## 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.

1. zepdrix

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

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

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

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

Solve for k! :)

6. anonymous

Okay, so I got 1998.03 as k, but that is not the correct answer.. :(

7. zepdrix

It's not? :[ Hmmmm

8. zepdrix

@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

principal* amount... bahh I think I've been spelling that wrong the entire time lol

10. anonymous

11. ganeshie8

@zepdrix shouldn't the interest be calculated on the previous year's outstanding balance ?

12. anonymous

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

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

I dunno how interest works :) lol Is it always based off of the current balance? I guess that makes sense :p

15. anonymous

This is the solution I found to a similar problem elsewhere, but it didn't make much sense to me

16. ganeshie8

in that solution, $$S(t)$$ represents the outstanding balance at any given time

17. anonymous

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

Firstly, notice that two things affect the change in outstanding balance $$S(t)$$ : 1) outstanding balance itself (S) 2) payments made (k)

19. ganeshie8

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

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

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

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

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

I can try, but just so you know, zepdrix is the master of differential equations around here..

25. anonymous

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

so the air resistance in the interval $$(0, 12)$$ is $$0.77\overline{v}$$ afterwards, the air resistance is $$14\overline{v}$$

27. anonymous

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

28. ganeshie8

Thats okay, vertical lines just indicate the magnitude

29. ganeshie8

so the air resistance in the interval $$(0, 12)$$ is $$0.77|v|$$ afterwards, the air resistance is $$14|v|$$

30. ganeshie8

Notice that part $$a$$ and $$b$$ concern with the interval $$(0, 12)$$ so simply solve the differential equation $$mv' = mg - 0.77v$$

31. ganeshie8

For parts $$a$$ and $$b$$ : $232\dfrac{dv}{dt} = 232*32-0.77v$

32. ganeshie8

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

This is the general solution that I got. Is this correct? $v=9696.96+\frac{ c }{ e^{.0033t} }$

34. ganeshie8

solve $$c$$ using the fact that the body is "free falling"

35. ganeshie8

what do you know about the initial velocity of a free falling body ?

36. anonymous

It's 0 initially

37. ganeshie8

right, $$v(0) = 0$$ use that to find the value of $$c$$

38. anonymous

So C=-9696.96 where the negative indicates the downward orientation

39. ganeshie8

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

That doesn't look like the same thing that I got though.. =/

41. ganeshie8

they are same, you have converted everything to decimals, wolfram didn't

42. anonymous

Okay, so then plug in 12 for t and that will give me the velocity needed for a

43. ganeshie8

Yes. use the wolfram answer, looks you have rounded a lot, so your numbers are way off..

44. ganeshie8

not a big deal if you hadn't rounded, you would have gotten the exact same answer as wolfram..

45. anonymous

Okay, so I got is 376.45

46. ganeshie8

v(12) = 376.45 looks good

47. anonymous

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

Distance*

49. ganeshie8

what do you know about the relation between "velocity" and "displacement" ?

50. anonymous

$\frac{ dx }{ dt }=v$

51. ganeshie8

$x(12) - x(0) = \int\limits_0^{12} v(t)\, dt$

52. ganeshie8

plugin $$v(t)$$ and evaluate the definite integral

53. ganeshie8

$x(12)-x(0) = \int\limits_0^{12}\frac{742400}{77} \left(1-e^{-77t/23200}\right)\, dt$

54. ganeshie8

evaluating is trivial don't let the messy numbers confuse you

55. anonymous

Okay, I got 115698.2487

56. anonymous

Which doesn't make sense since he starts at an altitude of 6,000ft...

57. ganeshie8

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

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

They probably don't expect you to put it in decimals

60. anonymous

That's also true.

61. anonymous

So far, (a) 376.45 and (b) 2274 are wrong :(

62. ganeshie8

both wrong ?

63. anonymous

Yeah..

64. ganeshie8

how many more attempts do u have

65. anonymous

2

66. Astrophysics

Check v(12) again

67. Astrophysics

Hmm...maybe they want a rounded answer? Significant digits?

68. anonymous

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

Yes, I don't see how part a could be anything different from $$376.4536197750639610724810777669453859565565074584767314312593...$$

70. anonymous

lol

71. anonymous

Oh, maybe it's suppose to be negative? But then again, part (b) is still wrong too..

72. ganeshie8

speed cannot be negative

73. anonymous

True -- I was thinking of velocity.

74. ganeshie8

part a was about speed right

75. anonymous

Find the speed of the skydiver when the parachute opens.

76. ganeshie8

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

Was the force of air resistance0.77|v|?

78. anonymous

Yes, when the parachute is closed and 14|v| when it opens.

79. Astrophysics

$\huge v(t) = Ce^{-(0.77/m)t}+\frac{ mg }{ 0.77 }$

80. Astrophysics

Correct?

81. Astrophysics

Solve for C with the initial conditions

82. anonymous

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

So when I solved it using separable equations, I got $v(t)=Ce^{-\frac{ m }{ .77 }t} +\frac{ mg }{ .77 }$

84. anonymous

Using your equation, I still get 376.45 when I plug in 12 for t

85. Astrophysics

Hmm yeah, me too

86. Astrophysics

Can you take a screenshot of the question and post it?

87. anonymous

88. Astrophysics

The first one should be right, the second you should add two decimal places

89. Astrophysics

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

So you have to find a dif equation for that to but now you'll have 14|v|

91. Astrophysics

But I really don't see what's wrong with part a) :\ ganeshie you see anything?

92. Astrophysics

Maybe write 376.50 lol

93. anonymous

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

I moved the m here mdv/dt to the right side

95. Astrophysics

∫1/m dt

96. Astrophysics

Well it's late, I'm off to bed, good luck, maybe email your prof and ask...

97. anonymous

Thanks for your help and time! :)