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Where can ı fiind thomas calculus solution manual

  1. Nov 5, 2006 #1
    i am searching that book thomas calculus solution manual. is there anybody who can tell me from where can i doenload. thanks
     
  2. jcsd
  3. Nov 5, 2006 #2
    you really should do your homework....
     
  4. Nov 6, 2006 #3
    it is not for hw

    i would like to check my answers and look some solutions that i couldnt solved
     
  5. Nov 6, 2006 #4

    arildno

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    In your head. If you work hard enough
     
  6. Nov 6, 2006 #5
    thanks

    very very thanks for your useful tips.
     
  7. Nov 6, 2006 #6

    arildno

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    Well, how about this, then:
    BUY the damn manual instead of acting like a thief trying to filch it from the net somewhere. :grumpy:

    Shame on you!
     
  8. Nov 6, 2006 #7
    thanks for your useful tips. i am not a short of person that will make his hw's with such a illegal way.ok it could be stay.
     
  9. Nov 6, 2006 #8
    i couldnt find it in my country and i have not a credit card to buy it fron net
     
  10. Nov 6, 2006 #9
    ok i changed my opininon. could you help m with only one problem. i havent solve it. that is in the chapter 3.additional problem 28 ;
    i would try to write a summary of the problem : assume an ice cube retains its cubical shape as it melts. if we call edge lenth s its vlume is v=s^3 and the surface area is = 6*s^2. we also assume that the cube's volume decreases at a rate that is proportional to its surface area. in math terms : dv/dt=-k(6*s^2) ; assume that the cube lost 1/4 of its volume during the first hour and that the volume is Vo at t=0. how long will it take the ice cube to melt...

    i tried to solve that with integrating(as t goes to 0 V goes to 3V/4) and tried to find k but then k is going to be a strange value.also the chapter is about derivative applications i can not solve it with integral. then what should i do please help me with this problem...
     
  11. Nov 6, 2006 #10

    arildno

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    Okay, first of all:
    Here, it is smart to express the surface S in terms of the volume V:
    [tex]S=6V^{\frac{2}{3}}[/tex]
    Thus, the differential equation for the rate of change of the volume is:
    [tex]\frac{dV}{dt}=-6kV^{\frac{2}{3}}[/tex]
    This is a separable equation:
    [tex]\frac{dV}{V^{\frac{2}{3}}}=-6kdt[/tex]
    or, integrating both sides from t=0 and and t=T:
    [tex]3(V(T)^{\frac{1}{3}}-V(0)^{\frac{1}{3}})=-6kT[/tex]
    or simply, for arbitrary T:
    [tex]V(T)=(V(0)^{\frac{1}{3}}-2kT)^{3}[/tex]
    Now you should be able to do the last steps on your own!
     
  12. Nov 6, 2006 #11
    yes i know first put V(t)=3v/4 t=1 find k than put v(t)=0 put k and find t. this what i must do isn't it?. but this is a problem from derivative chapter. but we found the answer with integration is it true?
     
  13. Nov 6, 2006 #12
    one more question: i have found the answer t=1/(1-(3/4)^1/3))) is it true?
     
  14. Nov 6, 2006 #13

    mathwonk

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    i will solve your problems for $50 apiece.
     
  15. Nov 6, 2006 #14

    chroot

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    Come on, guys.

    - Warren
     
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