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Motion Graph and how it acts. Please read.

  1. May 17, 2013 #1
    So I have a function which represents motion. This function is x^2 which is also the area of a square. Now, what I know from the calculus is that the derivative is 2x, no arguing there but the derivative of a squares area is the perimeter, which is 4x? 4x and 2x are not the same at all! Another thing I wanted to bring up is that if this function is increasing two feet per second( if you do the math and anylitically look at the points you will see that you gain an extra two feet per second) or accelerated two feet per second per second then why at point 1,1 is my velocity or derivative two feet per second if at 1,1 you haven't even traveled two feet, you have only traveled one foot in one second? Please answer both questions for me somebody. Thank you.
     
  2. jcsd
  3. May 17, 2013 #2

    berkeman

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    The derivative of a constant square area is zero. If it is not changing, the derivative with respect to time is zero. Where did you come up with this?

     
  4. May 17, 2013 #3
    I did because the actual change of the square is dependent on the perimeter??? I think this is right but I still don't get how the derivative of the function x^2 at point 1,1 is 2? Wouldn't you have to have that distance of 2 for this to be true?
     
  5. May 17, 2013 #4

    berkeman

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    No.


    You correctly stated that the derivative of x^2 is 2x. So 2x=2 implies that x=1. That seems pretty straightforard. Are you taking calculus now?
     
  6. May 17, 2013 #5

    Mark44

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    Correct, so the derivative of the square's area is NOT the perimeter. Why do you think that?
    You are confusing average velocity with instantaneous velocity. At (1, 1) the instantaneous velocity is 2 ft/sec, but that doesn't mean that the velocity has been constant between 0 and 1. At t = 0, v = 0 ft/sec, and at t = .5, v = 1 ft/sec.

    To help you understand the difference between average and instantaneous velocity, a car speedometer measures instantaneous velocity. If you drive a car for a distance of 60 miles over the course of an hour, your average velocity is 60 mph (miles/hour). During that trip there could be a time where your instantaneous velocity was 85 mph, and 35 mph at other times.
     
  7. May 17, 2013 #6
    Yes, I am taking calculus but what isn't making sense is what is happening before the point 1,1. You see, I took the distances differences and found that difference which led to 2, the acceleration I think. But, if this function is accelerating at two feet per second then at the first second why isn't the distance 2? I know 1times 1 is 1 but doesn't this acceleration hold true for every single second? And I absolutely don't get how to find the derivative of a squares area without time. How can you measure that change without time???
     
  8. May 17, 2013 #7
    0,0
    1,1 1
    2,4 3 2
    3,9 5 2
    This is how I did it anylitically, the distance that must be obtained from 1 to 4 is 3 because 3 plus 1 equals 4.
     
  9. May 17, 2013 #8

    Office_Shredder

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    jason, compare the derivative of a circle's area with respect to its radius, and the derivative of a circle's area with respect to its diameter. Now meditate on the square example some more
     
  10. May 17, 2013 #9

    Mark44

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    You need to explain what you're doing with the numbers above. I recognize that 1, 1 represents the point (1, 1) on the graph of y = x2, but what does the 3rd 1 represent?
    I have no idea what the 3nd and 4th rows are supposed to mean? If you want to talk about points on a graph, put the coordinates inside parentheses.
     
  11. May 17, 2013 #10
    It just doesn't make sense that you can have a velocity of 2 ft per second at 1,1 on the function y=x^2. How can you have this velocity if the distance doesn't even add up to 2 ft per second? Please help me. I really love math and would like to understand this. I will not stop until I figure it out.
     
  12. May 17, 2013 #11

    Mark44

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    Did you read post #5? I think I answered the question you're asking. Please reread what I wrote. If you still don't understand, I'll try to explain it a different way.
     
  13. May 17, 2013 #12
    I understand that you can have different velocities at different point but what I really don't understand if the math behind how you can have this velocity at this point. Could you maybe show me an example with how the accelerations have something to do with this or just anyway would be find. Maybe im thinking to deeply about it.
     
  14. May 17, 2013 #13
    So, I think this is how it is. At a certain time you will have the amount that you accelerated but at that time all of those accelerations add up to the velocity at that exact moment! Now all I have to do is do some math and connect it. I thought of this by thinking of a speedometer in my head and that it accelerated 60mph in one second. Everything is intertwined with eachother. Thanks everybody and if it doesn't end up making sense again I will post another post. Thank you guys. This really does help me.
     
  15. May 17, 2013 #14

    Mark44

    Staff: Mentor

    Let's get away from the velocity/acceleration business and just look at the graph of y = x, between x = 0 and x = 1.

    The two endpoints of the graph are at (0, 0) and (1, 1). The slope of the line that connects these points is 1, right? Pick a few points along the arc of the curve. At the point (1/4, 1/16) the slope of the tangent line is 2(1/4) = 1/2. The slope of the line from (0, 0) to (1/4, 1/16) is (1/16)/(1/4) = 1/16 * 4 = 1/4. Notice that these are different.

    If you were a bug crawling along the curve, the slope of the tangent line would be pointing straight ahead for you.

    The tangent line slope corresponds to the instantaneous velocity, the speed that a speedometer shows. The other slope, which is the slope of a secant line, corresponds to the average velocity, which is obtained by distance/time.
     
  16. May 17, 2013 #15
    That makes sense. So the following would hold true, If I had a speedometer for my motion of y=x^2, I accelerated 2 feet per second at the point where I have reached 1 foot at one second. This makes sense but how would I mathematically conclude what distance Im at if I know I have reached this velocity at one second? Or can I even do that? I really hope I don't sound ignorant because I feel like im not getting the full picture. I understand the math behind finding the derivative very well especially graphically with limits. That makes full sense to me but the conceptual part is kind of hard for me.
     
  17. May 17, 2013 #16

    Mark44

    Staff: Mentor

    Since you want to bring motion in, let's write this as s = t2, for 0 ≤ t ≤ 1.
    s'(t) = v(t) = 2t
    s''(t) = v'(t) = a(t) = 2
    The acceleration is constant at 2 feet per second per second. This means that the velocity increases by 2 ft/sec each second.
    Sure. s(t) (or y as I was calling it earlier) gives you the distance at time t. s(1) = 1, meaning you have gotten to 1 foot after 1 second.
     
  18. May 17, 2013 #17
    It is very interesting that you brought in inequalities which I happen to understand quite well. I like the fact by using the inequality it LIMITS you to what you can find, haha limits are quite fun as well. This has made a lot more sense now. This really does help me connect the graphical interpretation of the derivative and the mathematical one. But what I meant by finding the distance traveled is that you only know the velocity at a given time and do not know the distance. I tried using r=d/t but this only works for average speeds or velocities. Can you show me how to do this with using an instantaneous method?
     
  19. May 17, 2013 #18
    Also what I meant by post 7 is that if you take the differences of the distances then take that difference I think it gives you an acceleration which you can add and find the velocity at that second. I think this is the same thing as the equation of the derivative where you plug in the time and multiply. That example used the function y=x^2.
     
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