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A sample of tritium- 3 decayed to $ 94.5\% $ of its original amount after a year.(a) What is the half-time of tritium- 3?(b) How long would it take the sample to decay to $ 20\% $ of its original amount?

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a) $T_{1 / 2}=12.25$ yearsb) 28.45 years

02:26

Wen Zheng

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 8

Exponential Growth and Decay

Derivatives

Differentiation

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

04:40

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.

44:57

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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02:41

Calculate the time require…

we're doing a DK problem. So we're using the model M equals M, not each of the K T, which is the same as the population growth model. We just are using em instead of P. And the value of K is going to be negative instead of positive since it's not growth. Okay, so what we want to do is find the half life and the given information is that it takes one year to decay to 94.5% of its original amount. So we're going to use that piece of information to first find a model and then once we have the model, we confined the half life. So we're going to substitute 0.945 of the initial amount 0.945 times and not and for the final amount, will substitute am knockin for em, not keep it the same on will substitute one year and for the time now we divide both sides by I am not and we have 0.945 equals each of the K and we take the natural log of both sides and we get the value of K so we can substitute that into our model and we have em of tea equals m not times e to the natural log of 0.945 times time. So we'll use that model for the rest of the problem. Now we're gonna use it to find the half life and half Life is the time it takes to decay to half the amount. So if we're starting with them, not we would be ending with 1/2 am not when half am not equals. M not times e to the natural log of 0.945 t divide both sides by I am not, and we're left with 1/2 and I'm just gonna write it as 0.5 and then we take the natural log on both sides and then to get t by itself, we're going to divide by the natural log of 0.945 and we put this in the calculator on. We get approximately 12.25 years. So the half life of this substance is about 12.25 years. Now we want to do part B, and it's very similar to party. But instead of having half of it left, we have 20% of it left. So 20% of the original mount would be 200.2 times m. Not so we'll substitute that into our equation for em. 0.2 am not is equal to am not times e to the natural log of 0.945 times time and we're solving for time again because we want to know how long will it take? Divide both sides by I am not. And we have your point too. And then we take the natural log of both sides and then we divide both sides by the natural log of 0.945 to get t. Then we put it in the calculator and we get approximately 28.4 point 45 years. That's the time it takes to decay to 20% of what it started with

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