# Homework Help: Plotting Derivatives

1. Jan 19, 2010

### jayeffarr

The problem statement.

Suppose x''(t) = 1 for $$1\leq(t)\leq2$$, and x''(t) = 0 for all other (t)

(a) Plot x''(t) for $$0\leq(t)\leq3$$
(b) Plot x'(t) for $$0\leq(t)\leq3$$. Assume x'(0) = 0
(c) Plot x(t) for $$0\leq(t)\leq3$$. Assume x(0) = 0

The attempt at a solution

I assumed 'x' being the vertical axis and 't' being the horizontal axis.

For (a) I know that there are going to be two points at 1 and two points at 0.

My main question is when I graph these plots should I treat points 0 and 3 on the t-axis as discontinuities, and just put a point of where they're at and not include them when connecting the non-zero points, or should I connect all the points together, despite the discontinuity?

2. Jan 19, 2010

### tiny-tim

Hi jayeffarr!

(have a ≤ )
Personally, I'd leave them disconnected …

x''(t) has only one value for each t, so why pretend it has more?

3. Jan 19, 2010

### jayeffarr

Would you say to do the same thing when it comes to plotting x' and x...

Since x'(t) = t

and the plot will be (0,0), (1,1), (2,2), (3,0).

And x(t) = (1/2)t²

and the plot will be (0,0), (1,0.5), (2,2), (3,0).

...and just draw a line between 1 & 2, while having solid circles and 1 & 3?

Or are you saying just to plot the points without connecting ANY of them?

Last edited: Jan 19, 2010
4. Jan 20, 2010

### tiny-tim

Not following you

x'(t) is the area under the graph of x''(t).​

5. Jan 20, 2010

### HallsofIvy

I don't know what you mean by "two points". The graph is the horizontal straight line at y= 1 between 1 and 2, the horizontal straight line at y= 0, the x-axis, for all other t.

"at 3 on the t-xis"? x"(3)= 0 and is 0 for every point near t= 3. Did you mean t= 2? In any case, it really doesn't matter. Technically, the vertical line is not part of the graph but if you can notice that this forms a rectangle under the graph, that will help.