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ParthKohli Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Spacelimbus Group Title
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    @ParthKohli, without a doubt.

    • 2 years ago
  2. nitz Group Title
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    ya............mr.intelligent......

    • 2 years ago
  3. hoya Group Title
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    you are! haha-

    • 2 years ago
  4. ParthKohli Group Title
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    Hmm, thanks for flattering. I want more answers with a detail.

    • 2 years ago
  5. ParthKohli Group Title
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    And I doubt if you guys know my past experience :P

    • 2 years ago
  6. Compassionate Group Title
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    Parth, I will tutor you on Calculus for free if you want a good introduction and help.

    • 2 years ago
  7. Spacelimbus Group Title
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    calculus is only an expansion of algebra, guess that includes your answer already.

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

    • 2 years ago
  9. mathdumbo Group Title
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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

    • 2 years ago
  10. ParthKohli Group Title
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    Thank you. Yes, @ash2326

    • 2 years ago
  11. agentx5 Group Title
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    Calculus is MOSTLY Algebra, just FYI. ;-)

    • 2 years ago
  12. agentx5 Group Title
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    Speaking from direct experience.

    • 2 years ago
  13. ash2326 Group Title
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    Then you are ready:D

    • 2 years ago
  14. Compassionate Group Title
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    Parth, do you have knowledge in Trig, Algebra I-II, and working with functions?

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

    • 2 years ago
  16. hoya Group Title
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    math is hard keke ayiyi

    • 2 years ago
  17. ParthKohli Group Title
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    @Compassionate I do.

    • 2 years ago
  18. mathdumbo Group Title
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    It depends on the school. At MIT, we started with Integrals first and defined limits based on that.

    • 2 years ago
  19. vikrantg4 Group Title
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    Calculus will come next year in my syllabus lol :P

    • 2 years ago
  20. ash2326 Group Title
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    Functions->Limits->Continuity->DIfferentiability->Differentiation->Integration

    • 2 years ago
  21. Compassionate Group Title
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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."

    • 2 years ago
  22. agentx5 Group Title
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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|

    • 2 years ago
  23. ParthKohli Group Title
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    Derivative is small change in \(y\) - known as \(dy\) - over the small change in \(x\) - known as \(dx\).

    • 2 years ago
  24. agentx5 Group Title
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    You see this stuff over and over in Calculus

    • 2 years ago
  25. ash2326 Group Title
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    @mathdumbo That's a nice way:D

    • 2 years ago
  26. ParthKohli Group Title
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    Integral is a number of which something is a derivative of(as far as I know).

    • 2 years ago
  27. vikrantg4 Group Title
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    not a number but function.. :P

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

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

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

    • 2 years ago
  31. ParthKohli Group Title
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    Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.

    • 2 years ago
  32. ParthKohli Group Title
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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 :)

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

    • 2 years ago
  34. Compassionate Group Title
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    ^ You need to have a background with working on radicals.

    • 2 years ago
  35. ParthKohli Group Title
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    I believe that we should stick to the question, and yes I do know that :)

    • 2 years ago
  36. Compassionate Group Title
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    There is a lot of really small division in calculus.

    • 2 years ago
  37. ParthKohli Group Title
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    I don't think we can simplify that.

    • 2 years ago
  38. Compassionate Group Title
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    And by small, I mean integrals.

    • 2 years ago
  39. Compassionate Group Title
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    It was just an example.

    • 2 years ago
  40. vikrantg4 Group Title
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    I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?

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

    • 2 years ago
  42. 91 Group Title
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    Parth, don't forget the C

    • 2 years ago
  43. ParthKohli Group Title
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    Yep, because the derivative of any constant is 0, and \(x + 0 = x\).

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

    • 2 years ago
  45. agentx5 Group Title
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    You got it lol :-D

    • 2 years ago
  46. ParthKohli Group Title
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    I am just starting it ;)

    • 2 years ago
  47. ParthKohli Group Title
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    Thank you, all! You are a great help!

    • 2 years ago
  48. ParthKohli Group Title
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    What concept should I start with?

    • 2 years ago
  49. agentx5 Group Title
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    Limits.

    • 2 years ago
  50. ParthKohli Group Title
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    Any resource?

    • 2 years ago
  51. 91 Group Title
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    parth,I learn it on MIT , and Khanacedemy

    • 2 years ago
  52. agentx5 Group Title
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    Yes: http://tutorial.math.lamar.edu/sitemap.aspx

    • 2 years ago
  53. agentx5 Group Title
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    That is a phenomenal reference.

    • 2 years ago
  54. 91 Group Title
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    also , this awesome book

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

    • 2 years ago
  56. ParthKohli Group Title
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    P.S. I learned my logs and trig on there. :)

    • 2 years ago
  57. vikrantg4 Group Title
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    Try E-book "Calculus for dummies"

    • 2 years ago
  58. 91 Group Title
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    http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

    • 2 years ago
  59. 91 Group Title
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    That book help me a lot

    • 2 years ago
  60. agentx5 Group Title
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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)

    • 2 years ago
  61. agentx5 Group Title
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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?

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

    • 2 years ago
  63. ParthKohli Group Title
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    You may do it visually by taking the numbers closer to the number which it is approaching.

    • 2 years ago
  64. agentx5 Group Title
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    Aww come on draw it! ^_^ Make a sketch

    • 2 years ago
  65. ParthKohli Group Title
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    |dw:1342969380728:dw|

    • 2 years ago
  66. ParthKohli Group Title
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    Hmm. There's a thing that I am missing: CONIC SECTIONS.

    • 2 years ago
  67. ParthKohli Group Title
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    Would I be able to do Calculus without conic sections?

    • 2 years ago
  68. agentx5 Group Title
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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.

    • 2 years ago
  69. agentx5 Group Title
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    This is why sketching/graphing things visually is so helpful in Calculus :-D

    • 2 years ago
  70. ParthKohli Group Title
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    Hmm—you should have included that I am a beginner too :)

    • 2 years ago
  71. agentx5 Group Title
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    I had to get lunch, sry about the delay

    • 2 years ago
  72. agentx5 Group Title
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    Hey no worries, we all have to learn to fly sometime right? ^_^

    • 2 years ago
  73. ParthKohli Group Title
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    Isn't 'Does NOT Exist' just infinity?

    • 2 years ago
  74. agentx5 Group Title
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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|

    • 2 years ago
  75. ParthKohli Group Title
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    Oh!

    • 2 years ago
  76. agentx5 Group Title
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    \(\huge \lim_{x \rightarrow 1} | \frac{1}{x-1} | = \infty\) |dw:1342971089127:dw|

    • 2 years ago
  77. agentx5 Group Title
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    ^_^ Make more sense now?

    • 2 years ago
  78. ParthKohli Group Title
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    I, for once, believe in visual techniques :)

    • 2 years ago
  79. agentx5 Group Title
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    Amen brotha! ;-P

    • 2 years ago
  80. ParthKohli Group Title
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    Heh! That was so clear :')

    • 2 years ago
  81. agentx5 Group Title
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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 :-)

    • 2 years ago
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