Question

PLEASE HELP WITH THIS CALCULUS PROBLEM!



A smoke detector contains about 0.2mg of Americium 241, a radioactive element that decays over t years according to the relation m = 0.2(0.5)^(t/432.2) where m is the mass in milligrams, after t years.

a) The smoke detector will no longer work when the amount of Americium 241 drops below half its initial value. Is it likely to fail while you own it? Justify your answer.

b) If you buy a smoke detector today, how much of the americium 241 will remain after 50 years?

c) How long will it take for the amount of americium 241 to drop to 0.05mg?

Answer 1

It's an exponential decay model, it seems. So, for a) you will want to solve this equation for t:
0.1 (half the Americium 241) = 0.2(0.5)^(t/432.2)
For part b, you will want to solve this:
m = (0.2)*(0.5)^(50/432.2)
And for part c:
0.05 = (0.2)(0.5)^(t/432.2)

Now, for the first part, notice that we have 0.5^(t/432.2). When will this be equal to 0.5? You can get your answer to a) by inspection only.

Answer 2

Im sorry, but I dont know how to rearrange to solve for t :P

Answer 3

Well, since a) you can solve by inspection, I will work on the last one. First, divide both sides by .2, we are left with: \[\Large \frac{0.05}{0.2} = 0.05^{t/432.2} \]Take the natural log on both sides: \[ \ln{\frac{0.05}{0.2}} = \frac{t}{432.2} \ln{0.05}\]Dos this make it any easier?

Answer 4

not really

Answer 5

so t = 432.2?

Answer 6

Now it should be simple algebra, no? \[t = 432.2 \frac{ \ln{ \frac{0.05}{0.2} } }{\ln{0.05}}\]For the first part, yeah, t = 432.2 years, so you can answer it :-)

Answer 7

ok, so how do we rearrange for part b)?

Answer 8

It's already solved. We need to solve for m. You will only need a calculator now :-)

Answer 9

t = 50 for b), so, solve for m :-)

Answer 10

oo you just input it into the calculator right?

Answer 11

Yup.

Answer 12

I got 0.18459

Answer 13

Correct :-)

Answer 14

what about part c)?

Answer 15

oo part a basically right?

Answer 16

It's what I've done up there. You take the natural log of both sides to use the property: \[ \log_b a ^c = c log_ba \]

Answer 17

?

Answer 18

Yeah, you have to answer the question and explain a bit, but the time is 432.2, because we want t/432.2 = 1.

Answer 19

isnt it the exact same as part a) except for the fact that the m is 0.05 instead of 0.1?

Answer 20

You don't remember log equations? How do we solve: \[ 2^x = 3 \]It's, but now we can't solve by inspection, right? You have to use a calculator.

Answer 21

I dont understand

Answer 22

We have \[ 0.05 = (0.2)(0.5)^{t/432.2} \]Right?

Answer 23

yes

Answer 24

How do we solve for t? Do you remember logarithm properties?

Answer 25

nope

Answer 26

Well, the one we have to use here is the following: \[ \ln{b^c} = c \ln{b} \]Where b, c are real numbers.

Answer 27

Following so far?

Answer 28

can i do this? http://www.wolframalpha.com/input/?i=solve+0.05%3D0.2%280.5%29%5E%28t%2F432.2%29

Answer 29

Yeah, you can use Wolfram Alpha if it suits you. But you will have to learn to solve log equations sooner or later if you have a differential equations course.

Answer 30

Ok, thanks for your help :)

Answer 31

No problem :-)