Answer:
The arc-length is about 1.1953.
Step-by-step explanation:
We are given the equation:
[tex]y=3-x^2[/tex]
And we want to find the length of its arc from:
[tex]0\leq x\leq \sqrt3/2[/tex]
Recall that arc-length is given by the formula:
[tex]\displaystyle S = \int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2}\, dx[/tex]
By differentiating and substituting into the arc-length formula, we will acquire:
[tex]\displaystyle S=\int\limits^{\sqrt{3}/2}_0 {\sqrt{1+4x^2} \, dx[/tex]
To evaluate, we can use trigonometric substitution. Note that:
[tex]\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{(1)^2+(2x)^2} \, dx[/tex]
Since this is in the form a² + u², we will make the substitution u = atan(θ).
In this case, a = 1 and u = 2x. Thus:
[tex]\displaystyle 2x = \tan \theta[/tex]
Differentiating both sides with respect to x:
[tex]\displaystyle 2\, dx = \sec^2 \theta \, d\theta[/tex]
So:
[tex]\displaystyle dx = \frac{1}{2}\sec^2 \theta \, d\theta[/tex]
Additionally, we must rewrite our bounds. Hence:
[tex]\displaystyle 2(0) = \tan \theta \Rightarrow \theta = 0[/tex]
And:
[tex]\displaystyle 2\left(\frac{\sqrt{3}}{2}\right) = \tan\theta \Rightarrow \theta = \frac{\pi}{3}[/tex]
Thus:
[tex]\displaystyle S = \int_{0}^{\pi /3}\sqrt{1+(\tan\theta)^2}\cdot \frac{1}{2}\sec^2\theta\, d\theta[/tex]
Simplify:
[tex]\displaystyle S = \frac{1}{2}\int_{0}^{\pi /3}\sqrt{1+\tan^2\theta} \cdot \sec^2\theta d\theta[/tex]
Using trigonometric identities:
[tex]\displaystyle S = \frac{1}{2}\int_{0}^{\pi /3}\sqrt{(\sec^2\theta)} \cdot \sec^2\theta d\theta[/tex]
Simplify:
[tex]\displaystyle S = \frac{1}{2}\int_{0}^{\pi /3} \sec^3\theta \, d\theta[/tex]
We can apply the reduction formula:
[tex]\displaystyle \int \sec^nu \, du = \frac{\sec^{n-2}u\tan u}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2} u \, du[/tex]
Hence:
[tex]\displaystyle S = \frac{1}{2}\left(\frac{\sec \theta \tan\theta}{2}\Big|_{0}^{\pi /3} + \frac{1}{2} \int_0^{\pi /3} \sec \theta\, d\theta\right)[/tex]
This is a common integral:
[tex]\displaystyle S = \frac{1}{2}\left(\frac{\sec \theta \tan\theta}{2}+ \frac{1}{2} \left(\ln \left(\sec \theta + \tan\theta\right)\right)\Bigg|_{0}^{\pi /3} \right)[/tex]
Evaluate. Hence:
[tex]\displaystyle \begin{aligned} S = \frac{1}{2}\left[\left(\frac{\sec\dfrac{\pi}{3}\tan\dfrac{\pi}{3}}{2}+\frac{1}{2}\ln\left(\sec\dfrac{\pi}{3}+\tan\dfrac{\pi}{3}\right)\right) \\ - \left(\frac{\sec0\tan0}{2}+\frac{1}{2}\ln\left(\sec 0 +\tan 0\right)\right)\right] \end{aligned}[/tex]
Evaluate:
[tex]\displaystyle \begin{aligned} S &=\frac{1}{2} \left[ \left(\frac{(2)\left(\sqrt{3}\right)}{2} + \frac{1}{2}\ln \left(2 + \sqrt{3}\right) \right)-\left(\frac{(1)(0)}{2} + \frac{1}{2}\ln \left(1+0\right) \right) \right] \\ \\&= \frac{1}{2}\left(\sqrt{3} + \frac{1}{2} \ln \left(2+\sqrt{3}\right)\right) \\ \\ &\approx 1.1953 \end{aligned}[/tex]
Thus, the length of the arc is about 1.1953 units.
