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caseclosed

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Posts posted by caseclosed

  1. This is from the book, need help changing it into mathematical model.

     

    "Consider a tank used for certain hydrodynamic experiments. After one experiment the tank contains 200 litres of a dye solution with a concentration of 1 g/litre. To prepare for the next experiment, the tank is rinsed with fresh water flowing in at a rate of 2 litres/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank readches 1% of its original value."

  2. [math]\int{\sec{x}^3dx}=F[/math]

    [math]u=Sec(x)[/math]

    [math]du=Sec(x)Tan(x)dx[/math]

    [math]dv=Sec(x)^2dx[/math]

    [math]v=Tan(x)[/math]

    [math]F=[/math][math]Sec(x)Tan(x)[/math] [math]\int{\tan{x}^2sec{x}dx}[/math]

    [math]F=[/math][math]Tan(x)^2=Sec(x)^2-1[/math]

    [math]F=[/math][math]Sec(x)Tan(x)[/math] [math]-\int{\sec{x}^3dx}[/math][math]+\int{sec{x}dx}[/math]

    [math]F=[/math][math]Sec(x)Tan(x)[/math] [math]-F[/math][math]+ln(sec(x)+tan(x))[/math]

    [math]2F=[/math][math]Sec(x)Tan(x)[/math] [math]+ln(sec(x)+tan(x))[/math]

    [math]F=[/math][math](Sec(x)Tan(x))/2[/math] [math]+(ln(sec(x)+tan(x))/2[/math]

     

    thank you very much for giving me the hint. I so forgot how to do this.

  3. yes, I tried setting u=(x^2+1)^(1/2) and dv=dx

    but I am not getting any log in the answer which means wrong.

    using the parts on sec(theta)^3 gets me nowhere, I change to (tan(x)^2+1)sec(x) and then set u=tan(x)^2 and dv=sec(x) dx which lead me to very complicated. I know I did this before but I don't remember how I did it.

  4. Yea quite right tree. Done a/s-level mathematics but that didnt have any of this kinda stuff in. I decided to do a degree in computing (but it has several components to it including related mathematics) while working so they could pay for it but because i work im doing it through distance learning so there isnt as much support as it would be attending university full time.

    oh, because it looks like the Calculus 2 stuff I did this year, thats why I was wondering.

  5. heh, I just found out what I forgot, I forgot to balance out the e by adding natural log. so it would be e^L=e^(ln(e^x+x)*(1/x)) and then by using properties of log and change L to quotient to get (e^x+x)/x so that it is form infinity/infinity and now using L Hopital's rule to get (1+1)/1 so it is 2 but 2=L so the answer would be e^2. :) woot...

  6. I got (2)/(n!(n+2)) so as n increases to infinity, the thing approaches 0 so that is why it is decreasing right? and less than 1 means it converges.

    but I don't understand where the upper/lower bound, what is it.... suppose to be a finite number? or is it like lower bound 0 since it is approaching 0 and upper bound unbounded?

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