The concentration C, in parts per million, of a medication in the body t hours after ingestion i

The concentration C, in parts per million, of a medication...

Key Concepts

Function Evaluation

Function evaluation involves substituting given input values into a function to compute the corresponding outputs. In this context, it means calculating the concentration at specific times by plugging the values of t into the function, which helps in understanding the behavior of the medication concentration over time.

Graphing Functions

Graphing functions is a visual method to represent how a dependent variable changes with respect to an independent variable. By sketching the graph of the concentration function over a specified interval, one can see the overall trend, identify peaks, and understand the long-term behavior of the model.

Differentiation

Differentiation is a core concept in calculus that involves computing the derivative of a function. The derivative represents the instantaneous rate of change of the function with respect to the variable, and techniques like the product rule and chain rule are used to differentiate composite functions such as the one given.

Optimization in Calculus

Optimization in calculus involves finding the maximum or minimum values of a function, often by identifying critical points where the derivative equals zero. This process is used to determine the time at which the concentration reaches its maximum value and what that maximum concentration is.

Exponential and Polynomial Functions

Exponential functions and polynomial functions often appear together in mathematical models to describe real-world phenomena. Exponential functions can represent growth or decay processes, while polynomial functions can model acceleration or deceleration effects, and their combination can capture more complex behaviors observed in various applications.

Interpretation of the Derivative

The derivative of a function provides insight into the rate at which the function's value is changing at any given point. In practical terms, it indicates how quickly the medication concentration increases or decreases over time, which is critical for understanding the dynamics of absorption and elimination in the body.

Transcript

00:01 So we're given this concentration equation, which is this equation up here, c of t.

00:06 And what we're going to do is we're just going to find the value of c of t at different values of t.

00:11 So at t is equal to zero, we would have zero squared times 10 times e to the 0th, which would just be 0.

00:19 At t is equal to 1, we have 1 squared times 10 times e to the negative 1, so that would just be 10 divided by e.

00:27 At 2 we have 2 squared which is 4 times 10 would be 40 and then divided by e squared so 40 divided by e squared and then when t is equal to 3 we have 3 squared which is 9 times 10 would be 90 divided by e to the 3rd and then when t is equal to 10 you have 10 squared which is 100 times 10 would be a thousand divided by e to the 10th and so now we're just going to graph this c of t function from 0 to 10 from t is equal to 0 to 10 and so we're just going to use the values that we found in part a to sketch a quick graph so at t is equal to 0 c of t is also equal to 0 so we have this point at the origin at t is equal to 1 c of t is equal to 10 over e which if you plug into a calculator is about 3 .7 so it would be up here when t is equal to 2 we have 40 over e squared, which is equal to about 5 .4.

01:37 So it'd be up here.

01:39 And then when t is equal to 3, c of t is equal to 90 over e to the third, which is about 4 .48.

01:48 So that would be around here.

01:51 And then when t is equal to 10, we have that c of t is equal to a thousand over e to the 10, which is actually equal to about 0 .045.

02:02 So it's going to be all the way down here.

02:06 So if we were to draw a quick sketch just from these points, it would look something like this.

02:18 This isn't completely accurate.

02:19 We would have to find the values at 4, 5, 6, 7, 8, and 9 to be completely accurate.

02:25 But this is somewhat of a good sketch.

02:27 You can see that there's going to be a maximum point somewhere up here, and there's going to be a...

02:34 Slight drop off after that maximum point down to this point at 0 .45, or 10 comma 045.

02:46 And now for part c, what we're going to do is we're just going to find the derivative of the c of t function.

02:56 And so c prime of t, we're going to have to use the product rule to find this.

03:03 So we have the derivative of the first, which would be t squared.

03:06 So we can take this 10 out.

03:10 We can say the derivative of the first, which is 2t times the second, which is e to the negative t, plus the first times the derivative of the second, which would be negative 1 times e to the negative t.

03:27 So this is our derivative.

03:30 I'll simplify it a little bit.

03:31 So we have 10 times 2t, e to the negative t minus t squared.

03:38 Squared, e to the negative t.

03:47 And now what we're going to do is we're going to try and find a maximum value of concentration.

03:52 So the way we can do this is we can set our derivative equal to zero.

03:56 So we go back to c prime of t.

03:59 This is equal to 10 times 2 t, e to the negative t minus t squared.

04:09 E to the negative t.

04:11 So we can set this equal to zero.

04:13 And what we can do is we can actually factor out a t and an e to the negative t...