Question

A hawk flying at 19 m/s at an altitude of 165 m accidentally drops its prey.The parabolic trajectory of the falling prey is described by the equation y=165(x^2)/57
until it hits the ground, where y is its height above the ground and 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.

Accepted Answer

You need to evaluate the length of curve travelled by the prey from the time it is dropped until the time it hits the ground using the following formula, such that:
int_a^b sqrt(1 + (dy/dx)^2)dx
You need to find the limits of integration such that:
y = 0  when the pray hits the ground
y = 165  when the pray is dropped
You need to evaluate x  for y = 165  such that:
165(x^2)/57 = 165 => x^2/57 = 1 => x^2 = 57 => x = +-sqrt57
Hence, you need to evaluate the following definite integral, such that:
int_0^sqrt57 sqrt(1 + ((165(x^2)/57)')^2) dx
int_0^sqrt57 sqrt(1 + 5.7x^2) dx
You should come up with the following substitution such that:
5.7x = tan u => 5.7dx = 1/cos^2 u du = (1 + tan^2 u)du
dx = ((1 + tan^2 u)du)/5.7
Changing the variable yields:
int_0^(arctan 5.7sqrt57) sqrt(1 + tan^2 u)*(((1 + tan^2 u)du)/5.7)
int_0^(arctan 5.7sqrt57) sqrt (1/cos^2 u)*(((1 + tan^2 u)du)/5.7)
(1/5.7)int_0^(arctan 5.7sqrt57) 1/cos u*1/cos^2 u du
Since 1/cos u = sec u  yields:
(1/5.7)int_0^(arctan 5.7sqrt57) sec^3 u du = (1/5.7)(sec u*tan u + ln|sec u + tan u|)|_0^(arctan 5.7sqrt57)
Substituting back arctan(5.7x)  for u  yields:
int_0^sqrt57 sqrt(1 + 5.7x^2) dx = (1/5.7)(sec arctan(5.7x)*(5.7 x) + ln|secarctan(5.7x) + 5.7 x|)|_0^(sqrt57)
int_0^sqrt57 sqrt(1 + 5.7x^2) dx = (1/5.7)(sec arctan(5.7sqrt57)*(5.7sqrt57) + ln|sec arctan(5.7sqrt57) + 5.7 sqrt57|)
Hence, evaluating the length of curve travelled by the prey from the time it is dropped until the time it hits the ground yields int_0^sqrt57 sqrt(1 + 5.7x^2) dx = (1/5.7)(sec arctan(5.7sqrt57)*(5.7sqrt57) + ln|sec arctan(5.7sqrt57) + 5.7 sqrt57|).

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