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Differentiation Calclus

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Differentiate 00&eq=C%3D0.02\pi%20r^2%20%2B%20\frac{17}{r}%20%2B%200.06\pi%20r%20%2B\frac{12.75}{\pi%20r^2}%20.jpg

    2. Relevant equations

    see 1.

    3. The attempt at a solution

    2.php?z=100&eq=C'%3D0.04\pi%20x%20-\frac{17}{x^2}%20%2B%200.06\pi%20%2B%20\frac{25.5\pi}{r^3}%20.jpg
    btw the x in the second equation is meant to be an r
    how much did i fail? :(
     
    Last edited: Aug 24, 2010
  2. jcsd
  3. Aug 24, 2010 #2
    Why is [tex] \pi[/tex] in the numerator in the last term?
    The mislabelled "r" is just a mistake, let this be a lesson :wink: to keep an eye on what you're
    doing from now on :tongue: but please tell me why [tex] \pi[/tex] is in the numerator?
     
  4. Aug 24, 2010 #3
    i thought it was because original when it is brought up and r is put to the power of a negative it becomes a negative number...? probably wrong what is the right way to do it i always get mixed up with the order of how the terms and numbers go :(
     
  5. Aug 24, 2010 #4
    Lets find out for ourselves what actually happens to [tex] \pi[/tex] in this situation.
    For convenience lets set 12.75 = a so the last term in the equation becomes

    [tex] \frac{a}{\pi r^2} [/tex]

    Now, using the definition of the derivative [tex] f'(x) \ = \ \lim_{h \to 0} \ \frac{f(x \ + \ h) \ - \ f(x)}{h}[/tex]

    calculate the derivative of [tex] f(r) \ = \ \frac{a}{\pi r^2} [/tex]

    I'll start you off

    [tex] f(r \ + \ h) \ - \ f(r)} \ = \ (\frac{a}{\pi (r \ + \ h)^2} \ - \ \frac{a}{\pi r^2})[/tex]

    EDIT: Sorry, I messed up by skipping steps, divide through by "h" on both sides at the end :tongue:
     
  6. Aug 24, 2010 #5
    latex2png.2.php?z=100&eq=C'%3D0.04\pi%20r%20-(17r^-2)%20%2B%200.06\pi%20-%20(25.5\pi%20r^-3).jpg

    is that the correct answer? i'm sorry i havent answered your question, i haven't done the limits or the first principles in 2 years :(

    also note that the -2 and -3 are meant to be in superscript as powers but i dunno how to work latex like awesomesauce yet
     
  7. Aug 24, 2010 #6
    Everything except the [tex] \pi[/tex] is correct, you need to understand why what
    you've written is incorrect.

    If you just cross multiply the right hand side of the last latex message I wrote & then get it looking pretty, you're almost there. All you have to do then is divide both sides by h
    & then take a limit & you'll have an answer.
     
  8. Aug 24, 2010 #7
    okay i'll give it a shot
     
  9. Aug 24, 2010 #8
    latex2png.2.php?z=100&eq=\frac{\frac{a}{\pi%20r^2%20%2B%20\pi%20h^2}-\frac{a}{\pi%20r^2}}{h}%20.jpg

    this is what i got for the next step

    i can't seem to cross multiply or go any further ... ughhhhh
     
  10. Aug 24, 2010 #9
    [itex] f(r \ + \ h) \ - \ f(r)} \ = \ (\frac{a}{\pi (r \ + \ h)^2} \ - \ \frac{a}{\pi r^2}) \ = \ \frac{r^2}{r^2} \cdot \frac{a}{\pi (r \ + \ h)^2} \ - \ \frac{(r + h)^2}{(r + h)^2} \cdot \frac{a}{\pi r^2} \ = \ \frac{ar^2}{\pi r^2(r + h)^2} \ - \ \frac{a(r + h)^2}{\pi r^2 (r + h)^2}[/itex]

    [itex] f(r \ + \ h) \ - \ f(r)} \ = \ \frac{ar^2 \ - \ a(r + h)^2}{\pi r^2 (r + h)^2}[/itex]

    Can you keep going from here?

    If you look at the equation you'll see that I multiplied both of the crazy fractions by terms that are, in reality, only equal to 1. Doing this is smart because you're allowed to multiply a number by 1, even if it's written as [tex] \frac{2}{2} [/tex] or even [tex] \frac{(r + h)^2}{(r + h)^2} [/tex] :wink:

    It's the same thing as "cross multiplying" except this way makes sense as opposed to some magical trick :tongue:
     
  11. Aug 24, 2010 #10
    hmmm okay lemme see
     
  12. Aug 24, 2010 #11
    png.2.php?z=100&eq=\frac{a(r^2%20-%20(r%2Bh)^2)}{\pi%20r^2(r%2Bh)^2}%20%0A%3D%20\frac{a}{\pi}%20.jpg

    yeah? in this case a is 12.75 so pi stays underneath in the denominator? xD????
     
  13. Aug 24, 2010 #12
    lol, I don't think you remember this :tongue:

    www.khanacademy.org has great videos on how to do all of this, you should check them out.

    All you have to do here is to expand the a(r + h)² term on top of the fraction, then a lot
    of things cancel out by minus signs etc... Then, you're left with a new weird fraction.

    Set that up, then as the definition of the derivative implies you divide both sides by "h".

    More stuff will cancel out, this happens every time you take a derivative!!!

    All that is left is to take the limit as h-->0 on both sides & you'll have an answer.

    I really suggest going to watch the videos at that site &/or getting a good book like
    Swokowski's calculus (old edition is 20 cent on amazon & it's amazing!!!).
     
  14. Aug 24, 2010 #13
    latex2png.2.php?z=100&eq=\frac{-2ahr%20-%20ah^2}{\pi%20r^2(r%2Bh)^2}%20%0A.jpg

    so this is the "weird" new fraction?
     
  15. Aug 24, 2010 #14
    Yes, well done! Personally I think you should write everything out fully all the time so that
    you know exactly what you're doing & can see the logic to all of it.

    I'll just show you what I'm talking about


    [itex] f(r \ + \ h) \ - \ f(r)} \ = \ \frac{ar^2 \ - \ ar^2 \ - \ 2arh \ - ah^2}{\pi r^2 (r + h)^2}[/itex]


    [itex] f(r \ + \ h) \ - \ f(r)} \ = \ \frac{- \ 2arh \ - ah^2}{\pi r^2 (r + h)^2}[/itex]

    [itex] \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2arh \ - ah^2}{ h \pi r^2 (r + h)^2}[/itex]

    [itex] \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2ar \ - ah}{ \pi r^2 (r + h)^2}[/itex]

    You see, everything is logically laid out so that you can't make a mistake, try to get things
    written out this way until you can do all of this in your head.

    All you have to do is to take the limit of both sides now

    [itex] \lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \lim_{h \to 0} \ \frac{- \ 2ar \ - ah}{ \pi r^2 (r + h)^2}[/itex]

    See!!! On the left hand side you have the definition of the derivative clearly shown!
    On the right you need to send h to 0 & look, a term dissappears & the h in the denominator
    also dissappears so that you get a clean answer.

    [itex] \lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \lim_{h \to 0} \ \frac{- \ 2ar}{ \pi r^2 r^2}[/itex]

    [itex] \lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2ar}{ \pi r^4}[/itex]

    [itex] \lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 2a}{ \pi r^3}[/itex]

    a = 12.75

    [itex] \lim_{h \to 0} \ \frac{f(r \ + \ h) \ - \ f(r)}{h} \ = \ \frac{- \ 25.5}{ \pi r^3}[/itex]


    Check out the videos, they'll give you a lot of good explanations for all of this & I do advise
    getting that cheap book, it's better than Stewart, Thomas etc...
     
  16. Aug 24, 2010 #15
    you sir are a legend and yes actually funnily enough my modern history teacher recommended this site to me when i had spares today, she said it was epic with calculus etc so obviously she wasn't bsing lol xD thanks man i fully see it now (btw i did have it written down in lined book while i was doing it i just really couldn't remember xD)
     
  17. Aug 24, 2010 #16
    lol cool, no problem :wink:
     
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