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ParthKohli

  • 2 years ago

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

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  1. Spacelimbus
    • 2 years ago
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    @ParthKohli, without a doubt.

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

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

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

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

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

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

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

  9. mathdumbo
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

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

  21. Compassionate
    • 2 years ago
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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."

  22. agentx5
    • 2 years ago
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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|

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

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

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

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

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

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

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

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

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

  32. ParthKohli
    • 2 years ago
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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 :)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  49. agentx5
    • 2 years ago
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    Limits.

  50. ParthKohli
    • 2 years ago
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    Any resource?

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

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

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

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

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

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

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

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

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

  60. agentx5
    • 2 years ago
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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)

  61. agentx5
    • 2 years ago
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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?

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

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

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

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

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

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

  68. agentx5
    • 2 years ago
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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.

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

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

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

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

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

  74. agentx5
    • 2 years ago
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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|

  75. ParthKohli
    • 2 years ago
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    Oh!

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

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

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

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

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

  81. agentx5
    • 2 years ago
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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 :-)

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