A model for the concentration at time t of a drug injected...

A model for the concentration at time $ t $ of a drug injected into the bloodstream is $$ C(t) = K(e^{-at} - e^{-bt}) $$ where $ a $, $ b $, and $ K $ are positive constants and $ b > a $. Sketch the graph of the concentration function. What does the graph tell us about how the concentration varies as time passes?

A model for the concentration at time $ t $ of a drug injected into the bloodstream is
$$ C(t) = K(e^{-at} - e^{-bt}) $$
where $ a $, $ b $, and $ K $ are positive constants and $ b > a $. Sketch the graph of the concentration function. What does the graph tell us about how the concentration varies as time passes?

Transcript

00:01 So we're given this function to model concentration over time, and we're given these constraints.

00:07 A, b, and k are positive constants, and b is greater than k.

00:10 So i just picked these random numbers that do work for these constraints, and i plugged them into the function, and they got this function.

00:19 So if we want to graph it, the first thing i'm going to look at is the first derivative.

00:23 So c of t, c prime of t, so the first derivative, is going to be...

00:31 Negative e to the power of negative t plus 2e to the negative 2 t okay and i'm going to set this equal to 0 to find my critical points and i see that t is equal to ln of 2 all right so ln of 2 is approximately 0 .693 so if we test values so this is ln2 if we test values into our first derivative for the first derivative test for less than ln2, we're going to see that it's increasing and anything greater than is going to be decreasing.

01:12 So it's a local max.

01:16 All right, so this can be useful for later.

01:19 And then c double prime to look at concavity, c double prime of t, we're going to get e to the negative t minus 4e to the negative 2t.

01:33 If we set that equal to zero, we're going to see t, um, t, t, um, t equals 2ln2, which is approximately 1 .39.

01:48 And if we do the second derivative test, f double prime, and we test values around 1 .39, we see that it's going to be concave down and then concave up.

02:02 So this is an inflection point.

02:05 All right? so we're going to use this while we graph in just a moment.

02:10 But i'm going to look at one more thing, because our our graph actually does have an asymptote.

02:15 It has a horizontal asymptote.

02:18 So let's test the limit as our variable t approaches infinity of our function, which is 1 times e to the negative 1t, minus e to the negative 2t.

02:41 All right, and we're going to, so let's just say that we do the limit as t approaches infinity, i'm going to rewrite this in a way that looks easier to understand.

02:52 So this is the same thing as 1 over e to the t, because it's a negative exponent, we bring it to the denominator, so 1 over e to the t minus 1 over e to the 2t.

03:07 So this is the same thing.

03:11 If you take a look at this, and we substitute wherever we see t's, we're going to substitute infinity...

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