I have the answer can someone please see if I did this right....
Suppose we have partially filled water tank and water begins entering the tank. The amount of water, in gallons in the tank is given by w(t), where t is the number of minutes after water began entering the tank. The graph above is of
w'(t), the rate at which water is entering the tank in gallons/minute. Find and interpret

Question

I have the answer can someone please see if I did this right....
Suppose we have partially filled water tank and water begins entering the tank. The amount of water, in gallons in the tank is given by w(t), where t is the number of minutes after water began entering the tank. The graph above is of
w'(t), the rate at which water is entering the tank in gallons/minute. Find and interpret

anonymous · Answer

After intrepret is this$$\int\limits_{0}^{20} w'(t)  dt$$

anonymous · Answer

attachment

e.mccormick · Answer

So what answer did you get?

anonymous · Answer

I got $$\int\limits_{0}^{20} 60d20$$

anonymous · Answer

then 0 as a final answer but I think I might have messed something up

e.mccormick · Answer

Well, because it is a integral of the derivative, the goal is to find the original equation.

anonymous · Answer

The original equation is the one that I posted first

e.mccormick · Answer

That derivative seems to be piecewise, which makes it a little more complex.

It says the graph is of w'(t).

anonymous · Answer

right the left axis going vertical is that

e.mccormick · Answer

If you look, for $0\le t < 6$ you have one rate, $6\le t < 14$ is another, and $14\le t < 20$  is a third equation because while it is similar to the first, the base is different.

anonymous · Answer

ok then what is next I guess I am confussed on how to solve the problem then

anonymous · Answer

I understand where you got the last post from what do I do next?

e.mccormick · Answer

Well, find a set of equations that would result in that graph.  that will be w'(t).  Then integrate it.

anonymous · Answer

how do you do that

e.mccormick · Answer

Which part of that is the issue?

anonymous · Answer

finding the equatino the next step

e.mccormick · Answer

OK.  Well, those are linear representation, so we can use simple point slope.  

For $0\le t < 6$ you can put any points in the form of $(t,w'(t))=(x,y)$.  You have the points $(0,0)$ and $(2,10)$ that are on that line.  So what would the equation of the line be?

e.mccormick · Answer

Yes, we are reviewing point/slope formula.... welcome to calculus.

anonymous · Answer

so then is the equation y=10x=0

anonymous · Answer

sorryy=10x+0

anonymous · Answer

@e.mccormick is that close at all I havenot done the point slope formula in a while

anonymous · Answer

are you still there

e.mccormick · Answer

Got called away for a bit.... I am at work.  Hehe.

anonymous · Answer

oh ok sorry

e.mccormick · Answer

5x, not 10.  2t=10, so...

anonymous · Answer

y=5x+0

e.mccormick · Answer

So the first part of the piecewise function is:
5t for $0\le t < 6$
The second part is really simple:
30 for $6\le t < 14$
What would this leave for the last part?  It is similar to the first part.

anonymous · Answer

70

e.mccormick · Answer

Well, it would still have the 5t in there somehow, that is why I say it is similar, but the starting t is 14 to get 30.... and 30=6*5, not 14*5.... sooooo....

anonymous · Answer

so then 16

e.mccormick · Answer

5(t-??)=something.  That is the form it needs.  And the first value needs to be:
w'(14)=30

anonymous · Answer

ok would it be 2

e.mccormick · Answer

Lets try going back to point slope... you have these points:
(14,30) and (16,40)
Slope = $\Delta y/\Delta x=(40-30)/(16-14)=5$  That much we know. Then, we take one point and find the formula using:$y-y_1=m(x-x_1)$ where m=5, and $(x_1,y_1)$ is one of those two points.

e.mccormick · Answer

Yah, this is like Algebra 1 or 2, so by the time you are in calculus you have not seen these problems in a year or two.  So it catches a lot of people off guard.  Calculus is 90% doing algebra, trig, and so on to both set up and simplify the problems.

$y-30=5(x-14) \implies y-30=5x-70 \implies y=5x-40$

Remember doing those?

e.mccormick · Answer

This makes the piecewise function of:$$
   w'(t) = \left\{
\begin{array}{lcl}
       5t & \mathrm{for} &0\le t < 6\
       30 & \mathrm{for} &6\le t < 14\
       5t-40 & \mathrm{for} &14\le t \le 20
\end{array}
\right. 
$$Now you need to integrate that.  Remember that each piece will have its own integration rang based off of where it is on t.  For example, instad of doing one integral from 0 to 20, the first part will be an integral of 0 to 6 for the first piece. THe second will be defined by the second piece, and the third by the third.
I hope that gives you enough to finish whenever you get back.

anonymous · Answer

can someone please help me with this I have no idea what I am doing

e.mccormick · Answer

Ah ha!  You were back...

e.mccormick · Answer

Now, you see the w'(t) U wrote there?  That is what all the algebra was for.  That is what you need to integrate o get w(t).  But you can't do it in one shot.