# Is anyone familiar with Rate of change?

• Sep 10th 2008, 12:04 PM
Is anyone familiar with Rate of change?
The equation: H(t)=sqr root(3t+5)

The question: What is the average rate of change of H(t) with respect to t as t changes from 2 to 2.01?

Can anyone explain this to me?
• Sep 10th 2008, 12:07 PM
Moo
Hello,
Quote:

The equation: H(t)=sqr root(3t+5)

The question: What is the average rate of change of H(t) with respect to t as t changes from 2 to 2.01?

Can anyone explain this to me?

The average rate of change of a function f, between a and b is :

$\displaystyle \frac{f(b)-f(a)}{b-a}$

• Sep 10th 2008, 12:14 PM
So this means:

f(2.01) - f(2) / 2.01-2 ?
is this correct?
• Sep 10th 2008, 12:15 PM
Moo
Quote:

So this means:

f(2.01) - f(2) / 2.01-2 ?
is this correct?

Yes it is (Wink)

(here, it's not f, it's H but it doesn't matter, you've grasped the concept)
• Sep 10th 2008, 12:21 PM
Quote:

Originally Posted by Moo
Yes it is (Wink)

(here, it's not f, it's H but it doesn't matter, you've grasped the concept)

okay so now i just fill it in the function.
so it would be
1. f(2.01) - f(2) / 2.01-2
2. sqr root(3(2.01)+5) - sqr root(3(2)+5) / 2.01-2
3. im confused =/
• Sep 10th 2008, 12:24 PM
Moo
Quote:

$\displaystyle \frac{\sqrt{3 \cdot 2.01+5}-\sqrt{3 \cdot 2+5}}{2.01-2}=\frac{\sqrt{11.03}-\sqrt{11}}{0.01} \approx 0.45195907277229$ (use a calculator)