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The half-life of cesium- 137 is 30 years. Suppose we have a 100- mg sample.(a) Find the mass that remains after $ t $ years.(b) How much of that remains after $ t $ years.(c) After how long will only 1 mg remain?

a) $100 \cdot 2^{-t / 30}$b) $9.92 \mathrm{mg}$c) 199.3 years

02:27

Wen Z.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 8

Exponential Growth and Decay

Derivatives

Differentiation

Diego V.

April 14, 2019

Great, but only if we could understand your handwriting better

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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here we have 1/2 life problem and the model we use for half life is am a T equals m not eat the Katie. This is the exact same model that we use for population growth. We're just using an M instead of a P, and we can expect the value of K two b negative instead of positive, since it's going to be decay and not growth. So let's use the given information and do part A to find basically a general equation for the mass at Time t. So we know the initial masses 100 milligrams and we know that half life is 30 years. So if we want to substitute 30 40 we can put half the original amount, which would be 50 milligrams in for em. So we have 50 equals 100 times e to the 30 k, and we'll use that to self. Okay, so we'll divide both sides by 100 we get 1000.5 equals each of the 30 k and then we'll take the natural log of both sites. So natural log of 0.5 equals 30 k and then we'll divide both sides by 30. So k is the natural log of zero point 5/30. So we put that into our model and we get am of tea equals 100 times e to the natural log of zero point 5/30 times t. Now that would be fine to leave as our model. But there's also a way to simplify the expression that we have the power of eat. So let's take a look at how we might do that. So if you have e to the natural log of 0.5 over 30 times T, that could be rewritten as e to the natural log of 0.5 raised to the power t over 30. Because if you multiply the exponents is the Samos power to a power? Now eat the natural log of 0.5 simplifies to be just 0.5. So we have 0.5 to the tea over 30. And if you want to think of 0.5 is 1/2 and you if you want to think of 1/2 as to to the negative first we have to to the negative t over 30. So if we want to, we have a reason to weaken substitute the expression that we had in the model for two to the negative t over 30. So that gives us m of tea equals 100 times two to the negative t over 30. That's an alternative form of our model. Okay, now let's move on to part B and in part me, what we want to do is find the mass when the time is 100. I think it's 100 years, so we just take one of our two models. It doesn't matter which one we substitute 100 for the time, and we put it in the calculator and approximated and we get about 9.92 milligrams. And then for part C, What we want to do is find the time when we have one milligram, so find T for the mass equals one. So we can substitute one and for mass into either one of our two models and saw that for tea. So we're going to divide both sides of the equation by 100 we get 0.1 equals to raise to the power negative t over 30 and we're gonna make some more space here. So that was 0.1 equals to race to the power of negative T over 30 we're going to take the natural log of both sides. And then we can use the power property of lager thems to bring that power down to the front. So now we have the natural lock of 0.1 equals negative t over 30 times natural log to and what we want to do to get T by itself is we want to multiply both sides by 30 and divide both sides by negative natural octane. So that gives us t equals 30 natural log of 0.1 divided by negative natural log to. And then we put that in the calculator and we get an approximate value of 199.3 years. That's the time it would take for the substance to decay to one milligram

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