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anonymous

  • one year ago

A certain college graduate borrows 6038 dollars to buy a car. The lender charges interest at an annual rate of 12 %. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 7 years. Also determine how much interest is paid during the 7-year period.

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  1. zepdrix
    • one year ago
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    So we use that one equation, \(\large\rm A=Pe^{rt}\) for continuous compounding, ya? We start with a principle loan of \(\rm P=6038\) and an interest rate of \(\rm r=0.12\).\[\large\rm A=6038e^{.12t}\] At \(\rm t=7\) years, we want the amount \(\rm A\) to be fully paid off, so \(\rm A=0\). And we're making these payments of k... hmm how do we make that happen :) Thinking...

  2. zepdrix
    • one year ago
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    If we're making this payment \(\rm k\) once every year, then after the first year we will subtract \(\rm k\) from the total amount that we owe,\[\large\rm A_1=6038e^{.12(1)}-k\]But then when we go to calculate next years stuff... it's going to be based on this new amount :o Hmm this is tricky...

  3. zepdrix
    • one year ago
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    Err wait, is interest always calculated from the principle? I always forget how that works. Ya ya that would make a lot more sense.

  4. zepdrix
    • one year ago
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    So after the 7th year, we will have made 7 of these k payments.\[\large\rm 0=6038e^{.12* 7}-7k\]So I plugged in 0 for the amount owed. We've fully paid off the loan in 7 years. I plugged in t=7, because this is the specific time that we're interested in, 7 years after the start of the loan. And I subtracted 7 equal amounts, k, from the amount that we owe.

  5. zepdrix
    • one year ago
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    Solve for k! :)

  6. anonymous
    • one year ago
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    Okay, so I got 1998.03 as k, but that is not the correct answer.. :(

  7. zepdrix
    • one year ago
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    It's not? :[ Hmmmm

  8. zepdrix
    • one year ago
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    @ganeshie8 @zzr0ck3r @dan815 How do these compound interest questions work? Is interest calculated off of the principle amount? Or is it being recalculated after each time you make a payment?

  9. zepdrix
    • one year ago
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    principal* amount... bahh I think I've been spelling that wrong the entire time lol

  10. anonymous
    • one year ago
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    Hmm, your steps made sense, so I'm not sure what's wrong.

  11. ganeshie8
    • one year ago
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    @zepdrix shouldn't the interest be calculated on the previous year's outstanding balance ?

  12. anonymous
    • one year ago
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    I looked in my diff eq text and the model that they introduced involving interest rate is\[\frac{ dP }{ dt }=rP(t)+b\] where b is the constant rate of adding/withdrawing. But this didn't make much sense to me and where to go after that.

  13. ganeshie8
    • one year ago
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    Outstanding balance at the end of year 1 : \[\rm 6038e^{.12} -k\] Outstanding balance at the end of year 2 : \[\rm (6038e^{.12} -k)e^{.12} - k =6038e^{2*.12} -ke^{.12}-k \] Outstanding balance at the end of year 3 : \[\rm ((6038e^{.12} -k)e^{.12} - k)e^{.12} -k = 6038e^{3*.12} -ke^{2*.12}-ke^{.12}-k \]

  14. zepdrix
    • one year ago
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    I dunno how interest works :) lol Is it always based off of the current balance? I guess that makes sense :p

  15. anonymous
    • one year ago
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    This is the solution I found to a similar problem elsewhere, but it didn't make much sense to me

  16. ganeshie8
    • one year ago
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    in that solution, \(S(t)\) represents the outstanding balance at any given time

  17. anonymous
    • one year ago
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    I see that they used the method of integrating factor to solve the solution. But after that, I'm not sure what they did. And I'm not sure why they used t=3

  18. ganeshie8
    • one year ago
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    Firstly, notice that two things affect the change in outstanding balance \(S(t)\) : 1) outstanding balance itself (S) 2) payments made (k)

  19. ganeshie8
    • one year ago
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    If you're not making any payments, there will be a positive change in the outstanding balance (growth) : \[\dfrac{dS}{dt} = rS\] However if you do make payments "continuously" (practically impossible) so that "k" dollars is being paid yearly, the outstanding balance will be reduced : \[\dfrac{dS}{dt} = rS-k\]

  20. ganeshie8
    • one year ago
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    it is an ordinary "separable" differential equation, you can solve it using any of the tricks that you're familiar with. you don't really need integrating factor..

  21. ganeshie8
    • one year ago
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    Once you have the solution, the function S(t) represents the outstanding balance. For the loan to be completely paid off in \(3\) years, \(S(3)\) must equal \(0\). Set that equal to \(0\) and solve \(k\)

  22. ganeshie8
    • one year ago
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    And yes, in that solution, they want to pay off the loan in \(3\) years. thats the reason they are setting S(3) equal to 0

  23. anonymous
    • one year ago
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    Ah for some reason that didn't make sense at first when looking at it, but I understand it now! Do you mind if I ask you another question regarding a differential equation problem that I'm having difficulty with?

  24. ganeshie8
    • one year ago
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    I can try, but just so you know, zepdrix is the master of differential equations around here..

  25. anonymous
    • one year ago
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    A skydiver weighing 232 lb (including equipment) falls vertically downward from an altitude of 6 000 ft and opens the parachute after 12 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is 0.77 vertical line v vertical line when the parachute is closed and 14 vertical line v vertical line when the parachute is open, where the velocity v is measured in ft/s. Use g equals 32 ft/s2. Round your answers to two decimal places. (a) Find the speed of the skydiver when the parachute opens. v(12) = ?? (b) Find the distance fallen before the parachute opens. x(12) = ?? (c) What is the limiting velocity v Subscript Upper L after the parachute opens? VL = ?? I know that the model I should use is similar to\[m \frac{ dv }{ dt }=mg-\gamma v\] But I don't know where to go from there..

  26. ganeshie8
    • one year ago
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    so the air resistance in the interval \((0, 12)\) is \(0.77\overline{v}\) afterwards, the air resistance is \(14\overline{v}\)

  27. anonymous
    • one year ago
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    I apologize, it's suppose to say |v|, not vertical line. But when I copied some of the question, it transferred weird.

  28. ganeshie8
    • one year ago
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    Thats okay, vertical lines just indicate the magnitude

  29. ganeshie8
    • one year ago
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    so the air resistance in the interval \((0, 12)\) is \(0.77|v|\) afterwards, the air resistance is \(14|v|\)

  30. ganeshie8
    • one year ago
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    Notice that part \(a\) and \(b\) concern with the interval \((0, 12)\) so simply solve the differential equation \(mv' = mg - 0.77v\)

  31. ganeshie8
    • one year ago
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    For parts \(a\) and \(b\) : \[232\dfrac{dv}{dt} = 232*32-0.77v\]

  32. ganeshie8
    • one year ago
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    solving should be easy, once you have the solution, \(v(t)\), simply plugin \(t = 12\) for part \(a\) we will see how to work part \(b\) after that

  33. anonymous
    • one year ago
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    This is the general solution that I got. Is this correct? \[v=9696.96+\frac{ c }{ e^{.0033t} }\]

  34. ganeshie8
    • one year ago
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    solve \(c\) using the fact that the body is "free falling"

  35. ganeshie8
    • one year ago
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    what do you know about the initial velocity of a free falling body ?

  36. anonymous
    • one year ago
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    It's 0 initially

  37. ganeshie8
    • one year ago
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    right, \(v(0) = 0\) use that to find the value of \(c\)

  38. anonymous
    • one year ago
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    So C=-9696.96 where the negative indicates the downward orientation

  39. ganeshie8
    • one year ago
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    looks good, wolfram says \(v(t) = \frac{742400}{77} \left(1-e^{-77t/23200}\right)\) http://www.wolframalpha.com/input/?i=solve+232v%27%28t%29+%3D+232*32+-+77%2F100v%28t%29%2C+v%280%29%3D0

  40. anonymous
    • one year ago
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    That doesn't look like the same thing that I got though.. =/

  41. ganeshie8
    • one year ago
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    they are same, you have converted everything to decimals, wolfram didn't

  42. anonymous
    • one year ago
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    Okay, so then plug in 12 for t and that will give me the velocity needed for a

  43. ganeshie8
    • one year ago
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    Yes. use the wolfram answer, looks you have rounded a lot, so your numbers are way off..

  44. ganeshie8
    • one year ago
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    not a big deal if you hadn't rounded, you would have gotten the exact same answer as wolfram..

  45. anonymous
    • one year ago
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    Okay, so I got is 376.45

  46. ganeshie8
    • one year ago
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    v(12) = 376.45 looks good

  47. anonymous
    • one year ago
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    Okay, so from there how do I calculate height? I could use the kinematic equations from physics 1, but I think they want me to use their methods.

  48. anonymous
    • one year ago
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    Distance*

  49. ganeshie8
    • one year ago
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    what do you know about the relation between "velocity" and "displacement" ?

  50. anonymous
    • one year ago
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    \[\frac{ dx }{ dt }=v\]

  51. ganeshie8
    • one year ago
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    \[x(12) - x(0) = \int\limits_0^{12} v(t)\, dt\]

  52. ganeshie8
    • one year ago
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    plugin \(v(t)\) and evaluate the definite integral

  53. ganeshie8
    • one year ago
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    \[x(12)-x(0) = \int\limits_0^{12}\frac{742400}{77} \left(1-e^{-77t/23200}\right)\, dt\]

  54. ganeshie8
    • one year ago
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    evaluating is trivial don't let the messy numbers confuse you

  55. anonymous
    • one year ago
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    Okay, I got 115698.2487

  56. anonymous
    • one year ago
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    Which doesn't make sense since he starts at an altitude of 6,000ft...

  57. ganeshie8
    • one year ago
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    doesn't look correct wolfram says \( x(12)-x(0) \approx 2274\) http://www.wolframalpha.com/input/?i=+%5Cint_0%5E%7B12%7D+%5Cfrac%7B742400%7D%7B77%7D*%5Cleft%281-e%5E%7B-77t%2F23200%7D%5Cright%29%5C%2C+dt

  58. anonymous
    • one year ago
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    Damn. I'm trying to do it by hand because I know that we're not allowed to use a calculator but I don't think something like this will be expected with such odd numbers. That makes more sense though haha

  59. Astrophysics
    • one year ago
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    They probably don't expect you to put it in decimals

  60. anonymous
    • one year ago
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    That's also true.

  61. anonymous
    • one year ago
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    So far, (a) 376.45 and (b) 2274 are wrong :(

  62. ganeshie8
    • one year ago
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    both wrong ?

  63. anonymous
    • one year ago
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    Yeah..

  64. ganeshie8
    • one year ago
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    how many more attempts do u have

  65. anonymous
    • one year ago
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    2

  66. Astrophysics
    • one year ago
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    Check v(12) again

  67. Astrophysics
    • one year ago
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    Hmm...maybe they want a rounded answer? Significant digits?

  68. anonymous
    • one year ago
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    If I use wolfram alpha's equation \[v(t)=\frac{ 742400 }{ 77 }(1-e^{-77t/23200})\] then v(12) would give me 376.45

  69. ganeshie8
    • one year ago
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    Yes, I don't see how part a could be anything different from \(376.4536197750639610724810777669453859565565074584767314312593...\)

  70. anonymous
    • one year ago
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    lol

  71. anonymous
    • one year ago
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    Oh, maybe it's suppose to be negative? But then again, part (b) is still wrong too..

  72. ganeshie8
    • one year ago
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    speed cannot be negative

  73. anonymous
    • one year ago
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    True -- I was thinking of velocity.

  74. ganeshie8
    • one year ago
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    part a was about speed right

  75. anonymous
    • one year ago
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    Find the speed of the skydiver when the parachute opens.

  76. ganeshie8
    • one year ago
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    I'll stick to \(v(12) = 376.45\) for part \(a\) and \(x(12) =2273.71 \) for part \(b\) and blame the grader for now :)

  77. Astrophysics
    • one year ago
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    Was the force of air resistance0.77|v|?

  78. anonymous
    • one year ago
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    Yes, when the parachute is closed and 14|v| when it opens.

  79. Astrophysics
    • one year ago
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    \[\huge v(t) = Ce^{-(0.77/m)t}+\frac{ mg }{ 0.77 }\]

  80. Astrophysics
    • one year ago
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    Correct?

  81. Astrophysics
    • one year ago
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    Solve for C with the initial conditions

  82. anonymous
    • one year ago
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    That's weird. So it looks like you solved it using separable equations. But why is it different than solving it using integrating factor?

  83. anonymous
    • one year ago
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    So when I solved it using separable equations, I got \[v(t)=Ce^{-\frac{ m }{ .77 }t} +\frac{ mg }{ .77 }\]

  84. anonymous
    • one year ago
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    Using your equation, I still get 376.45 when I plug in 12 for t

  85. Astrophysics
    • one year ago
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    Hmm yeah, me too

  86. Astrophysics
    • one year ago
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    Can you take a screenshot of the question and post it?

  87. anonymous
    • one year ago
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    1 Attachment
  88. Astrophysics
    • one year ago
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    The first one should be right, the second you should add two decimal places

  89. Astrophysics
    • one year ago
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    For the last question you need to take the limit as t approaches infinity, that's where your initial velocity will be v(12)

  90. Astrophysics
    • one year ago
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    So you have to find a dif equation for that to but now you'll have 14|v|

  91. Astrophysics
    • one year ago
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    But I really don't see what's wrong with part a) :\ ganeshie you see anything?

  92. Astrophysics
    • one year ago
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    Maybe write 376.50 lol

  93. anonymous
    • one year ago
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    Yeah I have no idea what's wrong with it.. Though somehow in my notes I end up with something different. The general model that our professor gave us was \[v(t)=Ce^{\frac{ -m }{ .77 }t}+\frac{ mg }{ .77 }\] In yours, you have -.77/m, the reciprocal of what's in the exponent. But what's weird is that solving mine gave me some very high number. But solving yours gives me the exact same thing as before lol

  94. Astrophysics
    • one year ago
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    I moved the m here mdv/dt to the right side

  95. Astrophysics
    • one year ago
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    ∫1/m dt

  96. Astrophysics
    • one year ago
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    Well it's late, I'm off to bed, good luck, maybe email your prof and ask...

  97. anonymous
    • one year ago
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    Thanks for your help and time! :)

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