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tburn

  • 4 years ago

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

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

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

  4. saifoo.khan
    • 4 years ago
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    Perfect. ;D

  5. saifoo.khan
    • 4 years ago
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    exactly..

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

  7. saifoo.khan
    • 4 years ago
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    Lol.

  8. saifoo.khan
    • 4 years ago
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    It's not easy. haha

  9. Akshay_Budhkar
    • 4 years ago
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    u already did :P i am trying

  10. saifoo.khan
    • 4 years ago
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    haha.

  11. tburn
    • 4 years ago
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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.

  12. Akshay_Budhkar
    • 4 years ago
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    ok i get it

  13. Akshay_Budhkar
    • 4 years ago
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    lets start with A

  14. Akshay_Budhkar
    • 4 years ago
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    are you clear with A ?

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

  16. tburn
    • 4 years ago
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    Yes I understand what you said with A.

  17. Akshay_Budhkar
    • 4 years ago
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    @saifoo.khan yes

  18. Akshay_Budhkar
    • 4 years ago
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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

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

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

  21. Akshay_Budhkar
    • 4 years ago
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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

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

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

  24. tburn
    • 4 years ago
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    That is what I understand.

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

  26. tburn
    • 4 years ago
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    Yes, thats what I see.

  27. Akshay_Budhkar
    • 4 years ago
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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

  28. tburn
    • 4 years ago
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    Okay, that makes sense to me.

  29. Akshay_Budhkar
    • 4 years ago
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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

  30. Akshay_Budhkar
    • 4 years ago
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    is it unclear?

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

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

  33. tburn
    • 4 years ago
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    Okay

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

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

  36. Akshay_Budhkar
    • 4 years ago
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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'

  37. Akshay_Budhkar
    • 4 years ago
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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

  38. tburn
    • 4 years ago
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    Sorry I am just thinking. Okay, Thanks.

  39. Akshay_Budhkar
    • 4 years ago
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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

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

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