Question

A hawk flying at \({\rm{15\;m/s}}\)at an altitude of \({\rm{180\;m}}\)accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation\({\rm{y = 180 - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{45}}}}\) until it hits the ground, where\({\rm{y}}\)is its height above the ground and \({\rm{x}}\)is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

Short answer

Distance traveled by prey from time it is dropped until the time it hits the ground is\(209.1{\rm{ }}m\).

Step-by-step solution

Arc Length Function

Arc length function is useful to measure the arc length of a curve from a particular starting to any other point on the curve.

Arc length calculated as \(L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\)

Given parameters

Speed of a hawk is\(15m/s\), prey dropped from an altitude of\(180m\).

The parabolic trajectory of the falling prey is \(y = 180 - \frac{{{x^2}}}{{45}}\).

Calculating arc length

As prey dropped from\(180m\), i.e., \(x = 0\)

To find \(x\) for \(y = 0\), substitute \(y = 0\) in given equation

\(\begin{aligned}0 &= 180 - \frac{{{x^2}}}{{45}}\\\frac{{{x^2}}}{{45}} &= 180\\{x^2} &= 8100\\x &= \sqrt {8100} \\ &= 90\end{aligned}\)

Therefore, distance traveled by prey is the arc length from \(x = 0{\rm{ }}to{\rm{ }}90\).

As \(y = 180 - \frac{{{x^2}}}{{45}}\)

\(\frac{{dy}}{{dx}} = - \frac{2}{{45}}x\)

Compute arc length by formula \(L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\)

\(L = \int_0^{90} {\sqrt {1 + \frac{{{2^2}}}{{{{45}^2}}}{x^2}} } dx\)

Let \(x = \frac{{45}}{2}\tan t\;\;\; \to \;\;\;dx = \frac{{45}}{2}{\sec ^2}t{\rm{ }}dt\)

\(\begin{aligned}L &= \int_a^b {\sqrt {1 + \frac{{{2^2}}}{{{{45}^2}}}{{\left( {\frac{{45}}{2}\tan t} \right)}^2}} } \frac{{45}}{2}{\rm{ }}{\sec ^2}t{\rm{ }}dt\\ &= \frac{{45}}{2}\int_a^b {\sqrt {1 + \frac{{{2^2}}}{{{{45}^2}}}\left( {\frac{{{{45}^2}}}{{{2^2}}}{{\tan }^2}t} \right)} } {\sec ^2}t{\rm{ }}dt\\ &= \frac{{45}}{2}\int_a^b {\sqrt {1 + {{\tan }^2}t} } {\rm{ }}{\sec ^2}t{\rm{ }}dt\\ &= \frac{{45}}{2}\int_a^b {\sqrt {{{\sec }^2}t} } {\rm{ }}{\sec ^2}t{\rm{ }}dt\\ &= \frac{{45}}{2}\int_a^b {\sec {\rm{ }}} t{\rm{ }}{\sec ^2}t{\rm{ }}dt\\ &= \frac{{45}}{2}\int_a^b {{{\sec }^3}} t{\rm{ }}dt{\rm{ }}\_\_\_E{q^n}{\rm{ }}1\end{aligned}\)

Let \(I = \int_a^b {{{\sec }^3}} t{\rm{ }}dt\)

Integrating by parts,

\(\begin{aligned}u &= \sec t\\du &= \sec t \cdot \tan t{\rm{ }}dt\\and{\rm{ }}\\dv &= {\sec ^2}t{\rm{ }}dt\\v &= \tan t\end{aligned}\)

\(\begin{aligned}I &= \int_a^b {{{\sec }^3}} tdt\\I &= uv - \int v du\\I &= \sec t\tan t - \int {\tan } t\sec t\tan tdt\\I &= \sec t\tan t - \int {{{\tan }^2}} t\sec tdt\\I &= \sec t\tan t - \int {\left( {{{\sec }^2}t - 1} \right)} \sec tdt\\I &= \sec t\tan t - \int {{{\sec }^3}} tdt + \int {\sec } tdt{\rm{ }}\_\_\_E{q^n}{\rm{ }}2\end{aligned}\)

Equation 2 can be written as:

\(\begin{aligned}2I &= \sec t\tan t + \int {\sec } tdt\\I &= \frac{1}{2}\sec t\tan t + \frac{1}{2}\ln |\sec t + \tan t|\end{aligned}\)

Substitute value of I in equation 1,

\(L = \frac{{45}}{2}\left( {\frac{1}{2}\sec t\tan t + \frac{1}{2}\ln |\sec t + \tan t|} \right)_a^b\)

Calculating integral limits and rewriting arc length

Integral limits a and b written as,

\(\begin{aligned}a = \arctan \left( {\frac{2}{{45}}(0)} \right) = 0\\b = \arctan \left( {\frac{2}{{45}}(90)} \right) = \arctan 4\end{aligned}\)

Rewrite arc length by substituting values of a and b,

\(\begin{aligned}L &= \frac{{45}}{4}(\sec t\tan t + \ln |\sec t + \tan t|)_0^{\arctan 4}\\ &= \frac{{45}}{4}(\sec (\arctan 4)\tan (\arctan 4) + \ln |\sec (\arctan 4) + \tan (\arctan 4)| - (0 + 0))\\ &\approx 209.1\;m\end{aligned}\)

Therefore, distance traveled by prey from time it is dropped until the time it hits the ground is\(209.1{\rm{ }}m\).