Interpreting the Derivative of N = I/R - D

[musicgold]

Hi

I am not sure if my interpretation of the following derivative is correct.

N = I / R - D

Where , N, I and D are integers, while R is a fraction (1% to 15%).

If I differentiate the above equation with respect to R, I get the following equation.

dN/dR = I / R^2

The following is my interpretation of this derivative.

1. The lower the value of R, the higher the value of dN/dR, at a given I

2. At a given R, the higher the value of I , the higher the value of dN/dR

3. If I plot dN/dR against R, at various values of I, I will get exponentially declining curves, with curves with higher I values lying on the left of curves with lower I values.


Thank you,

MG.

 

[Mark44]

You should have dN/dR = -I/R². Also, when you differentiate a variable, you are assuming tacitly that it can take on all real values in some interval. You can evaluate the derivative at integer values.

The following is my interpretation of this derivative.

1. The lower the value of R, the higher the value of dN/dR, at a given I

2. At a given R, the higher the value of I , the higher the value of dN/dR

3. If I plot dN/dR against R, at various values of I, I will get exponentially declining curves, with curves with higher I values lying on the left of curves with lower I values.

1. The smaller R is the more negative dN/dR will be (assuming that I > 0).
2. For a given R, the larger I is, the more negative dN/dR will be (again assuming that I > 0).
3. Take into account that you had the wrong sign for your derivative.

 

[Mute]

If I plot dN/dR against R, at various values of I, I will get exponentially declining curves, with curves with higher I values lying on the left of curves with lower I values.

The highlighted part is incorrect. Exponential curves are of the form $b^{\pm a x}$, where x is the variable and so would be your R. The plus sign corresponds to exponential growth as x gets large and the minus corresponds to exponential decay as x gets large.

The behaviour of the derivative you give is that it varies inversely as a quadratic. (Inversely means 1/x and quadratic means x^2).

 

[musicgold]

thanks folks.

 

[paulfr]

I believe your derivative should be ...

dN/dR = - I / [ R - D ]^(- 2)

That just causes a shift of the graph D units to the right.

 

[Mark44]

I believe your derivative should be ...

dN/dR = - I / [ R - D ]^(- 2)

That just causes a shift of the graph D units to the right.

You are interpreting the original equation, N = I/R - D as if it had been written N = I/(R - D). I am working with the equation exactly as it was written, which is the same is if it had been written N = (I/R) - D.