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Hi guys, I'm new to here, and I need your help!

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html [Broken]
    this is the problem solution, after I saw it, I still don't understand how did they make the chart, which have f'(x) values.
    What equation should I use in this case?

    2. Relevant equations
    a=dv/dt


    3. The attempt at a solution
    I don't know how to do it at all.. can even find the right equation
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2009 #2

    Office_Shredder

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    Acceleration is the derivative of velocity. So by definition acceleration IS the slope of the velocity
     
  4. Sep 15, 2009 #3
    Thank you but could you look at my question again? I think this web site would explain my question better
     
  5. Sep 15, 2009 #4
    I am not spam!!!!!!!!!!!!!!!!!!

    Before I ask my question, I have to declare that I am not a spam! I am trying to get help here!! Who ever banned me: I AM NOT SPAM!!!
    I was hoping somebody answered my question already!! :grumpy:
    1. The problem statement, all variables and given/known data
    I post this web site is only because they have a good example! there is no viruses!!
    http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html [Broken]
    Or you can search "Graphing the Derivative
    miscellaneous on-line topics for
    Calculus Applied to the Real World" on Google, and click the first one.
    My question is: in the chart, they give me values of x and f'(x), but they didn't say how to get f'(x), could somebody tell me how?

    2. Relevant equations

    i don't know

    3. The attempt at a solution

    don't know.
     
    Last edited by a moderator: May 4, 2017
  6. Sep 15, 2009 #5

    sylas

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    Do you know what the derivative of a function is? What is the connection between the shape of a graph for a function, and the derivative of the function?

    Cheers -- sylas
     
    Last edited: Sep 15, 2009
  7. Sep 15, 2009 #6
    Is the slope of the tangent line.. But each point I only have one (x,y), who can I find the slope?
     
  8. Sep 15, 2009 #7
    Re: I am not spam!!!!!!!!!!!!!!!!!!

    They did say how they got f': "Remember that f'(x) is the slope of the tangent at the point (x, f(x)) on the graph of f."

    But just looking at the graph, you notice the grid pick two points that are on the tangent line and calculate the slope.
     
    Last edited by a moderator: May 4, 2017
  9. Sep 15, 2009 #8
    [tex]y=f(x)[/tex]

    [tex]\frac{dy}{dx}=f'(x) =\displaystyle\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

    If you don't know to differentiate that is going to be pretty useless and even when you can obtain derivatives it's not really used, you will use techniques to differentiate that are guided by general laws on how to do so. But that does show that it is in fact the change in y divided by the change in the x to infinite accuracy so you can find the slope at any given point on the curve.

    [tex]\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} = \displaystyle\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}=\tan\theta[/tex]
     
  10. Sep 15, 2009 #9

    sylas

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    I think you are making it too complicated, Gregg. Look at the link that was provided.

    To complete the exercise in that link, all you need to know is what "question!" has said... the derivative is the slope.

    So make a rough estimate of the slope. It only has to be very rough, and the page says that different people will get different estimates. Just eyeball the graph, and make a guess. The page has places where you can type in your guess.

    Cheers -- sylas
     
  11. Sep 15, 2009 #10
    To get the value for an estimate of the slope you need to draw a line tangent to the slope at a given x, then make a right-angled triangle and work out, or estimate, the ratio between the y side and the x side.

    http://musr.physics.ubc.ca/~jess/hr/skept/Math/img43.gif
     
    Last edited by a moderator: Apr 24, 2017
  12. Sep 15, 2009 #11

    sylas

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    Yes. In the http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html [Broken] you can mentally make this image in your head and guess the slope from that.

    Here's the graph for the "interactive exercise" provided as example two.
    http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/pics/fig3.gif [Broken]

    You can enter you guesses in the page for the slope at x = -3, -2, ... 1, 2, 3 and press a button to check your answers. My guesses were a bit off, but the "check your answers" button told me "close enough". If any guess is too far off, "check your answers" will tell you which one, and whether it is too large or too small (note that if the slope is negative, "too small" means too negative).

    Cheers -- sylas
     
    Last edited by a moderator: May 4, 2017
  13. Sep 15, 2009 #12
    My post wasn't a reply to yours but an answer for the poster that I wrote after actually looking at the problem.
     
    Last edited by a moderator: May 4, 2017
  14. Sep 15, 2009 #13

    sylas

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    Sure, I realize that. And I am saying that your method is good. You can use it as a mental image in your head required to answer the problem, and that is the intent of my post. I'll edit the first line of what you have quoted to make this more clear, by removing "although" and replacing "it will be enough to" with "you can".

    Cheers -- sylas
     
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