Find Max Current Given Charge

[jdawg]

Homework Statement

The expression entering the upper terminal (It is an ideal basic circuit element) is: q = 1/α₂ - (t/α +1/α²)e^-αt C
Find the maximum value of the current entering the terminal if α = 0.03679 s^-1

Homework Equations

The Attempt at a Solution

This is an example in my book that I need to understand to be able to to my homework, but I have no idea what's going on. Here is the solution:

i = dq/dt

i = 0 - [(1/α +1/α²)e^-αt(-α) +e^-αt(1/α)] Why did they subtract from 0?

i = -e^-αt[-t -1/α +1/α]

i = te^-αt

Current at max if di/dt=0 How do you know the current is at max at 0?

di/dt = te^-αt(-α) + e^-αt = 0

e^-αt(1-αt) = 0

t = 1/α

At this point I understand that they substituted for t in this equation to get the answer 10 Amps

Any help is greatly appreciated!

 

[Jeff Rosenbury]

• Why did they subtract from zero?

What is the derivative of a constant?

• How do they know current is maxed when the current's derivative = 0?

The flat points on a function's graph at typically either maximums or minimums. Since the function i(t) is negative for t < 0, and i(t) → 0 as t → ∞, any flat point above zero must be a max.

 

[rude man]

First, the given equation for q is dimensionally wrong: q has units of charge and 1/α² has units of T². BTW you copied the first term wrong. It's 1/α², not 1/α₂. Typo probably. Still, the equation has inconsistent dimensions.

Second, the point at which the slope = 0 could also be a point of inflection. In this case the second derivative would be 0 at that point. So there are 3 possibilities: max, min and inflection.
The test for a max is that d²i/dt² < 0.

The flat points on a function's graph at typically either maximums or minimums. Since the function i(t) is negative for t < 0, and i(t) → 0 as t → ∞, any flat point above zero must be a max.

I don't think i(t) is defined for t < 0. Or it would blow up. The test for a max. is the polarity of the second derivative.

 

[Jeff Rosenbury]

First, the given equation for q is dimensionally wrong: q has units of charge and 1/α² has units of T². BTW you copied the first term wrong. It's 1/α², not 1/α₂. Typo probably. Still, the equation has inconsistent dimensions.

Second, the point at which the slope = 0 could also be a point of inflection. In this case the second derivative would be 0 at that point. So there are 3 possibilities: max, min and inflection.
The test for a max is that d²i/dt² < 0.

I don't think i(t) is defined for t < 0. Or it would blow up. The test for a max. is the polarity of the second derivative.

You are mostly correct. I'm sure some clever mathematician could come up with a weird function that didn't obey even that rule. (Think of a disjoint function around an open point following some other function ... Math is funny that way.) That's why I used the word "typically".

This function would blow up, or rather down below zero. Yet pedantic equations often blow up. It's in their nature.

 

[jdawg]

I still don't understand... I know the derivative of a constant is 0, but why are we taking derivatives? I vaguely remember inflection points and all that, its been a little while since I've had calculus.

 

[Jeff Rosenbury]

You are given q. The derivative of q is i, which you want. Therefore you take the derivative.

Since the derivative is also the slope, we can use it to find maximums, minimums, and inflection points. (Where it will equal zero.)

 

[rude man]

I still don't understand... I know the derivative of a constant is 0, but why are we taking derivatives? I vaguely remember inflection points and all that, its been a little while since I've had calculus.

Not a good idea to forget that part of calculus. Set di/dt = 0, solve for i and check that d²i/dt² < 0.
A lot of calculus is best forgotten or probably should not be taught to begin with, like techniques of integration and the many ways to solve all kinds of differential equations. But finding a min or a max is of paramount importance. Review!