Tangent answer not agreeing with calculator?

Hi again, all! I worked out the velocity of this curve s= t^2 - 8t + 18 at t = 4 to be 2 m/s, but the graph on my calculator really doesn't look like the instant v at t = 4 is anywhere near 2 (looks like zero). So I tried to write out the general formula for this and am not sure my next step, any hint would be valued :)

lim h -> 0 (a^2 + 2ah + h^2 - 8a - 8h + 18) / h

There's nothing to cancel out so how can I make this into a workable limit?

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Originally Posted by DannyMath

Hi again, all! I worked out the velocity of this curve s= t^2 - 8t + 18 at t = 4 to be 2 m/s, but the graph on my calculator really doesn't look like the instant v at t = 4 is anywhere near 2 (looks like zero). So I tried to write out the general formula for this and am not sure my next step, any hint would be valued :)

lim h -> 0 (a^2 + 2ah + h^2 - 8a - 8h + 18) / h

There's nothing to cancel out so how can I make this into a workable limit?

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What you have is equivalent to saying $\displaystyle \lim\limits_{h\to 0}\dfrac{f(a+h)}{h}$ which is incorrect. You didn't subtract off that value of the function at a! If you evaluate $\displaystyle \lim\limits_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$, you will get the desired cancellations. Then plug in $\displaystyle a=4$ (you could actually do this from the start..makes calculations a tad easier to do).

Can you proceed?

I will first quote Homer Simpson: D'oh!

I will then quote Dr. Farnsworth: Huzzah!

I will then compliment both of your mini animations, which did indeed give me the chuckles. Good day to you, sir!