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ParthKohli

Hello! This is a general Math question. Based on my past experience, am I ready for Calculus?

  • one year ago
  • one year ago

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  1. Spacelimbus
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    @ParthKohli, without a doubt.

    • one year ago
  2. nitz
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    ya............mr.intelligent......

    • one year ago
  3. hoya
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    you are! haha-

    • one year ago
  4. ParthKohli
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    Hmm, thanks for flattering. I want more answers with a detail.

    • one year ago
  5. ParthKohli
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    And I doubt if you guys know my past experience :P

    • one year ago
  6. Compassionate
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    Parth, I will tutor you on Calculus for free if you want a good introduction and help.

    • one year ago
  7. Spacelimbus
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    calculus is only an expansion of algebra, guess that includes your answer already.

    • one year ago
  8. ash2326
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    @ParthKohli To learn calculus you should be well versed in Logarithms, Trigonometry, and Sets. Have you learned all these?

    • one year ago
  9. mathdumbo
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    I felt Calc I was like 75% Algebra and 20% Trig...pretty sure it will be easy for someone like you

    • one year ago
  10. ParthKohli
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    Thank you. Yes, @ash2326

    • one year ago
  11. agentx5
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    Calculus is MOSTLY Algebra, just FYI. ;-)

    • one year ago
  12. agentx5
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    Speaking from direct experience.

    • one year ago
  13. ash2326
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    Then you are ready:D

    • one year ago
  14. Compassionate
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    Parth, do you have knowledge in Trig, Algebra I-II, and working with functions?

    • one year ago
  15. ParthKohli
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    I went to a website and learned all these concepts very thoroughly. What is the starting in Calculus? Limits?

    • one year ago
  16. hoya
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    math is hard keke ayiyi

    • one year ago
  17. ParthKohli
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    @Compassionate I do.

    • one year ago
  18. mathdumbo
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    It depends on the school. At MIT, we started with Integrals first and defined limits based on that.

    • one year ago
  19. vikrantg4
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    Calculus will come next year in my syllabus lol :P

    • one year ago
  20. ash2326
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    Functions->Limits->Continuity->DIfferentiability->Differentiation->Integration

    • one year ago
  21. Compassionate
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    You're ready, Parth. If you wouldn't mind taking a few minutes out of your time today, and send me what you know and give me some examples. I can take that information and place you at a level. :) And like I said, "I can tutor you."

    • one year ago
  22. agentx5
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    And when I say that it is mostly, I mean that you'll REALLY need to be good with it. Everything from factoring, to trig identities, to partial fractions, to series and sequences, to things like this: |dw:1342968425801:dw|

    • one year ago
  23. ParthKohli
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    Derivative is small change in \(y\) - known as \(dy\) - over the small change in \(x\) - known as \(dx\).

    • one year ago
  24. agentx5
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    You see this stuff over and over in Calculus

    • one year ago
  25. ash2326
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    @mathdumbo That's a nice way:D

    • one year ago
  26. ParthKohli
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    Integral is a number of which something is a derivative of(as far as I know).

    • one year ago
  27. vikrantg4
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    not a number but function.. :P

    • one year ago
  28. ash2326
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    @ParthKohli You'll get a good start here http://ocw.mit.edu/high-school/calculus/

    • one year ago
  29. agentx5
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    Average slope: \[m=\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\] Turns into... \[m=\frac{d y}{d x}=\text{instantaneous slope}\]

    • one year ago
  30. ParthKohli
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    So, \[\int \limits {2xdx } = x^2\]?

    • one year ago
  31. ParthKohli
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    Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.

    • one year ago
  32. ParthKohli
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    And I believe that you are allowed to 'skip' some of the topics in Mathematics; that's how beautiful it is :)

    • one year ago
  33. Compassionate
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    Parth, how would you solve this: \[{2 - 5x \sqrt23}{-3x \sqrt2}\]

    • one year ago
  34. Compassionate
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    ^ You need to have a background with working on radicals.

    • one year ago
  35. ParthKohli
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    I believe that we should stick to the question, and yes I do know that :)

    • one year ago
  36. Compassionate
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    There is a lot of really small division in calculus.

    • one year ago
  37. ParthKohli
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    I don't think we can simplify that.

    • one year ago
  38. Compassionate
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    And by small, I mean integrals.

    • one year ago
  39. Compassionate
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    It was just an example.

    • one year ago
  40. vikrantg4
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    I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?

    • one year ago
  41. agentx5
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    You can move the constant out front: \(\large\int \limits {2x \ \ dx } = 2\int x \ \ dx = 2[\frac{x^{1+1}}{1+1}] + C = \cancel{2} * \frac{x^2}{\cancel{2}} + C = x^2 + C\)

    • one year ago
  42. 91
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    Parth, don't forget the C

    • one year ago
  43. ParthKohli
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    Yep, because the derivative of any constant is 0, and \(x + 0 = x\).

    • one year ago
  44. agentx5
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    The "C" means some constant. Why put it on indefinite integrals? Because \(\frac{d}{dx}\) constant = 0

    • one year ago
  45. agentx5
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    You got it lol :-D

    • one year ago
  46. ParthKohli
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    I am just starting it ;)

    • one year ago
  47. ParthKohli
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    Thank you, all! You are a great help!

    • one year ago
  48. ParthKohli
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    What concept should I start with?

    • one year ago
  49. agentx5
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    Limits.

    • one year ago
  50. ParthKohli
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    Any resource?

    • one year ago
  51. 91
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    parth,I learn it on MIT , and Khanacedemy

    • one year ago
  52. agentx5
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    Yes: http://tutorial.math.lamar.edu/sitemap.aspx

    • one year ago
  53. agentx5
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    That is a phenomenal reference.

    • one year ago
  54. 91
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    also , this awesome book

    • one year ago
  55. ParthKohli
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    Hah! Paul's Notes! I find it a little difficult to start with the basics on that.

    • one year ago
  56. ParthKohli
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    P.S. I learned my logs and trig on there. :)

    • one year ago
  57. vikrantg4
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    Try E-book "Calculus for dummies"

    • one year ago
  58. 91
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    http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

    • one year ago
  59. 91
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    That book help me a lot

    • one year ago
  60. agentx5
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    Parth try this one! ^_^ \[\huge \lim_{x \rightarrow 0} \frac{1}{x-1}\] and \[\huge \lim_{x \rightarrow 1} \frac{1}{x-1}\] and \[\huge \lim_{x \rightarrow \infty} \frac{1}{x-1}\] Do it visually :-D (draw a graph with arrows & stuff)

    • one year ago
  61. agentx5
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    Try it @ParthKohli :-D It's not a hard problem, it just makes you think and once you get the concepts down it just goes back to Algebra tricks for the most part. hint: What happens if the x = 1 to the denominator?

    • one year ago
  62. ParthKohli
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    I find these as the answers: 1) \(-1\) 2) \(\infty \) 3) \(\infty\) You use L'Hopital's Rule.

    • one year ago
  63. ParthKohli
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    You may do it visually by taking the numbers closer to the number which it is approaching.

    • one year ago
  64. agentx5
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    Aww come on draw it! ^_^ Make a sketch

    • one year ago
  65. ParthKohli
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    |dw:1342969380728:dw|

    • one year ago
  66. ParthKohli
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    Hmm. There's a thing that I am missing: CONIC SECTIONS.

    • one year ago
  67. ParthKohli
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    Would I be able to do Calculus without conic sections?

    • one year ago
  68. agentx5
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    Incorrect @ParthKohli |dw:1342970148488:dw| Correct answers are: \(\huge \lim_{x \rightarrow 0} \frac{1}{x-1} = -1\) \(\huge \lim_{x \rightarrow 1} \frac{1}{x-1} = \text{Does NOT exist.}\) \(\huge \lim_{x \rightarrow \infty} \frac{1}{x-1} = 0\) You cannot use the l'Hospital's rule unless after simplification you still have a \(\large \frac{\pm\ \infty}{\pm\ \infty}\) or \(\large \frac{0}{0}\) form when you go to substitute.

    • one year ago
  69. agentx5
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    This is why sketching/graphing things visually is so helpful in Calculus :-D

    • one year ago
  70. ParthKohli
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    Hmm—you should have included that I am a beginner too :)

    • one year ago
  71. agentx5
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    I had to get lunch, sry about the delay

    • one year ago
  72. agentx5
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    Hey no worries, we all have to learn to fly sometime right? ^_^

    • one year ago
  73. ParthKohli
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    Isn't 'Does NOT Exist' just infinity?

    • one year ago
  74. agentx5
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    Actually, no, take a closer look. The limit is as if you're taking your right and left hand and tracing the function from both points towards whatever the limit is approaching. In the case of the first one, it's clearly approaching -1 from both the right and left. But in the second one, it goes on way up to infinity, and the other way down to negative infinity. It creates a kind of a paradox, it can't be both and yet it is. So the limit simply doesn't exist. Here's an example where the limit is infinity: \[\huge \lim_{x \rightarrow \infty} x^2 = \infty\] |dw:1342970876711:dw|

    • one year ago
  75. ParthKohli
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    Oh!

    • one year ago
  76. agentx5
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    \(\huge \lim_{x \rightarrow 1} | \frac{1}{x-1} | = \infty\) |dw:1342971089127:dw|

    • one year ago
  77. agentx5
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    ^_^ Make more sense now?

    • one year ago
  78. ParthKohli
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    I, for once, believe in visual techniques :)

    • one year ago
  79. agentx5
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    Amen brotha! ;-P

    • one year ago
  80. ParthKohli
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    Heh! That was so clear :')

    • one year ago
  81. agentx5
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    And don't forget you can't just use l'Hostpital's rule whenever you feel like it, it's got to meet the prerequisites of the theorem that I mentioned above :-)

    • one year ago
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