PDA

View Full Version : Average rate of change


Rocketsurgeon
Sep 18, 2011, 12:34 PM
Can not figure out where I went wrong:
H(t)=t^2 = 5t; t=-1 and t=1

Began with the formula:
H(1)-h(-1)/1-(-1) = 1^2 +5(1) - (-1)^2 +5(-1) = (1+5) + (1-5) = 6-4 = 2 = 2
---------------------- ----------- ---- ---
1 1 1 1

I know I mess up with distributing negatives a lot so I tried it over 2 times to no avail. Could someone please tell me what I am doing wrong? (P.S. Sorry I do not know how to input mathematical notation. Any tips on that would be appreciated as well. For clarity's sake ^2 means squared)

FadedMaster
Sep 18, 2011, 08:44 PM
For future reference: https://www.askmehelpdesk.com/math-sciences/how-technical-scientific-documentation-formulas-50415.html


Also, can you double-check your formula. Is H(t) = t^2 or is H(t) = 5t?

Or did you mean something like H(t) = t^2 + 5t? I'm sure you meant the latter and just forgot to hold down the shift key when pressing the equals button. Your work is quite difficult to read.

Assuming the function is H(t) = t^2 + 5t and t_0=-1 is your initial time:

H(t) = t^2 + 5t

H(1) = (1)^2 + 5(1) = 1+5 = 6

H(-1) = (-1)^2 +5(-1) = 1-5 = -4

So we now have:

H(1) = 6

H(-1) = -4

\frac{\Delta H}{\Delta t} = \frac{H(1)-H(-1)}{(1)-(-1)} = \frac{6-(-4)}{2} = \frac{10}{2} = 5

For me it is always easiest to be methodical. When you combined everything you dropped your denominator, and you changed the subtraction between your two functions into addition.

It might help to try to solve your functions first, then plug them into your formulas.