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Strontium- 90 has a half-life of 28 days.(a) A sample has a mass of 50 mg initially. Find a formula for the mass remaining after $ t $ days.(b) Find the mass remaining after 40 days.(c) How long does it take the sample to decay to a mass of 2 mg?(d) Sketch the graph of the mass function.

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a) $m(t)=50 e^{\ln (0.5)(t / 28)} \quad$ or $\quad m(t)=50(0.5)^{t / 28}$b) $m(40)=18.5749 m g$c) $t=130$ daysd) (28,25) Sketch unavailable

04:13

Wen Zheng

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 8

Exponential Growth and Decay

Derivatives

Differentiation

Abdullah S.

March 12, 2021

find the side labeled x

Missouri State University

Oregon State University

Harvey Mudd College

Lectures

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In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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here we have 1/2 life problem and we're using the model M of T equals M, not A to the K t. Now this is the same model that we use for population. We just have a nem instead of a P, and we're going to find that the value of K will be negative instead of positive. But the concept is the same. So for part, a what we're going to do is use the given information to find our model. And what we know is the half life is 28 days and we know the initial amount is 50 milligrams and so we're going to substitute numbers into our model. So the initial value m not is 50. And then after 28 days, the final amount will be 25. It's half of that and so we can substitute 28 in for tea and we can sell for K. Solving for K is key to finding our model. So at this point, we want to divide both sides of the equation by 50 and we get 0.5 equals e to the 28 k. Then we take the natural log of both sides and then we divide both sides by 28. So K is the natural log of zero point 5/28. So that goes into our model and we have em of tea equals 50. The initial amount times E to the natural log 0.5 over 28 tens t and we'll use that model for the next part of the problem. So what we're doing in part B is we're going to find the mass at time 40 40 days. So we go ahead and substitute 40 into our model and we put it in the calculator and we get approximately 18.5749 milligrams. Alright. Now for part C, what we want to do is find the time for a given Mass so kind of the opposite The part B was given the time. Find the mass and part C is given the mass find the time. So in part c, we know the mass is two milligrams find the time. So let's put to infer rm of tea. We have our 50. We have our e to the natural log of point 5/28 times a time so Let's divide both sides of the equation by 50 and we get 0.4 equals e to the natural log of zero point 5/28 times t. Now let's take the natural log on both sides and then to get t by itself. What we want to do is multiplied by the reciprocal of this fraction. So we're going to end up multiplying by 28 dividing by the natural log of 280.5. And then we put that in the calculator and the approximate value we have for tea here is ah, 130. So it takes 130 days to get down to two milligrams. Never part C. We're going to sketch a graph and to sketch the graph, we can gather the information we have so far from the problem. So we know that at time zero there were 50 milligrams because the half life is 28 days. We know that at time 28 there were 25 milligrams. We know from part B that at time 40 there were 18.6 milligrams and we know from heart see that at time 1 30 there were two milligrams So let's gather all that data together. 0 50 28 25 40 and 18.6 and 1 30 And to and we know the basic shape of exponential decay, we know it's going to be exponential decreasing function so we can go ahead and plot these points. So we have the 0.0 50. We have the 0.28 25. It's gonna be roughly somewhere around here. We have the 0.40 18.6. That's gonna be right around there, and we have the 0.1 32 and that's gonna be somewhere around there. Okay, so here's our exponential decay sketch.

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