Question

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.

Answer 1

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...

Answer 2

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...

Answer 3

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

Answer 4

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.

Answer 5

Solve for k! :)

Answer 6

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

Answer 7

It's not? :[
Hmmmm

Answer 8

@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?

Answer 9

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

Answer 10

Hmm, your steps made sense, so I'm not sure what's wrong.

Answer 11

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

Answer 12

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.

Answer 13

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 :
\[\rm ((6038e^{.12} -k)e^{.12} - k)e^{.12} -k = 6038e^{3*.12} -ke^{2*.12}-ke^{.12}-k \]

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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 :
\[\dfrac{dS}{dt} = rS-k\]

Answer 20

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..

Answer 21

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 \(k\)

Answer 22

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

Answer 23

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?

Answer 24

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

Answer 25

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..

Answer 26

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

Answer 27

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

Answer 28

Thats okay, vertical lines just indicate the magnitude

Answer 29

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

Answer 30

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

Answer 31

For parts \(a\) and \(b\) :
\[232\dfrac{dv}{dt} = 232*32-0.77v\]

Answer 32

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

Answer 33

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

Answer 34

solve \(c\) using the fact that the body is "free falling"

Answer 35

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

Answer 36

It's 0 initially

Answer 37

right, \(v(0) = 0\)

use that to find the value of \(c\)

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

Okay, so I got is 376.45

Answer 46

v(12) = 376.45

looks good

Answer 47

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.

Answer 48

Distance*

Answer 49

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

Answer 50

\[\frac{ dx }{ dt }=v\]

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

Okay, I got 115698.2487

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

That's also true.

Answer 61

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

Answer 62

both wrong ?

Answer 63

Yeah..

Answer 64

how many more attempts do u have

Answer 65

2

Answer 66

Check v(12) again

Answer 67

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

Answer 68

If I use wolfram alpha's equation
\[v(t)=\frac{ 742400 }{ 77 }(1-e^{-77t/23200})\]

then v(12) would give me 376.45

Answer 69

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

Answer 70

lol

Answer 71

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

Answer 72

speed cannot be negative

Answer 73

True -- I was thinking of velocity.

Answer 74

part a was about speed right

Answer 75

Find the speed of the skydiver when the parachute opens.

Answer 76

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 :)

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

Correct?

Answer 81

Solve for C with the initial conditions

Answer 82

That's weird. So it looks like you solved it using separable equations. But why is it different than solving it using integrating factor?

Answer 83

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

Answer 84

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

Answer 85

Hmm yeah, me too

Answer 86

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

Answer 87

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

Answer 88

For the last question you need to take the limit as t approaches infinity, that's where your initial velocity will be v(12)

Answer 89

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

Answer 90

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

Answer 91

Maybe write 376.50 lol

Answer 92

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

Answer 93

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

Answer 94

∫1/m dt

Answer 95

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

Answer 96

Thanks for your help and time! :)