4. Initially 100 milligrams of a radioactive substance was...

4. Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 4%. If the rate of decay is proportional to the amount of the substance present at time t. (Round your answer to one decimal place.) (a) (12 pts) Find the amount remaining after 24 hours (b) (8 pts) What is the half life of this substance? (c) (5 pts) How fast is the substance decaying at t = 24 hours?

Transcript

00:01 For this problem, you are working with exponential growth and decay, and in particular, a decay problem.

00:12 And i know that because of what i have circled in blue, or boxed in blue, these two are connected to each other.

00:23 So if the rate of decay is proportional to the amount of substance, that is dpdt equals k times p.

00:35 All right, so that's telling me that that is an exponential model.

00:41 And so we can go ahead and substitute in what we have.

00:45 We know that the initial population is 100, and after six hours, the mass has decreased by 4%.

00:58 So that means that after six hours, which, decreasing by 4 % of 100 is going to give a value of 96.

01:14 All right, so, because you would have 100 minus 0 .04 times 100.

01:29 All right, and 0 .04 times 400, or times 100 is 4, so you would have 100 minus 4, or 96.

01:41 Remaining.

01:43 Okay, so that would be at six hours.

01:51 All right, so i know that's a point that i can use, and i also know the initial amount, so i can substitute in the p.

01:59 The ending amount is after six hours will be 96.

02:03 So what i'm going to do now, or what we need to do is find out what the k is.

02:07 We already know what c is, so we're working on what i have written in red.

02:16 To the 6t.

02:19 Nope, sorry, 6k.

02:21 We're looking for k.

02:23 All right, so we're going to take the ln of both sides that will allow us to bring that exponent, that variable exponent down.

02:32 So we'll have the ln of 96 equals, nope, that's not what we're doing first, i'm back up.

02:41 We're going to divide both sides by 100, and that will give us 0 .96 is equal to, to e to the 6k.

02:58 All right.

02:59 Now we'll take the ln of both sides to bring down that variable exponent.

03:04 Ln of 0 .96 is ln e to the 6k.

03:16 Bring that exponent down by using one of the laws of logarithms.

03:21 And the other thing we need to remember is that the ln of v is equal to 1.

03:26 So now we'll have the ln of 0 of 0.

03:29 0 .96 is equal to 6k and then solving for k k will equal ln of 0 .96 over 6.

03:54 Alright so now the model for the population here is going to be p is equal to 100 e to the ln of 0 .96 over 6 times so i'm going to answer letter a, and letter a will be p of 24.

04:30 So just substitute the 24 in.

04:38 Oops, let's write that correctly, e to the ln of 0 .96 over 6.

04:53 Put that line in there a little bit too early, times 24...

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