Differentiation of an exponential expression

[Missy]

Homework Statement

[/B]
I need to differentiate the exponential function i = 12.5 (1-e^-t/CR) and I need to plot a table so that I can do a graph of i against t but I'm not sure how. (CR is the equivelant of Capacitance 20 Micro Fards and Resistance 300 Kilo Ohms)

Homework Equations

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How do I differentiate this expression if it doesn'f follow the normal rules of differentiation (as follows)
1. y=ax^n becomes dy/dx = nax ^(n-1)
2. y = sinx becomes dy/dx = cosx
3. y = Ksinax becomes dy/dx = Kacosax
4. y = kcosax becomes kasinax
5. y = alnx becomes dy/dx = a/x

The Attempt at a Solution

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i = 12.5 (1-e^-1/cr)
i = 12.5 (1-e^-1/300000*0.00002)
i = 12.5 (1-e^-0.167)
i = 12.5 (1-e^-0.167/1)
i = 12.5 e^-0.167
i = 2.09e^0.154

 

[SteamKing]

There are special rules for differentiating exponential functions. Haven't you learned them yet?

http://en.wikipedia.org/wiki/Differentiation_rules

 

[Missy]

Our teacher didn't cover any rules past the 5 that I listed, however, it has come up in the distinction work that I am doing. I will look over the link that you posted now.

 

[Mark44]

Homework Statement

[/B]
I need to differentiate the exponential function i = 12.5 (1-e^-t/CR) and I need to plot a table so that I can do a graph of i against t but I'm not sure how. (CR is the equivelant of Capacitance 20 Micro Fards and Resistance 300 Kilo Ohms)

Homework Equations

[/B]
How do I differentiate this expression if it doesn'f follow the normal rules of differentiation (as follows)
1. y=ax^n becomes dy/dx = nax ^(n-1)
2. y = sinx becomes dy/dx = cosx
3. y = Ksinax becomes dy/dx = Kacosax
4. y = kcosax becomes kasinax
5. y = alnx becomes dy/dx = a/x

The Attempt at a Solution

[/B]
i = 12.5 (1-e^-1/cr)

No, this isn't right. You have left out t, the variable for time. For your table, find the current I for various values of t. For example, at t = 0.1 sec, you have
I= 12.5(1 - e^(-.1/(300000*0.00002)))
I get about .21 (A) for this.

The full problem description probably says what values to use for time.

i = 12.5 (1-e^-1/300000*0.00002)
i = 12.5 (1-e^-0.167)
i = 12.5 (1-e^-0.167/1)
i = 12.5 e^-0.167
i = 2.09e^0.154

 

[Missy]

No, this isn't right. You have left out t, the variable for time. For your table, find the current I for various values of t. For example, at t = 0.1 sec, you have
I= 12.5(1 - e^(-.1/(300000*0.00002)))
I get about .21 (A) for this.

The full problem description probably says what values to use for time.

Hi Mark, Thanks for your assistance. I did have 1 as my time constant in my workings. I am struggling to reach the same answer as you for the 0.1. I really wish that my teacher had covered this for us and maybe I would see it more clearly.

 

[Mark44]

Using R = 300,000 ohms and C = .00002 Farads, with t = .1 sec, you get
I = 12.5 (1 - e^(-.1/(300000 * 0.00002)))
= 12.5(1 - e^(-.1/6)) = 12.5(1 - .98347) = approx 0.207 (amp)

Do the same thing for whatever times are asked for in the problem. Or you could do the above calculation with t = 0.0, 0.1, 0.2, 0.3, and so on.

Part of the problem is in your calculations.

i = 12.5 (1-e^-1/300000*0.00002)
i = 12.5 (1-e^-0.167) <--- OK to here
i = 12.5 (1-e^-0.167/1) <--- why are you dividing by 1?
i = 12.5 e^-0.167 <--- No, this is wrong. You need to evaluate 1 - e^(-1/6) and then multiply by 12.5
i = 2.09e^0.154 <--- This works out to about 2.43. I get about 1.922

 

[Missy]

Using R = 300,000 ohms and C = .00002 Farads, with t = .1 sec, you get
I = 12.5 (1 - e^(-.1/(300000 * 0.00002)))
= 12.5(1 - e^(-.1/6)) = 12.5(1 - .98347) = approx 0.207 (amp)

Do the same thing for whatever times are asked for in the problem. Or you could do the above calculation with t = 0.0, 0.1, 0.2, 0.3, and so on.

Part of the problem is in your calculations.

I can see your answer, that's fantastic, thank you so so so much for helping me! I will plot the graph when I get in tomorrow and let you know how I got on :)

 

[Mark44]

Our teacher didn't cover any rules past the 5 that I listed, however, it has come up in the distinction work that I am doing. I will look over the link that you posted now.

The five that you listed don't apply to your current equation. You need the rule for differentiating exponential functions as well as the chain rule. Both are listed in that wiki article.

 

[Chestermiller]

If you manipulate your equation a little, you get:

[tex]ln(12-i)=ln(12)-\frac{t}{RC}[/tex]

What is the slope of the straight line you get if you plot the data as ln(12 - i) versus t? What is derivative of ln(12-i) with respect to t?

Chet

 

[Missy]

If you manipulate your equation a little, you get:

[tex]ln(12-i)=ln(12)-\frac{t}{RC}[/tex]

What is the slope of the straight line you get if you plot the data as ln(12 - i) versus t? What is derivative of ln(12-i) with respect to t?

Chet

The slope of the straight line works out at a gradient of 0.65 so it's increasing. However, when I attempted the differentiation of the entire formula I got 3.89 which indicates that either my gradients wrong (unlikely) or my differentation is wrong (likely). Please can i have more help differentiating i = 12.5e(-7/6) I am struggling with the negative AND fractional power. Just to be clear the power is (-7) over (6) not minus (7/6)

 

[Missy]

I know that the rule that is applying to my formula of i = 12.5 (1-e^(-t/cr) is:

"a(1-e^-bt)" which differentiates to abe^-bx

I have got as far as 12.5e^(-7/6) and I can see that I need to multiply 12.5 by 7 and then subtract 1 from the fraction but this gives me an overall answer that can't be right because the increase is too low, either that or my gradient is too high.

87.5e^-2.1666666667=10.02

 

[Chestermiller]

You said you know how to differentiate the natural log. So what is dln(12-i)/dt in terms of di/dt and i?

Chet

 

[Mark44]

I know that the rule that is applying to my formula of i = 12.5 (1-e^(-t/cr) is:

"a(1-e^-bt)" which differentiates to abe^-bx

I have got as far as 12.5e^(-7/6)

Your derivative should be written in terms of t, not x. So d/dt(a(1 - e^-bt) = abe^-bt.

and I can see that I need to multiply 12.5 by 7 and then subtract 1 from the fraction

No, don't subtract 1. That's not part of the derivative.
Also, since the current and the derivative are both functions of time t, both will have different values for different t. In your calculation above, what value of t are you using?

but this gives me an overall answer that can't be right because the increase is too low, either that or my gradient is too high.

87.5e^-2.1666666667=10.02