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# Arc Length and Surface Area

Finding the perimeter of arbitrary curves and the area of 3D shapes foils the traditional tools of Geometry, and calls for the help of integrals and derivatives to make these calculations.

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 \(\displaystyle \int \sqrt{a^2+u^2}\,du=\frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}\ln(u+\sqrt{a^2+u^2})+C.\)

(i) over any specified interval, the area between \(f(x)\) and the \(x\)-axis is equal to the arclength of the curve, and

(ii) \(f(0) = 1\).

If \(S = \displaystyle\int_{-1}^{2} f(x) dx\), then find \(\lfloor 1000S \rfloor\).

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