# Homework Help: Interpreting derivatives at a point

1. Dec 16, 2015

### musicgold

1. The problem statement, all variables and given/known data
How should I interpret the derivative of the following function at C=6%
$N = \frac {I} {C} - D$

2. Relevant equations
$\frac {dN} {dC} = \frac {-I} {C^2}$

At I =12 and C=6%, I am getting
$\frac {dN} {dC} = 3,333$

I am not sure what to make of this large number. I want to be able to say for every 1/100th increase in C, N changes by X dollars, or something like that. The following document, for example, shows one such example.
http://pblpathways.com/calc/C11_3_4.pdf

How can I interpret the 3,333 figure?

2. Dec 16, 2015

### Ray Vickson

You can think of $dN/dC = -3,333.333...$ (not +3,333) as the rate of change in $N$ per unit change in $C$. In other words, if you were to increase $C$ from $0.06 = 6 \%$ to $.06+1 = 1.06 = 106 \%$, $N$ would be expected to undergo a large decrease.

3. Dec 16, 2015

### musicgold

Thanks Ray.
I am confused. Let's say, originally I =12, C=6%, and D=75
As per the first equation N =125
Now If I changed C from 6% to 106%, N would be -64. Not sure how that change is related to -3,333.

4. Dec 16, 2015

### Ray Vickson

I said the change would be expected to be large; I did not say it would be exactly -3333.3333. The $r = -3333.333$ figure is a rate, which means that for small $\Delta C$ the change in $N$ will be approximately $r \Delta C$. However, the rate varies nonlinearly with $C$, so you cannot expect to be able to apply that rate over an interval that is not "small". You can still expect the result to be large, however.

To clarify: if you change $C$ from 6% to 6.1%, you are changing $C$ from 0.6 to .061, so $\Delta C = 0.01$. You would estimate the change in $N$ as approximately $-3,333.3333 \times 0.001 \doteq -3.333$. This is not too far from the actual change of $-3.279$.

The point (which you seem to have missed) is that the large value of the derivative is "misleading" in a way, and has to do with the units you are choosing (percent instead of the actual, numerial value).

5. Dec 16, 2015

### musicgold

Thanks Ray. Now I get the most of it.
I still however not clear on the 'misleading' part. I tried to manually calculate the change required in C to get -3,333 times decline in N, but couldn't find it.

If a 0.001 increase in C results in a 3.3 decline in N, what change in C would result in a 3,333 decline in N?

I also looked at this problem using different numbers, for example I=10,000, C=200.
n this case $\frac {dN} {dC} = -2.5$

So we can interpret that if C changes from 200 to 201, N would decline by 2.5, which makes sense. I am not sure how to interpret the answer for the original problem.

Last edited: Dec 16, 2015
6. Dec 16, 2015

### WWGD

Problem is that the rate of change is a local characteristic, although the rate may be a good approximation in a small neighborhood of the point (unless you have a linear function, in which case the rate of the change is constant). Still, for this approximation you can solve for DC/DN = -3,333.

7. Dec 16, 2015

### Ray Vickson

No increase in $C$ can lead to a 3,333 decrease in $N$. Plot the graph of $N = 12/C$ for $C > 0$ to see why.

8. Dec 16, 2015

### musicgold

Did you mean C < 0 ?
When I use less than 0 values, I get -ve N values.

9. Dec 16, 2015

### musicgold

Thanks WWGD
I am not sure what you mean here. Can you please explain a bit more?
Thanks.

10. Dec 16, 2015

### Staff: Mentor

If C > 0 but close to zero, a very small change in C leads to very large decrease in N.

I'm pretty sure Ray did not mean C < 0.

11. Dec 16, 2015

### Ray Vickson

No, I did not mean C < 0. We start from C = 0.06, and then look at increases (as I clearly stated). You will never get C < 0 that way.

Anyway, if we look at decreases (down from C = 0.06) the derivative predicts a large rate of increase in N. This will keep happening until the system goes crazy by passing through $C = 0$ and encountering singularities.