anonymous
  • 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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
dumbcow
  • dumbcow
is v velocity?
anonymous
  • 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
  • anonymous
I'm just confused.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dumbcow
  • dumbcow
correct you could find distance traveled as \[\int\limits_{0}^{3}v(t) dt\] do you mean the mean value theorem
anonymous
  • 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
  • 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
  • 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
  • dumbcow
i meant does not*
dumbcow
  • 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
  • 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
  • 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
  • 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
  • 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
  • dumbcow
use the method i used i got 41/6
anonymous
  • anonymous
ah, dumbcow I didn't even see that post! I'll try that! thank you!
anonymous
  • 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
  • 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
  • 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
  • dumbcow
sorry i didn't read that carefully enough you are right xavier its pos area + neg area
anonymous
  • 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
  • anonymous
thx guys, got it!!! you can give yourselves huge pats on the back!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.