Lise vs. Steve

The distance that Lise has jogged is given by

$\varrho = \int_{1}^{4} \sqrt{1 + (f'(x))^{2}} = \int_{1}^{4} \sqrt{x} = \frac{14}{3}.$

Therefore, the time she took to do it is given by $t = \varrho / v = \frac{2\sqrt{3}}{3}$ being $v$ her velocity. Also, the distance that Steve has jogged is simply $d = \sqrt{(4-1)^{2} + (f(4)-f(1))^{2}} = \sqrt{21}$ . If we want him to get to $(4, f(4)$ together with Lise, he must do it in the same time as she does. Therefore his velocity must be

$V = \frac{d}{t} = \frac{\sqrt{21}}{\frac{2\sqrt{3}}{3}} = \frac{3}{2} \times 7 \cong 3.97. \square$

The distance that Lise must travel is given by the arc length formula: $s= \int_a^b \sqrt{1+f'(x)^2} \, dx$ $=\int_1^4 \sqrt{1+\sqrt{x-1}^2} \, dx$ $=\int_1^4 \sqrt{x} \, dx$ $=\left[\frac{2}{3}x^{3/2} \right]_1^4 = \frac{14}{3}.$ Her speed is $7/\sqrt{3}$ , so it takes her $t=\frac{14/3}{7/\sqrt{3}}= \frac{2}{\sqrt{3}}$ seconds to reach the endpoint. The distance that Steve must travel is given by the distance formula to be $D=\sqrt{(4-1)^2+(2\sqrt{3}-0)^2}=\sqrt{21}.$ So, the speed he must travel at is $v=\frac{D}{t}=\frac{\sqrt{21}}{2/\sqrt{3}}=\frac{3\sqrt{7}}{2} \approx 3.969 \ \mathrm{units/s}.$