Time taken for travelling along a parabolic arc

The length of curve is given by

$\begin{aligned} s & = \int_a^b \sqrt {1+\left(\dfrac {dy}{dx}\right)^2} dx & \small \color{#3D99F6} \text{For }y=x^2 \text{ from }(-3,9) \to (3,9) \\ & = \int_{-3}^3 \sqrt{1+(2x)^2} dx & \small \color{#3D99F6} \text{Since the integral is even} \\ & = 2 \int_0^3 \sqrt{1+(2x)^2} dx & \small \color{#3D99F6} \text{Let }\tan \theta = 2x \implies \sec^2 \theta \ d\theta = 2 \ dx \\ & = \int_0^{\tan^{-1} 6} \sec^3 \theta \ d\theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = \tan \theta \sec \theta \bigg|_0^{\tan^{-1}6}- \int_0^{\tan^{-1}6} \tan^2 \theta \sec \theta \ d\theta \\ & = 6\sqrt{37} - {\color{#3D99F6}\int_0^{\tan^{-1}6} \sec^3 \theta \ d\theta} + \color{#D61F06} \int_0^{\tan^{-1}6} \sec \theta \ d\theta & \small \color{#3D99F6} \text{Note that }s = \int_0^{\tan^{-1}6} \sec^3 \theta \ d\theta \\ & = {\color{#3D99F6} \frac 12}\left(6\sqrt{37} + {\color{#D61F06} \int_0^{\tan^{-1}6} \frac {\sec^2 \theta + \tan \theta \sec \theta}{\sec \theta + \tan \theta} d\theta} \right) & \small \color{#D61F06} \text{Multiply up and down by }\sec \theta + \tan \theta \\ & = \frac 12 \left(6\sqrt{37} + \ln (\sec \theta + \tan \theta) \bigg|_0^{\tan^{-1}6} \right) \\ & = \frac 12 \left(6\sqrt{37} + \ln (\sqrt{37} + 6) \right) \\ & \approx {19.49417752} \end{aligned}$

Therefore the time to travel from $P$ to $Q$ is $t \approx \dfrac {19.49417752}3 \approx \boxed{6.498}$ .

The object traverses along a parabolic path from (-3,9) to (3,9). The total distance travelled by the object is the arc length of the parabola between (-3,9) and (3,9). The arc length can be calculated by the expression above.

The arc length evaluates to19.49 m, since the object is travelling at flxed speed of 3 m/s the time taken to cover a distance of 19.49 m is approc 19.49/3 = 6.5 seconds.