This is my very first question here. I am self-studying maths and I am currently doing pre-calculus from Gilbert Strang's textbook "Caculus 1". So I apologize in advance if this question comes across as basic.
I am puzzled by the following problem, which is a recommended problem (not mandatory) for pre-calc:
An up and down velocity has the following piecewise function
(1) $v(t) = 2t$ for $t \leq 3$
(2) $v(t) = 12 - 2t$ for $t > 3$
Draw the piecewise parabola $f(t)$ . Check that $f(6)$= area under the graph of $v(t)$.
Now, the area under the graph of $v(t)$ equals 18.
What I find difficult is to find the formula of $f(t)$ (the pieceswise parabola). At first I thought that the two velocity functions are simply the derivatives of the two parabola functions. By "reversing" the derivatives (sorry but I have no knowledge of integrals yet), I get:
(3) $f(t) = t^2$ for $x\leq3$
and
(4) $f(t) = 12t-t^2$ for $x > 3$
But this cannot be right. Concerning the area under (1) and (2) = 18, obviously if I plug $f(6)$ I get 36, which is wrong. In my understanding, 36 would be the area under (2) if it didn't start from $x > 3$, so without taking into account the fact that this is a piecewise function.
Moreover, if I plot functions (3) and (4), I get a "jump" at $t = 3$, from $f(t)=9$ (from (3)) to $f(t)=27$ (from (4)).
I don't know if one can get a discontinous function from a continuous derivative function, but it seems counterintuitive to me (we simply "changed" the velocity at $x=3$, why the heck should we "jump" from distance 9 to distance 27? Can we teleport?)
My guess is that function (4) is not quite right, and that I should get a continuous piecewise function of parabolas. But I don't know how to proceed mathematically, and I cannot find a worked out solution anywhere.
Most explanations of similar problems I found require knowing integrals.
Can anybody solve this without using integrals and provide a clear explanation to a newbie?
Thank you very much!
But, when $t>3$, $$ f(t)=\int_0^t v(s) ds = \int_0^3 v(s) ds + \int_3^t v(s) ds = 9+[12s-s^2]_3^t = -t^2+12 t-18. $$
If you wish to avoid integrals, you can just compute the area under the graphics of $v(t)$ up to a generic time $t$. Like before, when $t\leq 3$ the area under the graph (from 0 up to $t$) is the area of a triangle, given by $f(t) = \frac{t \cdot 2t}{2} = t^2$. When $t>3$, the cumulative area is all the area from $t = 0$ to $t=3$, which is 9, plus the area of a trapezoidal region, that you can compute interns of $t$.