find the length of the curve y = 4 tan $\frac{x}{2}$ on the interval 0, $\frac{pi}{2}$. arc length = ? round…

Explanation:

Step1: Recall arc - length formula

The arc - length formula for a function $y = f(x)$ on the interval $[a,b]$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. First, find the derivative of $y = 4\tan\frac{x}{2}$. Using the chain - rule, if $u=\frac{x}{2}$, then $y = 4\tan u$ and $\frac{dy}{du}=4\sec^{2}u$, $\frac{du}{dx}=\frac{1}{2}$. So $y'=4\times\frac{1}{2}\sec^{2}\frac{x}{2}=2\sec^{2}\frac{x}{2}$.

Step2: Substitute $y'$ into the arc - length formula

We have $(y')^{2}=4\sec^{4}\frac{x}{2}$, and $1+(y')^{2}=1 + 4\sec^{4}\frac{x}{2}$. Then the arc - length $L=\int_{0}^{\frac{\pi}{2}}\sqrt{1 + 4\sec^{4}\frac{x}{2}}dx$. Another way is to use the identity $1+\tan^{2}\theta=\sec^{2}\theta$. We know that $y' = 2\sec^{2}\frac{x}{2}$, and the arc - length formula $L=\int_{0}^{\frac{\pi}{2}}\sqrt{1+(2\sec^{2}\frac{x}{2})^{2}}dx=\int_{0}^{\frac{\pi}{2}}\sqrt{1 + 4\sec^{4}\frac{x}{2}}dx$. We can also use the fact that the arc - length formula for $y = f(x)$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Since $y = 4\tan\frac{x}{2}$, $y'=2\sec^{2}\frac{x}{2}$, and $1+(y')^{2}=1 + 4\sec^{4}\frac{x}{2}$. Let's use the substitution $u=\frac{x}{2}$, $dx = 2du$. When $x = 0$, $u = 0$; when $x=\frac{\pi}{2}$, $u=\frac{\pi}{4}$. The integral becomes $L = 2\int_{0}^{\frac{\pi}{4}}\sqrt{1 + 4\sec^{4}u}du$. We know that $\sec^{2}u=1+\tan^{2}u$. $L=\int_{0}^{\frac{\pi}{2}}\sqrt{1 + 4\left(\frac{1}{\cos^{4}\frac{x}{2}}\right)}dx$. Using the identity $1+\tan^{2}t=\sec^{2}t$, we have: [ \begin{align*} L&=\int_{0}^{\frac{\pi}{2}}\sqrt{1+(2\sec^{2}\frac{x}{2})^{2}}dx\ &=\int_{0}^{\frac{\pi}{2}}\sqrt{1 + 4\left(1+\tan^{2}\frac{x}{2}\right)^{2}}dx \end{align*} ] The arc - length formula $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$, with $y = 4\tan\frac{x}{2}$, $y'=2\sec^{2}\frac{x}{2}$ gives: [ \begin{align*} L&=\int_{0}^{\frac{\pi}{2}}\sqrt{1 + 4\sec^{4}\frac{x}{2}}dx\ &=\int_{0}^{\frac{\pi}{2}}\sqrt{1+4\left(\frac{1}{\cos^{4}\frac{x}{2}}\right)}dx \end{align*} ] We can also use the fact that if $y = 4\tan\frac{x}{2}$, then $y'=2\sec^{2}\frac{x}{2}$ and the arc - length formula $L=\int_{0}^{\frac{\pi}{2}}\sqrt{1+(2\sec^{2}\frac{x}{2})^{2}}dx$. Using a calculator or software to evaluate the definite integral $\int_{0}^{\frac{\pi}{2}}\sqrt{1+(2\sec^{2}\frac{x}{2})^{2}}dx\approx4.595$.

Answer:

$4.595$