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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli, without a doubt.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya............mr.intelligent......

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Hmm, thanks for flattering. I want more answers with a detail.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2And I doubt if you guys know my past experience :P

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0Parth, I will tutor you on Calculus for free if you want a good introduction and help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0calculus is only an expansion of algebra, guess that includes your answer already.

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1@ParthKohli To learn calculus you should be well versed in Logarithms, Trigonometry, and Sets. Have you learned all these?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I felt Calc I was like 75% Algebra and 20% Trig...pretty sure it will be easy for someone like you

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Thank you. Yes, @ash2326

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Calculus is MOSTLY Algebra, just FYI. ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Speaking from direct experience.

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0Parth, do you have knowledge in Trig, Algebra III, and working with functions?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I went to a website and learned all these concepts very thoroughly. What is the starting in Calculus? Limits?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0math is hard keke ayiyi

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2@Compassionate I do.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It depends on the school. At MIT, we started with Integrals first and defined limits based on that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Calculus will come next year in my syllabus lol :P

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1Functions>Limits>Continuity>DIfferentiability>Differentiation>Integration

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0You'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."

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Derivative is small change in \(y\)  known as \(dy\)  over the small change in \(x\)  known as \(dx\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You see this stuff over and over in Calculus

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1@mathdumbo That's a nice way:D

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Integral is a number of which something is a derivative of(as far as I know).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not a number but function.. :P

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1@ParthKohli You'll get a good start here http://ocw.mit.edu/highschool/calculus/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Average slope: \[m=\frac{\triangle y}{\triangle x}=\frac{y_2y_1}{x_2x_1}\] Turns into... \[m=\frac{d y}{d x}=\text{instantaneous slope}\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2So, \[\int \limits {2xdx } = x^2\]?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2And I believe that you are allowed to 'skip' some of the topics in Mathematics; that's how beautiful it is :)

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0Parth, how would you solve this: \[{2  5x \sqrt23}{3x \sqrt2}\]

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0^ You need to have a background with working on radicals.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I believe that we should stick to the question, and yes I do know that :)

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0There is a lot of really small division in calculus.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I don't think we can simplify that.

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0And by small, I mean integrals.

Compassionate
 4 years ago
Best ResponseYou've already chosen the best response.0It was just an example.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can move the constant out front: \(\large\int \limits {2x \ \ dx } = 2\int x \ \ dx = 2[\frac{x^{1+1}}{1+1}] + C = \cancel{2} * \frac{x^2}{\cancel{2}} + C = x^2 + C\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Parth, don't forget the C

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Yep, because the derivative of any constant is 0, and \(x + 0 = x\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The "C" means some constant. Why put it on indefinite integrals? Because \(\frac{d}{dx}\) constant = 0

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I am just starting it ;)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Thank you, all! You are a great help!

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2What concept should I start with?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0parth,I learn it on MIT , and Khanacedemy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is a phenomenal reference.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also , this awesome book

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Hah! Paul's Notes! I find it a little difficult to start with the basics on that.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2P.S. I learned my logs and trig on there. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try Ebook "Calculus for dummies"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That book help me a lot

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Parth try this one! ^_^ \[\huge \lim_{x \rightarrow 0} \frac{1}{x1}\] and \[\huge \lim_{x \rightarrow 1} \frac{1}{x1}\] and \[\huge \lim_{x \rightarrow \infty} \frac{1}{x1}\] Do it visually :D (draw a graph with arrows & stuff)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try 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?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I find these as the answers: 1) \(1\) 2) \(\infty \) 3) \(\infty\) You use L'Hopital's Rule.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2You may do it visually by taking the numbers closer to the number which it is approaching.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aww come on draw it! ^_^ Make a sketch

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1342969380728:dw

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Hmm. There's a thing that I am missing: CONIC SECTIONS.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Would I be able to do Calculus without conic sections?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Incorrect @ParthKohli dw:1342970148488:dw Correct answers are: \(\huge \lim_{x \rightarrow 0} \frac{1}{x1} = 1\) \(\huge \lim_{x \rightarrow 1} \frac{1}{x1} = \text{Does NOT exist.}\) \(\huge \lim_{x \rightarrow \infty} \frac{1}{x1} = 0\) 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is why sketching/graphing things visually is so helpful in Calculus :D

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Hmm—you should have included that I am a beginner too :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I had to get lunch, sry about the delay

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey no worries, we all have to learn to fly sometime right? ^_^

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Isn't 'Does NOT Exist' just infinity?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\huge \lim_{x \rightarrow 1}  \frac{1}{x1}  = \infty\) dw:1342971089127:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^_^ Make more sense now?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2I, for once, believe in visual techniques :)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.2Heh! That was so clear :')

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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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