# Finding instantaneous rate of change.

1. Jan 16, 2012

### anonymous12

1. The problem statement, all variables and given/known data
After you eat something that contains sugar, the pH level in our mouth changes. This can be modelled by the function $L(m)=(\frac{-20.4m}{m^2 + 36}) + 6.5$ where L is the pH level and m is the number of minutes that have elapsed since eating. Find the instantaneous rate of change in pH level at 2 minutes.

2. Relevant equations
iroc = instantaneous rate of change
$$iroc= \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$$

3. The attempt at a solution

$$iroc= \lim_{h\to 0} \frac{f(2+h) - f(2)}{h}$$
$$f(2) = 1.02$$
$$f(2+h)=(\frac{-20.4(2+h)}{(2+h)^2 + 36}) + 6.5$$
$$f(2+h) = \frac{-20.4h-40.8}{h^2 + 4h + 4 + 36} + 6.5$$
$$f(2+h) = \frac{-20.4h-40.8}{h^2 + 4h +40} + 6.5$$
$$iroc= \lim_{h\to 0} \frac{(\frac{-20.4h-40.8}{h^2 + 4h +40} + 6.5) - 1.02}{h}$$
I'm really not sure what to do next. I was thinking about subtracting 1.02 from 6.5 which will give me 5.48 and then attempt to add $\frac{-20.4h-40.8}{h^2 + 4h +40} + 5.48$. If anyone could help me I would really appreciate it.

2. Jan 16, 2012

### Redbelly98

Staff Emeritus
When you evaluated f(2) you forgot to add 6.5, which is part of the function.

3. Jan 16, 2012

### anonymous12

Ok so after I fix the silly mistake, this is how far I get.

$$iroc= \lim_{h\to 0} \frac{(\frac{-20.4h-40.8}{h^2 + 4h +40} + 6.5) - 5.48}{h}$$
$$iroc= \lim_{h\to 0} \frac{(\frac{-1.02^2 - 24.48h-81.6}{h^2 + 4h +40})}{h}$$

I'm not sure what I should do next.

4. Jan 16, 2012

### Redbelly98

Staff Emeritus
Notice that, when h=0, you get -1.02+6.5-5.48=0 in the numerator.
Notice that, when h=0, you no longer get 0 in the numerator. There is an algebra or arithmetic mistake somewhere.
If you can fix the algebra, it should become clearer what to do. If not, post back for more help.

p.s. I am moving this problem to the Calculus & Beyond forum.