# Math Help - Could someone please explain... (exponential decrease..)

1. ## Could someone please explain... (exponential decrease..)

When a certain medicine enters the bloodstream, ir gradually dilutes, decreasing exponentially with a half-life of three days. If the patient took 500mg of the medicine, what amount remain in his bloodstream one week later? At what rate (per day) is the medicine deluting?

Are there any formulas to figure it out?

Thank you!

2. Originally Posted by Polyxendi
When a certain medicine enters the bloodstream, ir gradually dilutes, decreasing exponentially with a half-life of three days. If the patient took 500mg of the medicine, what amount remain in his bloodstream one week later? At what rate (per day) is the medicine deluting?

Are there any formulas to figure it out?

Thank you!
Do you know the model for exponential function, namely $y=ab^t$? If so, just find the corresponding peremeters a and b, then you can obtain a formula. To do so, note that the statement "with a half-life of three days" will provide "points" which you can work with.

Then just plug in t to find the first part.

And note that $b=1+r$, so the 2nd part is actually asking for "r".

3. Hello, Polyxendi!

When a certain medicine enters the bloodstream,
it gradually dilutes, decreasing exponentially with a half-life of three days.

If the patient took 500 mg of the medicine,
(a) what amount remain in his bloodstream one week later?
(b) At what rate (per day) is the medicine deluting?

Are there any formulas to figure it out? . . . . yes!

The formula for "exponential decay" is: . $A \:=\:A_oe^{-kt}$

. . where: . $\begin{Bmatrix}A_o &=& \text{original amount} \\ \\[-4mm] k &=& \frac{\ln 2}{\text{half-life}} \end{Bmatrix}$

We have: . $\text{half-life} = 3 \quad\Rightarrow\quad k \:=\:\frac{\ln 2}{3} \:\approx\:0.231$
. . and: . $A_o = 500$

Hence: . $A(t) \;=\;500e^{-0.231t}$

"(a) One week later": . $A(7) \:=\:500e^{-0.231(7)} \;\approx\;99.25\text{ mg}$

Differentiate: . $\frac{dA}{dt} \:=\:-0.231(500)e^{-0.231t}$

(b) The daily rate of dilution is: . $\frac{dA}{dt} \;=\;-115.5e^{-0.231t}$