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Calculus1 7 Online
OpenStudy (anonymous):

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OpenStudy (anonymous):

OpenStudy (study_buddy99):

I believe it is 1200 because all you have to do is read the graph :)

OpenStudy (anonymous):

R is the amount of people joining the line, so I need to find the integral from 0 to 5 of the whole graph

OpenStudy (anonymous):

But I do not know exactly how to find the integral- do you know how?

OpenStudy (study_buddy99):

no, I don't I' sorry. Good luck!

OpenStudy (anonymous):

Since the function is piece wise, you'll need to integrate each part separately.

OpenStudy (anonymous):

We will let \(t\) be in hours, and say noon is hour \(0\).

OpenStudy (anonymous):

We'll find the number of people to show up by 2 P.M

OpenStudy (anonymous):

\(p\) will the function for the number of people at any time.

OpenStudy (anonymous):

\[ p(2)-p(0) = \int_0^2 R(t)~dt \]Between noon and 2 P.M. we have \(R(t) = \frac{2100-800}{2-0}t+800\). I used the slope point formula to get this.

OpenStudy (anonymous):

Keep in mind they already tell us \(p(0) = 300\).

OpenStudy (anonymous):

ah okay so then it would be the integral from 0 to 2 of 650t+800

OpenStudy (anonymous):

Whoops, I should have put \(\frac{1200-800}{2-0}\). Do you understand how I got that?

OpenStudy (anonymous):

Ah yes using the formula y=mx+b

OpenStudy (anonymous):

Okay, so you should be able to get the other \(R\) functions for other intervals.

OpenStudy (anonymous):

so then it would be integral of 0 to 2 from 200t+800

OpenStudy (anonymous):

and then the next integral would be from 2 to 3 of (1200-800) t + 1200 Am I correct?

OpenStudy (anonymous):

no it would be (800-1200)t +1200

OpenStudy (anonymous):

integral from 2 to 3 of (-400) t + 1200 right?

OpenStudy (anonymous):

The intercept is not going to be \(1200\). before it was a special case that the intercept and initial point were the same.

OpenStudy (anonymous):

Wait so how would I go about finding the intercept?

OpenStudy (anonymous):

Well, we start with \(y=mx+b\).

OpenStudy (anonymous):

We have the equations: \[ y_2=mx_2+b \\ y_1=mx_1+b \]Two linear equations, two variables.

OpenStudy (anonymous):

Basically, when you have \(m\), you can use \(y_2-mx_2=b\)

OpenStudy (anonymous):

Wait so

OpenStudy (anonymous):

When I'm trying to find the equation for 2pm to 3pm

OpenStudy (anonymous):

I know the slope is -400

OpenStudy (anonymous):

y=-400x + b what x and y values do i plug in

OpenStudy (anonymous):

Any point on the line.

OpenStudy (anonymous):

and I do need to create two distinct equations? or just one

OpenStudy (anonymous):

Well, you have solve for \(m\) the normal way and just use one, I guess

OpenStudy (anonymous):

So y=-400x + b, then plugging in a point on the line that i'm trying to find 800=-400(3)+b

OpenStudy (anonymous):

\((x,y) = (2,1200)\)

OpenStudy (anonymous):

Or \((x,y)=(3,800)\)

OpenStudy (anonymous):

so b = -400

OpenStudy (anonymous):

no no it would be b=2000

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

So this will allow you to solve for \(p(3)\)

OpenStudy (anonymous):

then I do, 4, 5, add them all up, and then subtract the number of people who have gotten on the ride, correct?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

We can use the fact\[ p(5) - p(0) = (p(2)-p(0)) + (p(3)-p(2)) + (p(4)-p(3)) + (p(5)-p(4)) \]

OpenStudy (anonymous):

I don't understand, I thought I'm just using integrals for each one then adding up each integral

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