Question

Ariel is taking a medicine for a common cold. The table below shows the amount of medicine f(t), in mg, that was present in Ariel's body after time t:
t(hours): 1 | 2 | 3 | 4 | 5
f(t) (mg): 236 | 223.73 | 211.65 | 200.22 | 189.41

Christina was administered 250 mg of the same medicine. The amount of medicine in her body f(t)after time t is shown by the equation below:

f(t) = 250(0.946)^t

Which statement best describes the rate at which Ariel's and Christina's bodies eliminated the medicine?

Ariel's body eliminated the antibiotic faster than Christina's body.
Ariel's body eliminated the antibiotic at the same rate as Christina's body.
Ariel's body eliminated the antibiotic at half of the rate at which Christina's body eliminated the antibiotic.
Ariel's body eliminated the antibiotic at one-fourth of the rate at which Christina's body eliminated the antibiotic.

Answer 1

        $\text{Solution}$ 

$\dfrac{d}{dt}\left(250\cdot \:0.946^t\right)=-13.87818\dots \cdot \:0.946^t$ 

$\dfrac{d}{dt}\left(250\cdot \:0.946^t\right)$ 

$250\dfrac{d}{dt}\left(0.946^t\right)$ 

$250\dfrac{d}{dt}\left(e^{t\ln \left(0.946\right)}\right)$ 

$250\dfrac{d}{du}\left(e^u\right)\dfrac{d}{dt}\left(t\ln \left(0.946\right)\right)$ 

$\dfrac{d}{du}\left(e^u\right)=e^u$ 

$\dfrac{d}{dt}\left(t\ln \left(0.946\right)\right)=-0.05551\dots$ 

$250e^u\left(-0.05551\dots \right)$ 

$250e^{t\ln \left(0.946\right)}\left(-0.05551\dots \right)$ 

$\mathrm{Simplify\:}250e^{t\ln \left(0.946\right)}\left(-0.05551\dots \right):\quad -13.87818\dots \cdot \:0.946^t$ 

$=-13.87818\dots \cdot \:0.946^t$ 

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

\text{t(hours):}   1         2           3          4            5 
$f(t) (mg): 236 | 223.73 | 211.65 | 200.22 | 189.41$ 

$236-223.73 = 12.27$ 

$\text{ Ariel's body eliminated the antibiotic faster than Christina's body. }$