# Using deritative to calc irc

• Jan 21st 2008, 03:14 PM
crazydaizy78
Using deritative to calc irc
Use the deritative to calculate the instanteous rate of change of the given input value. (You can eliminate the h algebraically.

13. s(t)=16r^2+64 where t=2

I haven't a clue where to start...limits are foreign to me. Please help if you can. (Tmi)
Thanks a million!!
• Jan 21st 2008, 07:45 PM
mr fantastic
Quote:

Originally Posted by crazydaizy78
Use the deritative to calculate the instanteous rate of change of the given input value. (You can eliminate the h algebraically.

13. s(t)=16t^2+64 where t=2 Mr F edit in red.

I haven't a clue where to start...limits are foreign to me. Please help if you can. (Tmi)
Thanks a million!!

You've got to overcome your xenophobia!

First find $\displaystyle \frac{s(t+h) - s(t)}{h}$:

1. $\displaystyle s(t + h) = 16(t + h)^2 + 64 = 16(t^2 + 2th + h^2) + 64= 16t^2 + 32 th + 16h^2 + 64$.

2. $\displaystyle s(t + h) - s(t) = 16t^2 + 32 th + 16h^2 - (16t^2 + 64)$

$\displaystyle = 16t^2 + 32 th + 16h^2 - 16t^2 - 64 = 32 th + 16h^2$.

3. $\displaystyle \frac{s(t+h) - s(t)}{h} = \frac{32 th + 16h^2}{h} = 32t + 16h$.

Now let h --> 0: $\displaystyle \lim_{h \rightarrow 0}\frac{s(t+h) - s(t)}{h} = 32t + 16(0) = 32t$.

Now sub t = 2. You get 64.