## anonymous 5 years ago Suppose that you take out an unsubsidized Stafford loan on September 1 before your junior year for \$4500 and plan to begin paying it back on December 1 after graduation and grace period 27 months later. The interest rate is 6.8%. How much of what you will owe will be interest?

1. anonymous

i believe its \$661.76

2. myininaya

cantorset can we use induction on your problem? do you know what that is?.

3. anonymous

yes we can

4. anonymous

how did you find me, lol

5. myininaya

i went to the homepage and you can see what everyone is doing

6. anonymous

the chat program doesnt work for me

7. anonymous

ok lets do this one for all n>=5 , n^3 < 3^n

8. myininaya

ok

9. myininaya

for n=5, we have 5^3=125<243=3^5. Sold the inequality holds for n=5

10. anonymous

right, basis case

11. myininaya

now we should assume it is true for integers k>=5 Then we show it is true for k+1 the induction step is the hardest so let me think about it

12. myininaya

it seems that you are going over logs so i'm pretty sure we may have to involved that somehow

13. anonymous

hmmm, they may be unrelated though. here is what i got ,

14. anonymous

that was a different question

15. anonymous

is there a way to post without having to click my mouse on post?

16. myininaya

no i guess not

17. anonymous

show that (k+1)^3 < 3^(k+1)

18. anonymous

k^3 + 3k^2 + 3k + 1 < 3^(k+1)

19. myininaya

omg i got it

20. myininaya

i got so excited i clicked out sorry. so we have (k+1)^3<K^3+1^3=k^3+1<3^k+1

21. anonymous

k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)

22. anonymous

(k+1)^3 = k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)

23. myininaya

i got excited ignore what i wrote

24. anonymous

(k+1)^3<K^3+1^3=k^3+1<3^k+1 this is false

25. anonymous

my proof isnt elegant, but i think it works, might have to do some side induction arguments for the intermediate steps

26. myininaya

you are right i got excited

27. anonymous

also you better use parenthesses :) more often

28. anonymous

the log proof is A LOT easier

29. myininaya

yep

30. anonymous

what did you get? log_3 n < n if n < 3^n , assuming we raise both sides to 3

31. anonymous

or you can start, if we can show that n < 3^n , for n>=1, then taking the log base 3 of both sides we have our result. what was your logic?

32. myininaya

ok lets skip the case for n=1 that is easy to do and we know it holds. so assume for integer k, log3(k)<k (3is base) now we want to show log3(k+1)<k+1 so we have ...

33. anonymous

so that proof goes, 1 < 3^n, assume k < 3^k then add 1 to both sides of that inequality we have k + 1 < 3^k + 1 < 3^k + 3 = 3^(k+1)

34. myininaya

yea maybe we should base 3 on both sides to get rid of log

35. anonymous

you mean exponentiate?

36. anonymous

you cant get rid of log with log :)

37. anonymous

can you help me with surface area of integral

38. myininaya

surface area of a solid?

39. anonymous

surface area of a torus

40. anonymous

so you cant finish the proof with log, no problemo

41. myininaya

one thing at a time

42. anonymous

how do you find the question i posted earlier

43. anonymous

this website is so confusing

44. myininaya

are you given the equation for the torus or is it a parametric curve/

45. anonymous

Calculus question: surface area of a torus. it is generated by circle with radius r , off center. Suppose the circle is centered at (x - R)^2 + y^2 = r^2, and R> r. so the distance from (0,0) to the center of the circle is (R,0) which has radius r. Ok the integral i got is int 2pi y sqrt [ 1+ (dy/dx)^2] = int 2pi sqrt ( r^2 - ( x-R)^2) [ 1 + (-2(x-R)/(2sqrt( r^2 - ( x-R)^2) ^2 ]

46. anonymous

we are given a circle , and its going to generate the torus by revolving about the y axis

47. anonymous

the final answer should be 4pi^2 R*r

48. myininaya

so that last thing you have the square cancels the sqrt

49. myininaya

ok im just going to do it on paper and scan it and you can open it give a few min

50. anonymous

right, good :) ok here is what i got

51. anonymous

integral 2pi y dS, remember ds can be sqrt 1 + dx/dy ^2 or sqrt 1 + dy/dx ^2

52. myininaya

we need another pi in there somewhere

53. myininaya

54. myininaya

this is what i got using the integral you found we can get times another pi it would be correct

55. myininaya

don't look at the name of the file. i know we are not finding volume

56. anonymous

right, its surface area, here lets use twiddla

57. anonymous
58. anonymous

if you dont mind, we can draw on whiteboard