqrt{\frac{1}{2t}+2\cosh(18t)}dt\ &=9\left[\frac{1}{9}\sinh(18t)-\sqrt{2}\sqrt{t}\right]_0^1\ &=\sinh(18)-\sqrt{2} \end{align*} ]