# Exponential functions and graphing them.

• Sep 22nd 2009, 07:17 AM
Rumor
Exponential functions and graphing them.
I'll just get right to the point and give the problem.

"If a 100 mg (milligram) tablet of an asthma drug is taken orally and if none of the drug is present in the body when the tablet is first taken, the total amount of A in the bloodstream after t minutes is modeled by the function:

A=100(1-e^(-0.05t)).

a) How many minutes are needed for 70mg of the drug to enter the bloodstream? Find the answer in decimal form.

b) Draw the graph of the function for the first two hours (120 minutes). Label the axes and indicate the value of any asymptotes for your graph."

• Sep 22nd 2009, 07:22 AM
Prove It
Quote:

Originally Posted by Rumor
I'll just get right to the point and give the problem.

"If a 100 mg (milligram) tablet of an asthma drug is taken orally and if none of the drug is present in the body when the tablet is first taken, the total amount of A in the bloodstream after t minutes is modeled by the function:

A=100(1-e^(-0.05t)).

a) How many minutes are needed for 70mg of the drug to enter the bloodstream? Find the answer in decimal form.

b) Draw the graph of the function for the first two hours (120 minutes). Label the axes and indicate the value of any asymptotes for your graph."

Let $\displaystyle A = 70$.

$\displaystyle 70 = 100\left(1 - e^{-0.05t}\right)$

$\displaystyle 0.7 = 1 - e^{-0.05t}$

$\displaystyle e^{-0.05t} = 0.3$

$\displaystyle -0.05t = \ln{0.3}$

$\displaystyle t = -20\ln{0.3}$

$\displaystyle t = -20\left(\ln{\frac{3}{10}}\right)$

$\displaystyle t = -20\left(\ln{3} - \ln{10}\right)$

$\displaystyle t = 20\ln{10} - 20\ln{3}$

$\displaystyle t \approx 24.0795\,\textrm{sec}$.

As for the graphing, do you own a graphing calculator? Do you understand how to apply transformations to the graph of $\displaystyle y = e^x$?
• Sep 22nd 2009, 07:31 AM
Rumor
Quote:

Originally Posted by Prove It
Let $\displaystyle A = 70$.

$\displaystyle 70 = 100\left(1 - e^{-0.05t}\right)$

$\displaystyle 0.7 = 1 - e^{-0.05t}$

$\displaystyle e^{-0.05t} = 0.3$

$\displaystyle -0.05t = \ln{0.3}$

$\displaystyle t = -20\ln{0.3}$

$\displaystyle t = -20\left(\ln{\frac{3}{10}}\right)$

$\displaystyle t = -20\left(\ln{3} - \ln{10}\right)$

$\displaystyle t = 20\ln{10} - 20\ln{3}$

$\displaystyle t \approx 24.0795\,\textrm{sec}$.

As for the graphing, do you own a graphing calculator? Do you understand how to apply transformations to the graph of $\displaystyle y = e^x$?

I do own one. When I put the original equation in, it shows up fine on the graph. But when I attempt to plug the number in for t, it gives me an error. So I must not understand it completely.
• Sep 22nd 2009, 07:43 AM
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Quote:

Originally Posted by Rumor
I do own one. When I put the original equation in, it shows up fine on the graph. But when I attempt to plug the number in for t, it gives me an error. So I must not understand it completely.

Remember to put in the multiplication symbol.

• Sep 22nd 2009, 07:54 AM
Rumor
Quote:

Originally Posted by Prove It
Remember to put in the multiplication symbol.

When entering it into y= on the calculator, it's supposed to look like this:

100(1-e^(-0.05*24.0795)) , right?

Whenever I try to graph it, it gives me a window error because when I press ZoomFit, it places the Ymin and Ymax at the same value. But even when I change the values manually, all the graph shows me is a straight line.
• Sep 22nd 2009, 07:57 AM
Prove It
Quote:

Originally Posted by Rumor
When entering it into y= on the calculator, it's supposed to look like this:

100(1-e^(-0.05*24.0795)) , right?

Whenever I try to graph it, it gives me a window error because when I press ZoomFit, it places the Ymin and Ymax at the same value. But even when I change the values manually, all the graph shows me is a straight line.

No.

You need to enter it as:

y = 100*(1 - e^(-0.05*x))
• Sep 22nd 2009, 08:12 AM
Rumor
Quote:

Originally Posted by Prove It
No.

You need to enter it as:

y = 100*(1 - e^(-0.05*x))

Ah, okay. I apologize for being so slow with this. D: But thank you for your patience! :]
• Sep 22nd 2009, 08:25 AM
pacman
graph of A=100(1-e^(-0.05t)).
A=100(1-e^(-0.05t)). for t = -30 to 30
• Sep 22nd 2009, 08:29 AM
pacman
t = -30 to 150
A=100(1-e^(-0.05t)). for t = -30 to 150
• Sep 22nd 2009, 08:40 AM
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Quote:

Originally Posted by pacman
A=100(1-e^(-0.05t)). for t = -30 to 150

Note that $\displaystyle t \geq 0$ as it is impossible to have negative time.