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tburn Group Title

Can some help with question 2 and 4? I will post document.

  • 2 years ago
  • 2 years ago

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  1. Akshay_Budhkar Group Title
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    A] Propose a value for the following definite integral and give an interpretation for the result : is nothing but the area under the curve from the values x=6 to x=8

    • 2 years ago
  2. saifoo.khan Group Title
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    @Akshay_Budhkar , we can find that by simply using area of a square? right?

    • 2 years ago
  3. Akshay_Budhkar Group Title
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    yes 8^2=64 =D

    • 2 years ago
  4. saifoo.khan Group Title
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    Perfect. ;D

    • 2 years ago
  5. saifoo.khan Group Title
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    exactly..

    • 2 years ago
  6. Akshay_Budhkar Group Title
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    aha the y-axis is degrees / hour! not degrees!

    • 2 years ago
  7. saifoo.khan Group Title
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    Lol.

    • 2 years ago
  8. saifoo.khan Group Title
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    It's not easy. haha

    • 2 years ago
  9. Akshay_Budhkar Group Title
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    u already did :P i am trying

    • 2 years ago
  10. saifoo.khan Group Title
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    haha.

    • 2 years ago
  11. tburn Group Title
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    Amistre64 was helping me on this question but I was still lost so I was hoping if someone else also explained it maybe I would understand.

    • 2 years ago
  12. Akshay_Budhkar Group Title
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    ok i get it

    • 2 years ago
  13. Akshay_Budhkar Group Title
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    lets start with A

    • 2 years ago
  14. Akshay_Budhkar Group Title
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    are you clear with A ?

    • 2 years ago
  15. saifoo.khan Group Title
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    @Akshay_Budhkar idk but i feel like we have to find some derivatives?

    • 2 years ago
  16. tburn Group Title
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    Yes I understand what you said with A.

    • 2 years ago
  17. Akshay_Budhkar Group Title
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    @saifoo.khan yes

    • 2 years ago
  18. Akshay_Budhkar Group Title
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    ok now let degrees be d and time be t so on y axis we have d/t and x axis we have t Now till x= 1 it is a straight line, with a slope -8 and intercept -8 so the equation is y=mx+c that is y=-8x-8 that is d/t=-8t-8 d=-8t^2=8t

    • 2 years ago
  19. Akshay_Budhkar Group Title
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    *d=-8t^2-8

    • 2 years ago
  20. Akshay_Budhkar Group Title
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    Now for minima or maxima the first derivative of the function is o therefore\[{d(d) \over dt }=0=-16t-8 \therefore t= 1/2\]

    • 2 years ago
  21. Akshay_Budhkar Group Title
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    Now to find if it is a minima or a maxima we take the second derivative. if it is negative it is a maxima if positive minima ( mostly i maybe mixing them ) so in this case it is -16 which is negative. therefore it is a maxima at 1/2

    • 2 years ago
  22. tburn Group Title
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    Thank you so much I understood that! Would you be able to help me with question 4 by chance?

    • 2 years ago
  23. Akshay_Budhkar Group Title
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    ok lemme see, is it one unit for each square right?

    • 2 years ago
  24. tburn Group Title
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    That is what I understand.

    • 2 years ago
  25. Akshay_Budhkar Group Title
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    ok lets start with the simplest C. f(3) = 5 any doubt?

    • 2 years ago
  26. tburn Group Title
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    Yes, thats what I see.

    • 2 years ago
  27. Akshay_Budhkar Group Title
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    ok now A. f'(-2) Now i believe that at a given point \[f'(x)={dy \over dx} = { y \over x}\] if we estimate. so f'(-2) =-2/-2 = 1

    • 2 years ago
  28. tburn Group Title
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    Okay, that makes sense to me.

    • 2 years ago
  29. Akshay_Budhkar Group Title
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    for D its the same logic, area under the curve. i believe you can assume the graph to be a triangle and calculate the area

    • 2 years ago
  30. Akshay_Budhkar Group Title
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    is it unclear?

    • 2 years ago
  31. tburn Group Title
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    So you think D would be =-2 to 4?

    • 2 years ago
  32. Akshay_Budhkar Group Title
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    the area under the graph , from -2 to 4

    • 2 years ago
  33. tburn Group Title
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    Okay

    • 2 years ago
  34. Akshay_Budhkar Group Title
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    Can you solve it? For the other questions i would personally plot a graph for f'(x)

    • 2 years ago
  35. tburn Group Title
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    f"(x)=d/dx*f'(x)

    • 2 years ago
  36. Akshay_Budhkar Group Title
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    yes i know. Now to plot the graph u must know that it is zero at minima and maxima, and u can fidn approximate values by using the approximation of y/x, and plot the graph of f'

    • 2 years ago
  37. Akshay_Budhkar Group Title
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    then after i get the graph, f"' will be obtained by using y/x again and the last one by area under curve in that graph

    • 2 years ago
  38. tburn Group Title
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    Sorry I am just thinking. Okay, Thanks.

    • 2 years ago
  39. Akshay_Budhkar Group Title
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    yup give it a thought, and lemme know if ustuck =D and u can tag me in some other question if u get some similar doubt =D

    • 2 years ago
  40. tburn Group Title
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    Okay, thank you so much for your help!

    • 2 years ago
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