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- anonymous

Need help with the step-by-step of using the Midpoint Rule to find the distance traveled in the following scenario: v(t) = 3t-5, 0

Mathematics
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- anonymous

Need help with the step-by-step of using the Midpoint Rule to find the distance traveled in the following scenario: v(t) = 3t-5, 0

Mathematics
- jamiebookeater

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- dumbcow

is v velocity?

- anonymous

yes :D, therefore, the displacement is just the antiderivative of that function, right?
My question is, how do I utilize the Midpoint Rule to find the total distance traveled, as opposed to the displacement?

- anonymous

I'm just confused.

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- dumbcow

correct you could find distance traveled as
\[\int\limits_{0}^{3}v(t) dt\]
do you mean the mean value theorem

- anonymous

well, for the displacement, I used the integral and the evaluation theorem.. but I'm stumped for the distance. do I used the mean value theorem? I was given a clue, the "midpoint rule," but I'm not sure how to approach it..

- anonymous

I must say, I actually don't think the midpoint rule has anything to do with this, haha.. do you guys agree? I am totally lost then

- dumbcow

Since the displacement does equal total distance then there must be a change in direction. find where v(t) = 0 as the turnaround point

- dumbcow

i meant does not*

- dumbcow

If t* is the time of direction change then you are looking for
x(t*) + (x(t*)-x(3))
where x(t) is antiderivative of v

- anonymous

To find th distance traveled you would normally use the arc length equation:
\[\int\limits_{0}^{3}\sqrt{1+[v'(t)]^{2}}dt\] which yields 3sqrt(10)
However this time you will be doing an approximation using the midpoint rule. Imagine the line y=3t-5 and draw three rectangles under, spanning from 0 to 1, then 1 to 2, then 2 to 3. Then take the midpoint of each of these intervals. 1/2, 3/2 etc. If you name the midpoints t1, t2, t3 you can drive the following formula for the length:
You know the distance between the sucessive t values is always 1 since this is what i described in the drawing above. So by pythagoren theorem L=sqrt(1+[v(t)-v(t-1)]^2). Evaluate this for t=3/2 and t=5/2 normally but half the values when you evaluate for n=1/2 and t=7/2 because the interval they measure is too big (for example with the t=1/2 you are calculating the length of the interval -1/2 to 1/2 when you actually want the lower bound to be 0) Sum all these values and you get 3sqrt(10)

- anonymous

Xavier, that's what I originally got! But the answer is actually "41/6 meters." I have an idea- if the displacement is the total area under the graph between 0 and 3 (i.e. the positive area - the negative area), would the distance traveled be just the positive area??

- anonymous

I need to go to bed.. I'm going to try this in the morning. Thanks so much for the help guys. I'm a fan. :)

- anonymous

I thought of the distance travelled as the length of the line not the area under. I don't know why its 41/6. I mean if you take advantage of the fact that its linear and use the distance formula you get sqrt[(3-0)^2(4-(-5))^2]=3sqrt(10).

- dumbcow

use the method i used i got 41/6

- anonymous

ah, dumbcow I didn't even see that post! I'll try that! thank you!

- anonymous

Wow...v(t) is the velocity. I feel dumb for missing that for this long....thought it was just any old function. Got 41/6 too. Sorry

- dumbcow

"f the displacement is the total area under the graph between 0 and 3 (i.e. the positive area - the negative area), would the distance traveled be just the positive area??"
that is correct

- anonymous

If i think about that one logically i'd say no. If i go forward then walk backwards the same distance, the displacement is 0. So is the area under the graph. But the distance i walked is the sum of the positive and abs(negative area)

- dumbcow

sorry i didn't read that carefully enough
you are right xavier
its pos area + neg area

- anonymous

Well the positive value of the neg area. I checked using triangles:
.5*5*(5/3)+.5(3-(5/3))*4=41/6.
I guess since you know the area is going to be negative on that on that certain interval you can do:
\[-\int\limits_{0}^{5/3}(3x-5 )dx + \int\limits_{5/3}^{3} (3x-5) dx = 41/6\] except with all the xs replaces with ts >__<
Also you need to use the midpoint rule instead of definite integrals: http://tutorial.math.lamar.edu/Classes/CalcII/ApproximatingDefIntegrals.aspx

- anonymous

thx guys, got it!!! you can give yourselves huge pats on the back!!

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