Hatrack River Forum: Math Help

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Posted by dantesparadigm (Member # 8756) on :

Edit: This thread is now the place to ask any math related questions you may have, I'm recycling.

I considered posting this to a math forum, but then I remembered Jatraqueros were the smartest people in the world.

Anyway, part of a calculus lab I'm working on involves finding the point of inflection of an equation by finding where the second derivative is equal to zero. I understand how to find the derivatives and set the second one to zero, the problem is the algebra involved just gets so gorram turgid (did I use that right?) that I end up coming up with a wrong answer, so, as much as I hate to do this I'm asking Hatrack to bail me out, or perhaps guide me with subtle hints.

The equation in question is C(t) = 14.6te^(-5.36t)

I really appreciate any help that might be offered, as at this point I’ve gone through half a dozen sheets of paper and I’m about ready to smash my head in with a natural log.

[ January 20, 2007, 08:55 AM: Message edited by: dantesparadigm ]

Posted by Tatiana (Member # 6776) on :

Is an exponential function going to have an inflection point? Does't it just curve smoothly upwards? I think there's no inflection point, but let's try taking the derivative to be sure.

I'm going to parse this as u = 14.6t and v = e^(-5.36t)

Then d (u*v) / dt = u dv + v du = 14.6t (-5.36) e^(-5.36t) + 14.6 * e^(-5.36t)

Posted by Tatiana (Member # 6776) on :

So then you combine like terms, and get (-78.256t + 14.6)*e^(-5.36t)

For the second derivative of that, this time make
u = -78.256t + 14.6 and
v = e^(-5.36t)

So d(u*v)/dt = udv + vdu = (-78.256t + 14.6)*-5.36*e^(-5.36t) + e^(-5.36t)(-78.256) =
(419.4521t -78.256 -78.256)e^(-5.36)t =
(419.4521t -156.512)e^(-5.36)t

where does that equal zero?

clearly, the only place is where 419.4521t = 156.512.

so t = 156.512 / 419.4521
t = .3731343

Did I do that right?

Posted by Mathematician (Member # 9586) on :

quote:
Originally posted by Tatiana:
Is an exponential function going to have an inflection point? Does't it just curve smoothly upwards? I think there's no inflection point, but let's try taking the derivative to be sure.

First, yes, you are correct that a plain old exponential function wouldn't have an inflection point - but this isn't a plain old exponential - it's got that pesky t being multiplied by it.

Second, just because something "curves smoothly upward" doesn't mean it doesn't have an inflection point. As an example, f(x) = -x^3 + 3x + 4 has an inflection point at 0, but it "curves smoothly upward" around x = 0.

quote:
Originally posted by Tatiana:

Did I do that right?

I didn't check the multiplication/division of the numbers, so I'm not certain the answer is correct, but the general method followed over the last two posts seems perfect to me.

PS, Keep the math questions coming! (Though I'm probably done answering them for tonight - already had a 2 hour exam today).

PPS - What does "mayfly" in the topic mean?

Posted by Tatiana (Member # 6776) on :

Kyya got the same answer, so I guess I did. =) Does it make sense to you what I did, dp? I just used the product rule both times.

Posted by SoaPiNuReYe (Member # 9144) on :

Mayfly means it the topic will be deleted after a short time.

Posted by SenojRetep (Member # 8614) on :

C(t) = 14.6te^(-5.36t)
C'(t) = 14.6t*(-5.36)e^(-5.36t)+14.6*e^(-5.36t)
..... = 14.6(-5.36t+1)e^(-5.36t)
C''(t) = -5.36*14.6(-5.36t+1)e^(-5.36t)+14.6*(-5.36)e^(-5.36t)
..... = (419.45216t-156.512)e^(-5.36t)

Setting C''(t) = 0, I get 419.45216t*e^(-5.36t) = 156.512*e^(-5.36t); dividing both sides by the exponential (which will never be zero so you can do that) results in
419.45216t = 156.512, or
t = 0.373134...

And now, looking at Tatiana's post I see we have confirmation.

Posted by Tatiana (Member # 6776) on :

Oooh I posted my last post before I saw your post, m.

Mayfly means it's subject to immediate deletion once the person gets the answer to their question. In this case I trusted dp that he had been working on the problem for a long time and couldn't quite wrap his head around it.

Posted by SenojRetep (Member # 8614) on :

Often I find that in problems with ugly coefficients, it's best to replace them with something simple.

So here, for instance, I would start by letting a=14.6 and b=-5.36; your work is much faster when you don't have to keep track of such complicated representations.

C(t) = a*te^(bt)
C'(t) = (a*b)te^(bt)+a*e^(bt)
..... = a(bt+1)e^(bt)
C''(t) = ab(bt+1)e^(bt)+(ab)e^(bt)
..... = (abb)te^(bt)+(2ab)e^(bt)

Then substitute back in and solve.

Posted by dantesparadigm (Member # 8756) on :

Perfect, I was starting it by pulling out the 14.6 and not doing anything with it until I was done, I guess it caused more problems than it solved.

I'm going to go ahead and edit the topic to make this a general math help thread.

Thanks a bunch.

Posted by SenojRetep (Member # 8614) on :

You can do that, too. Using a and b you have
C(t) = a(te^(bt))
C'(t) = a((bt+1)e^(bt))
..... = a(b(te^bt)+e^(bt))
C''(t) = a(b(bt+1)e^(bt)+be^(bt))
...... = (abb)te^(bt)+(2ab)e^(bt)