Finding instantaneous rate of change.

[anonymous12]

Homework Statement

After you eat something that contains sugar, the pH level in our mouth changes. This can be modeled 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.

Homework Equations

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

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.

 

[Redbelly98]

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

 

[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.

 

[Redbelly98]

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

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

Notice that, when h=0, you get -1.02+6.5-5.48=0 in the numerator.

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

Notice that, when h=0, you no longer get 0 in the numerator. There is an algebra or arithmetic mistake somewhere.

I'm not sure what I should do next.

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.