## anonymous 5 years ago can someone help me with my problem?

1. anonymous

and what is your problem jmfranco?

2. anonymous

2) We often use radio-isotopes to estimate the age of things (such as the Earth). Assuming the half-life of some isotope is 1000 years, and you found a rock of unknown age near the Mid Atlantic Ridge. You measured the amount of isotope in the rock and determined that 12.5% of the original amount when the rock formed was still present. How old is the rock

3. anonymous

Let's have a chat here franco. This is a growth and decay problem. Is that the topic you're studying?

4. anonymous

ok..yes it is

5. anonymous

great. now, sometimes growth and decay have been modeled by exponential functions. This is the model common model and the one you'll be using here. Do you know the form of the exponential function that is used to model growth and decay? is so, type it in here...

6. anonymous

I hope you don't mind that I'm not just giving you the answer, but you'll be better able to solve all the other problems you'll have this way. it will only take a few minutes.

7. anonymous

okay i don't really know exponential. i understand

8. anonymous

r us still there?

9. anonymous

yes, i'm here

10. anonymous

my connection just failed me for a bit, but i'm back

11. anonymous

this website is having a lot of traffic and seems slow, check out this site has a good explanation: http://serc.carleton.edu/quantskills/methods/quantlit/expGandD.html

12. anonymous

take a look and let me know if you have any questions

13. anonymous

please let me know once you have briefly looked at the page...

14. anonymous

then we can choose between those two equations and use the information you were given to solve the problem. it's not hard at all and you'll see.

15. anonymous

ok looked at it its sort of complicated can u go step by step

16. anonymous

sure

17. anonymous

so the two functions are n(t) = n0*e^(kt)

18. anonymous

now, the exponent of e is that decides whether we have a growth or decay. If it is negative, then the graph starts high and comes down (decay) and the opposite is true if the exponent is positive.

19. anonymous

with me so far?

20. anonymous

yes

21. anonymous

by the way, what grade you in? so I know what you sort of should know...

22. anonymous

college lol

23. anonymous

hehe, no worries. I just graduated a year ago myself. any how...i hope you're not rushed cuz I'm definitely not moving very quickly...let me know if you're rushed.

24. anonymous

so let's go over the equation in more detail...

25. anonymous

the site gives good description for each variable involved.

26. anonymous

lol cool. well i got this question and another, these questions r complicated and due at 4

27. anonymous

today sorry don't rush its okay.

28. anonymous

I'm gonna ask you to rephrase them for me here: what does N(t) mean? and the N_0(t) (that is the subscript zero), the k? the t?

29. anonymous

once you have those, we can replace the variables with the information in your problem. I just wanna make sure you're with me...

30. anonymous

Nt is quatity in time, t is time,n0 is initial quantity, k is a constant and eX is exponential function

31. anonymous

good

32. anonymous

so, it's N(t) and it is a function of time. It's very important that you understand that. and eX is e^(x) which is the mathematically constant e (approx 2.71..) to power x.

33. anonymous

okay

34. anonymous

so what the formulate says is N(t) [the amount as a function of time] = n0*e^(k*t) [the original amount times the exponential function to power k times t]

35. anonymous

k here is the rate of change. the larger is it, the faster the amount grows or, in this case, decays.

36. anonymous

What is the question asking of us? what do you think will be the variable that will be the final answer?

37. anonymous

stay with me franco, we're almost there

38. anonymous

not sure ive never done a prblem like this one

39. anonymous

the question asks you how old is the rock when only 12.5% of it is left..

40. anonymous

so we're looking for time, which will be contained in our variable t.

41. anonymous

a clue that is give to you without knowing it is the term "half-life"

42. anonymous

how long is the half-life of this rock?

43. anonymous

Franco?

44. anonymous

1000

45. anonymous

thank you.

46. anonymous

so that means that in a 1000 years, we'll only have 50% of the original material left. that IS what half life means.

47. anonymous

okay so what do i do??

48. anonymous

so let's set up the equation

49. anonymous

actually, I'll let you do it: in 1000 years, we will only have 50% of the original material. can you set that up?

50. anonymous

i'll prompt with good questions: first: what is t?

51. anonymous

nooo :( i don't kno how this is complicateddd!

52. anonymous

okay: time in this case is 1000 years. second: notice how you can divide both sides by n0 (the original material) - once we do this, what do we have on the left side of the equation?

53. anonymous

this is only solving the first component of the problem.any way, when we divide both sides by n0, we get the equation: N(t)/n0 = e^(kt)

54. anonymous

are you with me?

55. anonymous

in a thousand years, we know we will only have half left, so what will N(t)/n0 equal? think about this one cause it solves 50% of the problem!

56. anonymous

Franco? are you still with me?

57. anonymous

is n0 1000

58. anonymous

i didvide both sides by that?

59. anonymous

n0 is the original amount. was that given in the probem? 1000 is the number of years...

60. anonymous

before i loose you here, we divide the variable n0 by both sides so that we can remove it from the right side and put it on the left. you understand what this results in?

61. anonymous

I miss half life problems, what is that calculus 1

62. anonymous

no what is the asnwer/

63. anonymous

now the equation looks like this: $n(t) \div n _{0} = e ^{k*t}$

64. anonymous

no, there is no calculus involved. this is still pre-calc

65. anonymous

do you understand how we get that equation?

66. anonymous

how do you know so much about math thierryu

67. anonymous

I studied engineering in college, i just graduated last year

68. anonymous

are you here to help us Amanda?

69. anonymous

Franco, do you see how we got that last equation I put in?

70. anonymous

yes

71. anonymous

good

72. anonymous

Now, if it has been a 1000 years and we only have 50% of the original materia.

73. anonymous

so now if t=1000, we're gonna look at n0/N(1000). All we know about this fraction is that N(t) at t=1000 is 50% of n0, the original amount.

74. anonymous

digest that.

75. anonymous

therefore, N(t)/n0 when t=1000 years should give us 1/2. do you agree?

76. anonymous

I meant to say N(t)/n0 all along. sorry for the confusion.

77. anonymous

78. anonymous

should just got to wolfram mathmatica, and type in halflife

79. anonymous

Franco, are you with me? do you see why N(t)/n0 at the half-time (t=1000) will give us 1/2 ?

80. anonymous

so wats the answer??? yea i see

81. anonymous

i did but i will let you help him first

82. anonymous

alright. so now we have 1/2 = e^(k*t)

83. anonymous

im a girl

84. anonymous

and t= 1000 so we really have .5=e^(1000k)

85. anonymous

franco plug in your time and your values for n and No

86. anonymous

ok

87. anonymous

now we solve for k by first takiing the natural log of both sides: ln(.5) = ln(e^(1000k))

88. anonymous

we do this because we know that the ln(e^(x)) equals x.

89. anonymous

so ln(e^(1000k)) reduces to simply 1000k!

90. anonymous

now you can use your calculator to solve for k from teh equation: $\ln(.5) = 1000*k$

91. anonymous

what is k?

92. anonymous

once you find k, we go back to the board, but now we have everything we need expect time which is exactly what we're looking for!

93. anonymous

umm not sure..

94. anonymous

why? because we only have 12.5% left. so the ratio of N(t)/n0 = .125

95. anonymous

96. anonymous

there you'd just use a calculator; get the natural log of .5 and then divide it by 1000

97. anonymous

are you with me now?

98. anonymous

i know this was elongated by if you go over this chat and solve the same or a similar problem, you'll get a hang of it. don't be discouraged.

99. anonymous

is it negative .0003

100. anonymous

great. i didn't check but it looks about right

101. anonymous

so on to solving the problem

102. anonymous

i really havve to go can u just solve for me?

103. anonymous

we only have 12.5% left; which means the ratio of N(t)/n0 = 12.5% = .125.

104. anonymous

so since we now have k

105. anonymous

we end up with $.125 = e^(-.0003t)$

106. anonymous

remember how we solved for k? we do the same for t now

107. anonymous

ln(.125) = ln(e^(-.0003*t)) = -.0003*t

108. anonymous

so $\ln(.125) = -.0003*t$

109. anonymous

can you solve for t? and that's gonna be the number of years that has passed for us to only have 12.5% of the rock left.

110. anonymous

save this chat and go over it when you get a chance and you should be able to do this types of problems on your own.

111. anonymous

did you find t?

112. anonymous

I, for one, would love to know what it is...

113. anonymous

no

114. anonymous

divde ln(.125) by .0003

115. anonymous

yes, by negative .0003

116. anonymous

so how long will have passed franco?

117. anonymous

6900?

118. anonymous

119. anonymous

sounds about right. definitely way more than a thousand because it is way past the 50% mark. so that makes sense.

120. anonymous

ok thanks

121. anonymous

you're welcome.

122. anonymous

123. anonymous

nope, i am still lost

124. anonymous

would you care to chat on the other thread

125. anonymous

since you're not rushed and i just got home, give me about half an hour to grab a snack and start dinner and I'll be back. I'm putting on a timer for 30 minutes sharp.

126. anonymous

I may also have to review a little calculus ;)

127. anonymous

haha okay i will be around

128. anonymous

could you post the entire problem for me here again please?

129. anonymous

using lagrange multipliers find the maxium and minimum values of the function f(x,y,z)=xy^2z^3 subject to the contraints x^2+y^2+2z^2=25