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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