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Derivatives questions

  1. Oct 16, 2005 #1
    1) I'm not really sure how to get this, i know the answers but i can't get them
    for the first range, f(x) is obviously 0.15x, but i can't get the next ones. The second range answer is 0.28x - 3295.50. can someone tell me how to get this answer?
    http://i2.photobucket.com/albums/y15/seiferseph/2.jpg

    2) i'm not quite sure how to find the limit when t aproaches infinity. i don't know the rules of exponents w/ infinity, the answer is 6.5, how do i get it?
    http://i2.photobucket.com/albums/y15/seiferseph/3.jpg

    3) how do i solve b? some hints would be perfect, thanks!
    http://i2.photobucket.com/albums/y15/seiferseph/1.jpg

    thanks again!
     
  2. jcsd
  3. Oct 16, 2005 #2

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    For 1), if you read the table, the end bit, where it says that the 0.28% applies to the amount over x, the person's income.

    Tax = 3,802.50 + 0.28(X- 25,350)
    Tax = 0.28x + 3,802.50 - 0.28*25,350
    Tax = 0.28x - 3,2950.50
    ===================
     
  4. Oct 16, 2005 #3

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    For 2),

    [tex]e^{-kt} = \frac{1}{e^{kt}}[/tex]

    where k is a positive number.

    As t increases, how does [tex]e^{kt}[/tex] vary ?
     
  5. Oct 16, 2005 #4
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    for2):

    lim(t->infinity) [6.5 -2 exp(-.035*t)
    = lim(t->infinity) (6.5 - 2/exp(.035*t))

    the limit of a constant is the constant, correct?
    Then,
    lim(t->infinity) 6.5 = 6.5

    -2* lim(t->infinity) 1/exp(0.035*t)
    What is the exp(infinity), look at the graph of the exponetial function, it should be infinity.

    What is 1/(a large#)? It should be a small number.

    And infinitly large number inverted is an infinitely small number; ie zero

    So,
    -2* lim(t->infinity) 1/exp(0.035*t) = -2*1/(infinity) = -2*0 = 0

    Putting the pieces together,
    lim(t->infinity) [6.5 -2 exp(-.035*t)
    = 6.5 - 2*lim(t->infinity) 1/exp(0.035*t)
    = 6.5 -2*0 = 6.5
     
  6. Oct 16, 2005 #5

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    For 3),
    You have the differential eqn,

    [tex]\frac{dH}{dt} = 175 - 0.35H[/tex]

    this becomes,

    [tex]\frac{dH}{0.35H - 175} = -dt[/tex]

    giving,

    [tex]\int \frac{dH}{0.35H - 175} = - \int dt[/tex]

    Can you solve this now ?
     
  7. Oct 16, 2005 #6
    oh i see now, i didn't get how to get the minus sign, i didn't understand that it was the amount over 25350, so x - 25350. thanks!

    oh i see, thanks!, so its basically just 6.5 - 2/infnity which obviously simplifies to 6.5, ugh i feel so stupid.

    i'm not sure how to solve it, it is the same form as Newton's law of cooling right? but i'm not sure how to sovle it
     
  8. Oct 16, 2005 #7

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    The DE is the same form as Newton's law of cooling.

    As regards solving it, have you integrated,

    [tex]\int \frac{dH}{0.35H - 175}[/tex]

    yet ?

    What did you get ?
     
    Last edited: Oct 16, 2005
  9. Oct 16, 2005 #8
    i see, i got it, thanks for all the help!
     
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