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

1. Spacelimbus

@ParthKohli, without a doubt.

2. nitz

ya............mr.intelligent......

3. hoya

you are! haha-

4. ParthKohli

Hmm, thanks for flattering. I want more answers with a detail.

5. ParthKohli

And I doubt if you guys know my past experience :P

6. Compassionate

Parth, I will tutor you on Calculus for free if you want a good introduction and help.

7. Spacelimbus

8. ash2326

@ParthKohli To learn calculus you should be well versed in Logarithms, Trigonometry, and Sets. Have you learned all these?

9. mathdumbo

I felt Calc I was like 75% Algebra and 20% Trig...pretty sure it will be easy for someone like you

10. ParthKohli

Thank you. Yes, @ash2326

11. agentx5

Calculus is MOSTLY Algebra, just FYI. ;-)

12. agentx5

Speaking from direct experience.

13. ash2326

14. Compassionate

Parth, do you have knowledge in Trig, Algebra I-II, and working with functions?

15. ParthKohli

I went to a website and learned all these concepts very thoroughly. What is the starting in Calculus? Limits?

16. hoya

math is hard keke ayiyi

17. ParthKohli

@Compassionate I do.

18. mathdumbo

It depends on the school. At MIT, we started with Integrals first and defined limits based on that.

19. vikrantg4

Calculus will come next year in my syllabus lol :P

20. ash2326

Functions->Limits->Continuity->DIfferentiability->Differentiation->Integration

21. Compassionate

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

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

Derivative is small change in $$y$$ - known as $$dy$$ - over the small change in $$x$$ - known as $$dx$$.

24. agentx5

You see this stuff over and over in Calculus

25. ash2326

@mathdumbo That's a nice way:D

26. ParthKohli

Integral is a number of which something is a derivative of(as far as I know).

27. vikrantg4

not a number but function.. :P

28. ash2326

@ParthKohli You'll get a good start here http://ocw.mit.edu/high-school/calculus/

29. agentx5

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

So, $\int \limits {2xdx } = x^2$?

31. ParthKohli

Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.

32. ParthKohli

And I believe that you are allowed to 'skip' some of the topics in Mathematics; that's how beautiful it is :)

33. Compassionate

Parth, how would you solve this: ${2 - 5x \sqrt23}{-3x \sqrt2}$

34. Compassionate

^ You need to have a background with working on radicals.

35. ParthKohli

I believe that we should stick to the question, and yes I do know that :)

36. Compassionate

There is a lot of really small division in calculus.

37. ParthKohli

I don't think we can simplify that.

38. Compassionate

And by small, I mean integrals.

39. Compassionate

It was just an example.

40. vikrantg4

I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?

41. agentx5

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

Parth, don't forget the C

43. ParthKohli

Yep, because the derivative of any constant is 0, and $$x + 0 = x$$.

44. agentx5

The "C" means some constant. Why put it on indefinite integrals? Because $$\frac{d}{dx}$$ constant = 0

45. agentx5

You got it lol :-D

46. ParthKohli

I am just starting it ;)

47. ParthKohli

Thank you, all! You are a great help!

48. ParthKohli

49. agentx5

Limits.

50. ParthKohli

Any resource?

51. 91

parth,I learn it on MIT , and Khanacedemy

52. agentx5
53. agentx5

That is a phenomenal reference.

54. 91

also , this awesome book

55. ParthKohli

Hah! Paul's Notes! I find it a little difficult to start with the basics on that.

56. ParthKohli

P.S. I learned my logs and trig on there. :)

57. vikrantg4

Try E-book "Calculus for dummies"

58. 91
59. 91

That book help me a lot

60. agentx5

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

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

I find these as the answers: 1) $$-1$$ 2) $$\infty$$ 3) $$\infty$$ You use L'Hopital's Rule.

63. ParthKohli

You may do it visually by taking the numbers closer to the number which it is approaching.

64. agentx5

Aww come on draw it! ^_^ Make a sketch

65. ParthKohli

|dw:1342969380728:dw|

66. ParthKohli

Hmm. There's a thing that I am missing: CONIC SECTIONS.

67. ParthKohli

Would I be able to do Calculus without conic sections?

68. agentx5

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

This is why sketching/graphing things visually is so helpful in Calculus :-D

70. ParthKohli

Hmm—you should have included that I am a beginner too :)

71. agentx5

72. agentx5

Hey no worries, we all have to learn to fly sometime right? ^_^

73. ParthKohli

Isn't 'Does NOT Exist' just infinity?

74. agentx5

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

Oh!

76. agentx5

$$\huge \lim_{x \rightarrow 1} | \frac{1}{x-1} | = \infty$$ |dw:1342971089127:dw|

77. agentx5

^_^ Make more sense now?

78. ParthKohli

I, for once, believe in visual techniques :)

79. agentx5

Amen brotha! ;-P

80. ParthKohli

Heh! That was so clear :')

81. agentx5

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 :-)