Substance A decomposes at a rate proportional to the amount...

Substance $A$ decomposes at a rate proportional to the amount of $A$ present. a) Write an equation that gives the amount $A$ left of an initial amount $A_{0}$ after time $t$. b) It is found that $8 \mathrm{~g}$ of $A$ will reduce to $4 \mathrm{~g}$ in $3 \mathrm{hr}$. After how long will there be only 1 g left?

Substance $A$ decomposes at a rate proportional to the amount of $A$ present.
a) Write an equation that gives the amount $A$ left of an initial amount $A_{0}$ after time $t$.
b) It is found that $8 \mathrm{~g}$ of $A$ will reduce to $4 \mathrm{~g}$ in $3 \mathrm{hr}$. After how long will there be only 1 g left?

Key Concepts

Initial Value Problems

Initial value problems involve solving a differential equation given a set of initial conditions, which specify the system's state at the starting point. This practice is essential for determining the particular solution of a model, ensuring that the solution reflects the actual scenario from the known initial state.

Exponential Decay

Exponential decay describes processes where a quantity decreases at a rate proportional to its current value, leading to a mathematical model that typically involves an exponential function. This concept is widely used in contexts such as radioactive decay, chemical reactions, and natural cooling processes.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are used to express the relationship between a function and its rate of change. In decay problems, the differential equation dA/dt = kA is fundamental, as it mathematically describes how the quantity changes over time, with the constant k guiding the rate and direction (decay or growth).

Transcript

00:01 So we're told that the decomposition or the amount, the change in amount, is directly proportional to the amount present.

00:09 And we're to assume that at times zero, there are a sub -zero is the amount that's there.

00:17 We want to write a model for that.

00:18 So let's, first of all, split this up.

00:21 So we have d -a over a is equal to k times d -t, and we can integrate both sides.

00:31 That's going to give me natural log of a will equal k times t plus c and we know that we are in base e so that a will equal e to the k times t plus c and we know that i guess i should let's do one little change here let's why don't we take this equation before i rewrite it that way.

01:01 And why don't we figure out what c is? because we can use this point right here.

01:06 So let's do a little aside here.

01:08 So we know that at time zero, that if we plug in a sub zero, that that gives us a time of zero.

01:17 And so we know c is equal to that natural log of a sub zero.

01:22 So this c value we can actually replace right here and i'm going to erase it away.

01:29 This c value will end up being ln of a sub 0.

01:35 And so we'll have ln of a sub 0 here.

01:39 And then that's the answer.

01:41 However, let's clean it up.

01:42 We have a is equal to e to the k times t, times using an exponential property, becomes e to the ln of a sub zero.

01:55 And then this, e to the ln of a sub zero, becomes a sub zero.

01:59 So we end up getting a sub zero times e to the k, t.

02:06 So there's our model, our generic model.

02:09 Now we want to in part b, look, if we start with a grams and if it goes to four grams in a time of three hours, we want to know how much time will it take to become one gram.

02:27 So we want to know what's the time to become one gram.

02:32 So let's go back and let's plug this information in.

02:36 So if we start with eight grams, our model's going to look like this...

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