Thread: Integrals: Arc Length and hydrostatic force problem

1. Integrals: Arc Length and hydrostatic force problem

Alright my first is to simply find the length of the curve of:

y = ln(sec x) from 0 to pi/4

the steps I have taken is the usual dy/dx = tan x

then

integral of sqrt(1 + (tan x)^2)

from here is where I am stuck. I've tried using u-substitution but then du would've been more complicated to calculate out and I don't know how to integrate the expression sqrt(1 + (tan x)^2). what should I proceed with?

my next problem is: a swimming pool 20ft wide, 40ft long, bottom is an inclined plane. shallow end is 3 ft long, deep end is 9ft long and pool is full of water. find hydrostatic force on one of the sides.

I started by drawing a picture of said pool then I made a plane with the x axes with origin at top of pool going down
then i make a strip of $\Delta x$.

this is where I am stuck as I am not sure as to which side I should be working towards finding pressure of (if it matters).

how should I proceed with this one? I'm looking at a few other answers and I'm guessing I should find the area of the strip, if so, how to properly do this? cause at some points the strip would be a rectangle but at some it'd be a triangle.

2. Originally Posted by deltemis
Alright my first is to simply find the length of the curve of:

y = ln(sec x) from 0 to pi/4

the steps I have taken is the usual dy/dx = tan x

then

integral of sqrt(1 + (tan x)^2)

from here is where I am stuck. I've tried using u-substitution but then du would've been more complicated to calculate out and I don't know how to integrate the expression sqrt(1 + (tan x)^2). what should I proceed with?

[snip]
Know your identities: $1 + \tan^2 x = \sec^2 x$.

3. Originally Posted by deltemis
Alright my first is to simply find the length of the curve of:

y = ln(sec x) from 0 to pi/4

the steps I have taken is the usual dy/dx = tan x

then

integral of sqrt(1 + (tan x)^2)

from here is where I am stuck. I've tried using u-substitution but then du would've been more complicated to calculate out and I don't know how to integrate the expression sqrt(1 + (tan x)^2). what should I proceed with?

my next problem is: a swimming pool 20ft wide, 40ft long, bottom is an inclined plane. shallow end is 3 ft long, deep end is 9ft long and pool is full of water. find hydrostatic force on one of the sides.

I started by drawing a picture of said pool then I made a plane with the x axes with origin at top of pool going down
then i make a strip of $\Delta x$.

this is where I am stuck as I am not sure as to which side I should be working towards finding pressure of (if it matters).

how should I proceed with this one? I'm looking at a few other answers and I'm guessing I should find the area of the strip, if so, how to properly do this? cause at some points the strip would be a rectangle but at some it'd be a triangle.
$1 + \tan^2{x} = \sec^2{x}$

so

$\sqrt{1 + \tan^2{x}} = \sec{x}$.

So the arc length is $\int{\sec{x}\,dx}$.

Can you go from here?

4. ah forgot about that identity, yep got that first one solved

not sure about that second problem