Let Q represent a mass of carbon 14(14 C) (in grams), whose half-life is 5715 years. The quantit

Let Q represent a mass of carbon 14(14 C) (in grams), whose...

Let $Q$ represent a mass of carbon 14$(14 \mathrm{C})$ (in grams), whose half-life is 5715 years. The quantity of carbon 14 present after $t$ years is $Q=10\left(\frac{1}{2}\right)^{t / 5715}$ (a) Determine the initial quantity (when $t=0$ ) (b) Determine the quantity present after 2000 years. (c) Sketch the graph of this function over the interval $t=0$ to $t=10,00$ .

Let $Q$ represent a mass of carbon 14$(14 \mathrm{C})$ (in grams), whose half-life is 5715 years.
The quantity of carbon 14 present after $t$ years is $Q=10\left(\frac{1}{2}\right)^{t / 5715}$
(a) Determine the initial quantity (when $t=0$ )
(b) Determine the quantity present after 2000 years.
(c) Sketch the graph of this function over the interval $t=0$ to $t=10,00$ .

Key Concepts

Graphing Exponential Functions

Graphing exponential functions provides a visual representation of the decay process over time. The characteristic curve illustrates a rapid decrease initially, which then levels off as the quantity approaches zero, and it helps in understanding the long-term behavior of the decaying substance.

Evaluating Exponential Functions

Evaluating exponential functions involves substituting specific values of the independent variable, such as time, to obtain the corresponding quantity. This is done using the exponential decay formula, and it allows for determining the amount remaining after any given period.

Half-Life

The half-life of a decaying substance is the time required for its quantity to reduce to half of its initial amount. It is a key parameter in exponential decay problems, allowing the conversion of a physical decay process into a mathematical form by indicating how quickly the substance diminishes over time.

Exponential Decay

Exponential decay describes a process in which a quantity decreases at a rate proportional to its current value. This model is characterized by a constant percentage reduction over equal time intervals, making it ideal for representing processes such as radioactive decay, population decrease, or depreciation.

Initial Value in Exponential Functions

The initial value, often denoted as Q(0), represents the starting quantity before any decay has occurred. In an exponential decay model, this value is crucial as it sets the baseline from which all subsequent reductions are measured and calculated.

Transcript

00:01 All right, so we are provided an equation for the half life of carbon 14 in grams.

00:07 That equation is right here, q equals 10 times 1⁄2, raised to the t divided by 5 ,715, where t is our time in years.

00:19 So first it asks us to find out how much carbon 14 do we start with? so we're going to put zero in for our years, or our t equals zero.

00:27 Let's go ahead and plug that into our equation.

00:31 So we have q equals 10 times 1 half, one half.

00:45 And here we plug in our 0 for our t divided by 5 ,000, 115.

00:56 And i like solving these equations in step by step.

01:00 So first let's solve for our exponents.

01:02 Well, 0 divided by 5 ,715 is going to be 0, because 0 divided by any number is 0.

01:11 So that makes it a little easier for us because now we're going to have 10 times 1⁄2 raised to the 0.

01:21 Well, any number raised to 0 is actually just the number 1.

01:26 No matter what number you start out with, if it's raised to 0, it's going to end up being 1.

01:31 So it's actually going to be 10 times 1.

01:36 And that makes our answer 10.

01:39 And it says grams, so we're just going to put a little gram sign there.

01:44 So we start out with 10 grams of carbon 14.

01:49 There is our solution for part a.

01:52 Now part b is saying, well, how many grams of carbon 14 are we going to have after 2 ,000 years? so now let's just plug 2 ,000 in, just like how we did with our first 1st.

02:03 Part.

02:04 So we're going to do 10 times 1 half, raised to 2 ,000 divided by 5 ,715.

02:23 And you can solve this in parts again by dividing 2 ,000 by 5 ,715, then raising one half to that and timesing that by 10.

02:32 Or if you just plug it into your calculator the way it is, your answer should come out as 7 .8, 4, 6 grams.

02:47 And there's our solution for part 2.

02:51 Now, part b, part c asks us to graph this.

02:56 And we can make any graph because if we have two points and we do have two points...