Algebraisk løysinga av grenseverdiar

Hastigheita på ein sprettball

Hans Georg Schaathun

Mars 2016

$$ h(t) = 10 - 4.9t^2$$

$$ \bar v(t,\Delta t) = \frac{\Delta h}{\Delta t} = \frac{h(t+\Delta t) - h(t)}{\Delta t} $$

$$ \bar v(t,\Delta t) = \frac{[10 -4.9(t+\Delta t)^2] - [10-4.9t^2]}{\Delta t} $$

$$ \bar v(t,\Delta t) = 4.9\frac{ t^2 - (t+\Delta t)^2}{\Delta t} $$

$$ \bar v(t,\Delta t) = 4.9\frac{ t^2 - [t^2+2t\Delta t+(\Delta t)^2]}{\Delta t} $$

$$ \bar v(t,\Delta t) = - 4.9\frac{ 2t\Delta t+(\Delta t)^2}{\Delta t} $$

$$ \bar v(t,\Delta t) = - 4.9(2t+\Delta t) \quad\text{ if } \Delta t\neq 0 $$

$$ \bar v(t,\Delta t) = - 4{,}9(2t+\Delta t) \quad\text{ if } \Delta t\neq 0 $$

$$ v(t) = \lim_{\Delta t\to0} \bar v(t,\Delta t) $$

$$ v(t) = \lim_{\Delta t\to0} [- 4{,}9(2t+\Delta t)] $$

$$ v(t) = - 4{,}9\cdot 2t = - 9{,}8t $$