Delivery Drone

Calculus Level 2

A delivery drone flying at constant speed $15 \text{ m/s}$ and constant height $2700 \text{ m}$ toward a destination drops its goods. If the trajectory of the falling goods until it hits the ground can be described by the equation $y=2700-\frac{x^2}{75},$ where $x$ is the horizontal distance it travels and $y$ is its height above the ground, what is the distance (not horizontal displacement) traveled by the goods until it hits the ground?

Note: You can use $\displaystyle \int \sqrt{a^2+u^2}\,du=\frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}\ln(u+\sqrt{a^2+u^2})+C.$

225\sqrt{145}+\frac{75}{4}\ln(12+\sqrt{145})

450\sqrt{37}+\frac{75}{2}\ln(12+\sqrt{37})

450\sqrt{145}+\frac{75}{2}\ln(12+\sqrt{145})

225\sqrt{37}+\frac{75}{4}\ln(12+\sqrt{37})

1 solution

Andrew Ellinor
Nov 17, 2015

In order to determine the distance that this projectile traveled, we'll use the arclength formula, given by $S = \displaystyle\int_{a}^{b}\sqrt{1 + \left(f'(x)\right)^2}dx.$ Here, because the projectile starts out at $x = 0$ and lands at $x = 450$ , our integral will be:

$\int_{0}^{450}\sqrt{1 + \left(-\frac{2x}{75}\right)^2}dx = \int_{0}^{450}\sqrt{1 + \frac{4x^2}{5625}}dx = \int_{0}^{450}\frac{2}{75}\sqrt{\frac{5625}{4} +x^2}dx$

Using the given integral in the note, this evaluates to:

$\frac{x}{75}\sqrt{\frac{5625}{4} +x^2} + \frac{5625}{8}\ln \left(x + \sqrt{\frac{5625}{4} +x^2}\right) \bigg|_{0}^{450} = 225\sqrt{145} + \frac{75}{4}\ln(12 + \sqrt{145}).\square$

Delivery Drone

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