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Topic: CALCULUS_Velocity/Distance functions.
Replies: 2   Last Post: Jun 23, 2014 8:05 PM

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 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: CALCULUS_Velocity/Distance functions.
Posted: Jun 14, 2014 11:56 AM

>
> Could you please help me and explain how the book
> gets to the solution?
> - I do not get how the velocity (v) results 2. -
>
> Exercise.
> What functions have f (t + 1)=f (t)+ 2?
> Solution.
> The function increases by 2 in one time unit so the
> slope (velocity) is 2; f (t) =2t +C with constant C
> =f (0).
>

f(t) is a function of t, as its value changes as t changes.
velocity is the rate of change of f with respect to t,
written as df/dt
In this case f(t) = 2*t + C
When t=0 f(t)=C
To get v, which is df/dt, differentiate with respect to t
The differential of 2*t wrt t is 2, so v=2
If t increases by 1, f increases by 2
If t increases by 3, f increases by 6
Draw a graph with t on the x-axis, f on the y-axis
It is a straight line, goes thru (0,C)
and has a slope of +2
Saying f(t+1)=f(t)+2 is just another way of saying if t increases by 1 then f increases by 2
Hope this helps.
Regards, Peter Scales.

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