# Thread: Is anyone familiar with Rate of change?

1. ## 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?

2. Hello,
Originally Posted by jemadd2
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 :

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

Try to do it for your problem, please

3. So this means:

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

4. Originally Posted by jemadd2
So this means:

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

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

5. Originally Posted by Moo
Yes it is

(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 =/

6. Originally Posted by jemadd2
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 =/
It's correct !

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

7. wow thank you so much....im glad i understand now