# Differentiation Application

• May 20th 2010, 04:50 PM
kaiser0792
Differentiation Application
Thanks to those who responded yesterday to my question regarding the differentiation in this problem. As it turns out, I wasn't off as far as I thought I might be. I believed that correctly differentiating the function was the root of my problem, but as it turns out, I'm not sure how to apply the derivative correctly. Here is a restatement of the problem fleshed out:

The expression for the charge entereing the upper terminal of a basic,
2-terminal circuit element, is given by,

$q=\frac{1}{\alpha^2}-{(\frac{t}{\alpha}+\frac{1}{\alpha^2})}e^{-\alpha t}$ Coulombs.

Find the maximum value of the current entering the terminal if $\alpha=0.03679\ s^{-1}$.

I thought that finding $\lim_{t\rightarrow\infty}\frac{dq}{dt} = te^{-\alpha t}$ would give me the answer. I applied L' Hopital's Rule and ended up with zero, which obviously is incorrect. The book gives an answer of 10 Amperes. If there are any engineers out there, I could use some help. P.S. Amperes = Couloumbs/Second, which is what led me to differentiation in the first place.
• May 20th 2010, 05:19 PM
drumist
At a given time $t$, the current is equal to $\frac{dq}{dt} = I$. So to find the maximum current, we need to take another derivative to maximize $I$.

$\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$

Can you finish?
• May 20th 2010, 06:44 PM
kaiser0792
I assume that the second derivative is taken to find relative maximum.
But I still cannot figure out where to go.

If I take the limit as t approaches infinity, I still end up with 0. And I might be wrong to even take that approach.

$\lim_{t\rightarrow \infty}(\frac{1}{e^{\alpha t}}-\frac{\alpha t}{e^{\alpha t}})= 0 - \frac{\infty}{\infty}$

Applying L' Hopital's Rule to the indeterminate form: gives 0 - 0 = 0.

Where am I going wrong???
• May 20th 2010, 07:09 PM
kaiser0792
Quote:

Originally Posted by drumist
At a given time $t$, the current is equal to $\frac{dq}{dt} = I$. So to find the maximum current, we need to take another derivative to maximize $I$.

$\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$

Can you finish?

I assume that the second derivative is taken to find relative maximum.
But I still cannot figure out where to go.

If I take the limit as t approaches infinity, I still end up with 0. And I might be wrong to even take that approach.

http://www.mathhelpforum.com/math-he...853e5354-1.gif

Applying L' Hopital's Rule to the indeterminate form: gives 0 - 0 = 0.

Where am I going wrong???
• May 20th 2010, 07:58 PM
drumist
I'm not sure why you are taking a limit. There is no limit involved in the calculation.

You're correct that I was looking for relative maximums because that is usually the method used to solve for the absolute maximum.

Since we are looking for a maximum for $I$, we take its derivative and set it equal to zero, and solve for $t$:

$\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$

Since $e^{-\alpha t}$ can never equal zero, we know that:

$1-\alpha t = 0 \implies t = \frac{1}{\alpha}$

So the value of $t$ where the maximum occurs is $t=\frac{1}{\alpha}$. Therefore:

$I = t e^{-\alpha t} = \frac{1}{\alpha} e^{-1} = \frac{1}{\alpha e} \approx 10$
• May 20th 2010, 08:31 PM
kaiser0792
Quote:

Originally Posted by drumist
I'm not sure why you are taking a limit. There is no limit involved in the calculation.

You're correct that I was looking for relative maximums because that is usually the method used to solve for the absolute maximum.

Since we are looking for a maximum for $I$, we take its derivative and set it equal to zero, and solve for $t$:

$\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$

Since $e^{-\alpha t}$ can never equal zero, we know that:

$1-\alpha t = 0 \implies t = \frac{1}{\alpha}$

So the value of $t$ where the maximum occurs is $t=\frac{1}{\alpha}$. Therefore:

$I = t e^{-\alpha t} = \frac{1}{\alpha} e^{-1} = \frac{1}{\alpha e} \approx 10$

Thank you for your patience, and I'm sure that I'm trying it, but my reasoning was that since dq/dt represents current, I was thinking that if I differentiated "q" and took the limit as t approached infinity, that the current would be maximized. I follow the mathematics of your explanation, but I still don't understand why setting the 2nd derivative equal to zero leads to max current. I thougt that critical numbers came from the 1st derivative set equal to zero and that if the second derivative evaluated at a critical number was less than zero, that it indicated a relative max. If you would be so kind as to elaborate a little further. I really want to learn, I'm just disconnecting somewhere. Thanks
• May 20th 2010, 10:40 PM
drumist
It's important to realize that the problem is asking you to find the maximum value of current, not the maximum value of charge.

To maximize current, the first step is to find an expression for current. Since the problem told you what the charge is, you can take a derivative to find the expression for current, which is what we did in the other thread from yesterday.

If we were trying to find the maximum value of charge, then we would set $\frac{dq}{dt} = 0$ because $\frac{dq}{dt}$ is the first derivative of charge. However, that's not what the problem asks for!

Instead, we want the maximum value of current. The current is equal to $\frac{dq}{dt}$, so to find the maximum value, we must calculate the first derivative of the current, which is $\frac{d^2 q}{dt^2}$ and set that equal to zero.

If you wanted to verify that it was a maximum and not a minimum, you would want to calculate the second derivative of current (which is the third derivative of charge), i.e., $\frac{d^3q}{dt^3}$ and determine if it is positive or negative at the value of $t$ we found.

You'll notice in my work I went ahead and created a new variable $I$ to represent the current. I just did this to hopefully avoid the confusion between charge and current. You may notice that:

$I = \frac{dq}{dt}$ <-- we want to maximize this

$\frac{dI}{dt} = \frac{d^2q}{dt^2}$ <-- so we set this equal to zero

$\frac{d^2 I}{dt^2} = \frac{d^3q}{dt^3}$ <-- and make sure this is positive
• May 21st 2010, 12:14 AM
kaiser0792
Thank you Drumist, I was missing that fact that what I was considering to be the 2nd derivative (of charge) was the first derivative of current!

You have been a great help. I really appreciate it.