Here's the question you clicked on:
lexusbreon
If a stone is thrown down at 100 ft/s from a height of 1,250 feet, its height after t seconds is given by s = 1,250 − 100t − 16t^2 Estimate its instantaneous velocity at time t = 2
We know that the equation for instantaneous velocity is \( f'(x)=\large \frac{f(x+h)-f(x)}{(x+h)-x}\)
Are you familiar with this equation?
so \(f(t)=1250-100t-16t^2\) \(f(t+h)=1250-100(t+h)-16(t+h)^2\) so \(f'(x)= \large \frac{(1250-100(t+h)-16(t+h)^2)-(1250-100t-16t^2)}{(x+h)-x}\) \(f'(x)=\large \frac{1250-100t-100h-16t^2-32th-16h^2-1250+100t+16t^2}{h}\) \( f'(x)=\large \frac{-100h-32th-16h^2}{h}\) \(f'(x)= \large \frac{h(-100-32t-16h)}{h}\) \(f'(x)=(-100-32t-16h)\)
Ok Are you familiar with limits? Like what does f'(t) equal as h approaches 0?
Did ya follow so far?
Ok and keep in mind by accident I used X's in stead of T's so just ignore that
|dw:1373424544924:dw|
ok so the difference btwn x and x+h is just h
Basically we have 2 points; x and x+h and we want them to beee sooooo close together So we want h to be realllllly small So we want h to be like nearly 0 Like the number right next to 0 So there is this method called limits and we limit the equation meaning we want to see what the equation will equal as h APPROACHES 0 \(\large f'(t)_{\lim h \to 0} -100-32t-16h\) Now there is only one term that has h in it which is -16h Now as h approaches 0 then -16h approaches 0 since 16*.00000000001=.00000000016 So its basically 0 so we consider it as if it is 0 and just trash that whole term \(\large f'(t)_{\lim h \to 0} -100-32t-16h=-100-32t\) Now we need to find the Instantaneous velocity when t=2 So we plug into our equation t=2 \(f'(2)=-100-32(2)=-164
So at t=2 the stone is travelling -164 ft/s downwards
I hope you followed :)