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|).` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now