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Thread: Calculus Notes

  1. #91
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA

    Fluid pressure

    Now, let's start on an interesting topic....fluid pressure.

    If we submerge a flat piece of, say, metal horizontally in water the pressure on it is straightforward. The force on it is $\displaystyle F={\rho}hA$.

    Where $\displaystyle {\rho}=\text{weight density of the fluid}$ and

    h is the depth and A is the surface area of the object.

    Pressure is defined as force per unit volume. Therefore, the pressure p exerted at each point of a flat surface of area A submerged horizontally at a depth h is given by $\displaystyle p=\frac{F}{A}={\rho}h$

    In the British system weight density is measured in pounds per cubic foot and pressure is measured in pounds per square foot.

    In metric, weight density is measured in newtons per cubic meter and pressure in newtons per square meter.

    One Newton per square meter is called a pascal

    One Pa is approximately 0.02089 pounds per square foot.

    The weight density of water is approximately 9810 newtons per cubic meter and 62.4 pounds per cubic foot in the British system.

    Here is a very trivial case:

    If a flat circular piece of metal of radius 2 meters is submerged horizontally in water so that the top surface is at a depth of 3 meters, then the force on top of the plate is:

    Since we are using metric, we use Newtons.

    $\displaystyle F={\rho}hA={\rho}h({\pi}r^{2})=(9810)(3)(4\pi)=117 ,720{\pi} \;\ N\approx{369,828} \;\ N$

    And the pressure at each point on the plate surface is

    $\displaystyle p={\rho}h=(9810)(3)=29,430 \;\ \frac{N}{m^{2}}=29,430 \;\ Pa$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Suppose we want to find the force on a vertical submerged surface. That is a little more involved than a simple submerged horizontal surface.

    Let's introduce a vertical x-axis whose positive direction is downward and whose origin is at any convenient point.

    Let's say the submerged portion of the surface extends from x=a to x=b on the x-axis. Also, suppose that a point x on the axis lies h(x)

    units below the surface and that the cross section of the plate at x has width w(x). Next, divide the interval [a,b] into n subintervals with length

    $\displaystyle {\Delta}x_{1}, \;\ {\Delta}x_{2}, \;\ ......., \;\ {\Delta}x_{n}$

    and that each subinterval choose an arbitrary point $\displaystyle x_{k}$. We approximate the section of the plate along the kth subinterval by a rectangle

    of length $\displaystyle w(x_{k})$ and width $\displaystyle {\Delta}x_{k}$.

    Because the upper and lower edges of the plate are at different depths we can't use the simple formula $\displaystyle F={\rho}hA$.

    But, if $\displaystyle {\Delta}x_{k}$ is small enough the difference in depth between the upper and lower edges is small and we can reasonably

    assume the entire rectangle is concentrated at a single depth $\displaystyle h(x_{k})$ below the surface.

    Therefore, we get: $\displaystyle F_{k}={\rho}h(x_{k})w(x_{k}){\Delta}x_{k}$

    Where $\displaystyle h(x_{k})$ is the depth and $\displaystyle w(x_{k}){\Delta}x_{k}$ is the area of the rectangle.

    Looks like a Riemann sum, huh?. Well, it is and that's where the integration comes in. We add up all those rectangles to find the total pressure.

    Thus, the total force on the plate is $\displaystyle F=\sum_{k=1}^{n}F_{k}=\sum_{k=1}^{n}{\rho}h(x_{k}) w(x_{k}){\Delta}x_{k}$

    To find the exact value of the force we take the limit as $\displaystyle {\Delta}x_{k}\rightarrow{0}$.

    This gives us $\displaystyle F=\int_{a}^{b}{\rho}h(x)w(x)dx$

    So, we have the formula for fluid force:

    Assume that a flat surface is submerged vertically in a liquid that has weight density $\displaystyle {\rho}$ and that the submerged portion

    extends from x=a to x=b on a vertical x-axis. For $\displaystyle a\leq{x}\leq{b}$, let w(x) be the width of the surface at x and let h(x) be the

    depth at point x. Then the total fluid force on the surface is $\displaystyle F=\int_{a}^{b}{\rho}h(x)w(x)dx$

    In the next thread we will begin some actual problems.
    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  2. #92
    GAMMA Mathematics
    colby2152's Avatar
    Nov 2007
    Alexandria, VA

    Lesson 5.5 - Substitution

    Lesson 5.5 - Substitution

    Many integrals can be tough to solve from the get go. Most of the time, the Calculus student will have to manipulate such integrands using some algebraic techniques, namely substitution. Substitution, also known as U-Substitution, is one of the most simple techniques to solve for an integral.

    Initially, you start off with an integral such as $\displaystyle \int f(x) dx$. When you cannot solve it outright, you will need to make a substitution where $\displaystyle u = u(x)$.

    You may use any variable here, but popular notation is to use the variable "u". Upon choosing your substitution, you must derive both sides of the equation with their respected variables like so: $\displaystyle du = u'(x) dx$

    The next step is to take these actions upon the original integral.

    $\displaystyle \int f(x) dx = \int f(x(u)) x'(u) du$

    After solving for the new integral, the next step is to substitute back the original parameters and solve for the limits if is a definite integral. Substitution, like many topics in math, is best shown through a series of examples...

    Example 5.5-A

    $\displaystyle \int \frac{x}{x^2+3} dx$

    What you need to recognize here are the powers. The derivative of $\displaystyle x^2 + 3$ is $\displaystyle 2x$ which is of the same form ($\displaystyle ax$) as the numerator of the integrand. We know what to do, so let's put our new found substitution technique into action!

    $\displaystyle u = x^2 + 3$

    $\displaystyle du = 2x dx $

    $\displaystyle \frac{du}{2} = x dx$

    We can now substitute these into the integral...

    $\displaystyle \int \frac{du}{2u}$

    $\displaystyle \frac{\ln|u|}{2} + C$

    One last step: The original integral is in the form of x, so we need to back substitute our parameters...

    $\displaystyle u = x^2 + 3$

    $\displaystyle \frac{\ln|u|}{2} + C = \frac{\ln|x^2+3|}{2} + C$

    Example 5.5-B

    $\displaystyle \int xe^{x^2} dx$

    Once again, identity the powers. The derivative of $\displaystyle x^2$ is $\displaystyle 2x$ which is of the same form of $\displaystyle x$.

    $\displaystyle u=x^2$

    $\displaystyle du=2x dx$

    $\displaystyle \int \frac{e^u}{2} du$

    $\displaystyle \frac{e^u}{2} + C = \frac{e^{x^2}}{2} + C$

    Now let's try substitution that isn't involving powers upon x...

    Example 5.5-C

    $\displaystyle \int (sin(x)+4)^6 cos(x) dx$

    $\displaystyle u = sin(x) + 4$

    $\displaystyle du = cos(x) dx$

    $\displaystyle \int u^6 du$

    $\displaystyle \frac{u^7}{7}+C$

    $\displaystyle \frac{(sin(x)+4)^7}{7}+C$
    Last edited by colby2152; Jun 20th 2008 at 09:45 AM.
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  3. #93
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    I almost forgot to continue my lesson on fluid pressure. I will post several problems as I get time.

    Let's start with a circular window on a submarine.

    A circular observation window on a submarine has a radius of 1 foot, and the center of the window is 200 feet below the water surface. What is the fluid force.

    We take advantage of our symmetry. So we will locate the origin at the center of the window.

    The depth at y is then $\displaystyle h(y) = 200-y$

    The horizontal length of the window is 2x, and you can use the equation

    for the circle $\displaystyle x^{2}+y^{2}=1$, to solve for x as follows:

    Length = $\displaystyle 2x = 2\sqrt{1-y^{2}} = L(y)$

    Now, because y varies from -1 to 1, and using 64 pounds per cubic foot as

    the weight density of seawater.

    $\displaystyle F=w\int_{a}^{b}h(y)L(y)dy$

    $\displaystyle =64\int_{-1}^{1}(200-y)(2)\sqrt{1-y^{2}}dy$

    At first glance this looks tough to solve. But if we break it in two and apply our symmetry we get:

    $\displaystyle F=25600\int_{-1}^{1}\sqrt{1-y^{2}}dy-128\int_{-1}^{1}y\sqrt{1-y^{2}}dy$

    The second integral is 0 because the integrand is odd and the limits are symmetric about the origin.

    But the first integral is the area of a semicircle of radius 1, so we have:

    $\displaystyle 25600(\frac{\pi}{2})=12800{\pi}\approx{40212.4} \;\ pounds$

    So, the fluid force on the window is about 40,212.4 pounds.

    Not that bad, huh?.

    No doubt, math like this is used when designing windows on subs or any other underwater windows.

    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  4. #94
    Newbie leonidas0101's Avatar
    Jun 2015

    Re: Calculus Notes

    In addition to my Calculus I notes, Galactus has offered to post some notes & shortcuts on topics in Calculus II & Calculus III ranging from Integration by Parts, Partial Differentiation, Multiple Integrals and LaGrange Multipliers.
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  5. #95
    Feb 2016

    Re: Calculus Notes

    How do I access the figures in the lessons ?

    Tony James
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  6. #96
    MHF Contributor

    Apr 2005

    Re: Calculus Notes

    Quote Originally Posted by topsquark View Post
    I would like to point out another notation which crops up from time to time and ends up confusing people:
    $\displaystyle f(x) + g(x) \rightarrow (f + g)(x)$

    $\displaystyle f(x) + g(x) \rightarrow (f - g)(x)$
    Now, that's really going to confuse people!

    $\displaystyle f(x)g(x) \rightarrow (fg)(x)$

    $\displaystyle \frac{f(x)}{g(x)} \rightarrow \left ( \frac{f}{g} \right ) (x)$

    and the typical notation for composites:
    $\displaystyle f(g(x)) \rightarrow (f \circ g)(x)$

    Thanks from topsquark
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