# Need Some Calc. Help

• Nov 25th 2009, 02:24 PM
redsoxfan0825
Need Some Calc. Help
Find the value of k so that f(x)=xe^x/k has a critical value at x=10.

I am pretty sure I have to find the derivative here, using the product rule? But I'm not exactly how to find the derivative of e^x/k and after that I'm not sure what to do. Set the derivative equal to 10 perhaps? Someone help me clear this up, please.
• Nov 25th 2009, 02:32 PM
MathTooHard
$\displaystyle k$ is a constant, so we can pull it out, getting $\displaystyle \frac{dy}{dx}=\frac{1}{k}(\frac{d}{dx}(xe^x))$. Use the product rule to differentiate $\displaystyle xe^x$, set $\displaystyle \frac{dy}{dx}=0$, and then plug $\displaystyle 10$ in for $\displaystyle x$. Now, solve for $\displaystyle k$.
• Nov 25th 2009, 02:33 PM
skeeter
Quote:

Originally Posted by redsoxfan0825
Find the value of k so that f(x)=xe^x/k has a critical value at x=10.

I am pretty sure I have to find the derivative here, using the product rule? But I'm not exactly how to find the derivative of e^x/k and after that I'm not sure what to do. Set the derivative equal to 10 perhaps? Someone help me clear this up, please.

assuming you mean ...

$\displaystyle f(x) = xe^{\frac{x}{k}}$

$\displaystyle f'(x) = x \cdot \frac{1}{k}e^{\frac{x}{k}} + e^{\frac{x}{k}}$

$\displaystyle f'(x) = e^{\frac{x}{k}}\left(\frac{x}{k} + 1\right)$

critical values occur where x = 0 ...

$\displaystyle \frac{x}{k} = -1$

at $\displaystyle x = 10$ , $\displaystyle k = -10$
• Nov 25th 2009, 03:18 PM
redsoxfan0825
Thank you very much. There are a couple of other problems I'm having trouble with as well.

"The quantity of a drug in the bloodstream t hours after a tablet is swallowed is given by: q(t) = 351(e^-t - e^-2t) in milligrams
When is the quantity of the drug in the bloodstream maximized? That is, find the value of t that maximizes q(t). Round your answer to two decimal places."

This is a surge function problem, but it's not in the normal format of ate^-bt that I'm used to working with. I can do (1/b) to find t, but I don't know how to find b. My first thought would be to combine e^-t - e^-2t to find b but I'm not sure if I can combine those just because they have the same base.

Also, for part of a problem I need the derivative of e^-kt, would that just be -ke^-kt? Or is k positive?
• Nov 25th 2009, 03:27 PM
skeeter
Quote:

Originally Posted by redsoxfan0825
Thank you very much. There are a couple of other problems I'm having trouble with as well.

"The quantity of a drug in the bloodstream t hours after a tablet is swallowed is given by: q(t) = 351(e^-t - e^-2t) in milligrams
When is the quantity of the drug in the bloodstream maximized? That is, find the value of t that maximizes q(t). Round your answer to two decimal places."

This is a surge function problem, but it's not in the normal format of ate^-bt that I'm used to working with. I can do (1/b) to find t, but I don't know how to find b. My first thought would be to combine e^-t - e^-2t to find b but I'm not sure if I can combine those just because they have the same base.

Also, for part of a problem I need the derivative of e^-kt, would that just be -ke^-kt? Or is k positive?

you're making this much harder on yourself than it really is ...

$\displaystyle q(t) = 351(e^{-t} - e^{-2t})$

$\displaystyle q'(t) = 351(-e^{-t} + 2e^{-2t})$

$\displaystyle q'(t) = 0$ when $\displaystyle 2e^{-2t} - e^{-t} = 0$

$\displaystyle e^{-t}(2e^{-t} - 1) = 0$

solve the single correct factor than can equal 0 for t

next time start a new problem with a new post.
• Nov 25th 2009, 04:23 PM
redsoxfan0825
I solved for 2e^-t - 1 = 0 and got t = .69, but when I put that back into the equation I get error so I think I made a mistake somewhere.
• Nov 25th 2009, 07:03 PM
skeeter
Quote:

Originally Posted by redsoxfan0825
I solved for 2e^-t - 1 = 0 and got t = .69, but when I put that back into the equation I get error so I think I made a mistake somewhere.

that is the correct value of t.

the maximum value of $\displaystyle q(t)$ occurs at $\displaystyle t = \ln(2) \approx 0.693...$

$\displaystyle q(\ln{2}) \approx 87.75$ mg