ParthKohli
  • ParthKohli
Hello! This is a general Math question. Based on my past experience, am I ready for Calculus?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@ParthKohli, without a doubt.
anonymous
  • anonymous
ya............mr.intelligent......
anonymous
  • anonymous
you are! haha-

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More answers

ParthKohli
  • ParthKohli
Hmm, thanks for flattering. I want more answers with a detail.
ParthKohli
  • ParthKohli
And I doubt if you guys know my past experience :P
Compassionate
  • Compassionate
Parth, I will tutor you on Calculus for free if you want a good introduction and help.
anonymous
  • anonymous
calculus is only an expansion of algebra, guess that includes your answer already.
ash2326
  • ash2326
@ParthKohli To learn calculus you should be well versed in Logarithms, Trigonometry, and Sets. Have you learned all these?
anonymous
  • anonymous
I felt Calc I was like 75% Algebra and 20% Trig...pretty sure it will be easy for someone like you
ParthKohli
  • ParthKohli
Thank you. Yes, @ash2326
anonymous
  • anonymous
Calculus is MOSTLY Algebra, just FYI. ;-)
anonymous
  • anonymous
Speaking from direct experience.
ash2326
  • ash2326
Then you are ready:D
Compassionate
  • Compassionate
Parth, do you have knowledge in Trig, Algebra I-II, and working with functions?
ParthKohli
  • ParthKohli
I went to a website and learned all these concepts very thoroughly. What is the starting in Calculus? Limits?
anonymous
  • anonymous
math is hard keke ayiyi
ParthKohli
  • ParthKohli
@Compassionate I do.
anonymous
  • anonymous
It depends on the school. At MIT, we started with Integrals first and defined limits based on that.
anonymous
  • anonymous
Calculus will come next year in my syllabus lol :P
ash2326
  • ash2326
Functions->Limits->Continuity->DIfferentiability->Differentiation->Integration
Compassionate
  • 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."
anonymous
  • anonymous
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|
ParthKohli
  • ParthKohli
Derivative is small change in \(y\) - known as \(dy\) - over the small change in \(x\) - known as \(dx\).
anonymous
  • anonymous
You see this stuff over and over in Calculus
ash2326
  • ash2326
@mathdumbo That's a nice way:D
ParthKohli
  • ParthKohli
Integral is a number of which something is a derivative of(as far as I know).
anonymous
  • anonymous
not a number but function.. :P
ash2326
  • ash2326
@ParthKohli You'll get a good start here http://ocw.mit.edu/high-school/calculus/
anonymous
  • anonymous
Average slope: \[m=\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\] Turns into... \[m=\frac{d y}{d x}=\text{instantaneous slope}\]
ParthKohli
  • ParthKohli
So, \[\int \limits {2xdx } = x^2\]?
ParthKohli
  • ParthKohli
Yes, @agentx5. I saw a video on Khanacademy on the intuition of differentiation.
ParthKohli
  • ParthKohli
And I believe that you are allowed to 'skip' some of the topics in Mathematics; that's how beautiful it is :)
Compassionate
  • Compassionate
Parth, how would you solve this: \[{2 - 5x \sqrt23}{-3x \sqrt2}\]
Compassionate
  • Compassionate
^ You need to have a background with working on radicals.
ParthKohli
  • ParthKohli
I believe that we should stick to the question, and yes I do know that :)
Compassionate
  • Compassionate
There is a lot of really small division in calculus.
ParthKohli
  • ParthKohli
I don't think we can simplify that.
Compassionate
  • Compassionate
And by small, I mean integrals.
Compassionate
  • Compassionate
It was just an example.
anonymous
  • anonymous
I have heard that we can calculate "nth" root of any number by differentiation. Am I correct ?
anonymous
  • anonymous
You can move the constant out front: \(\large\int \limits {2x \ \ dx } = 2\int x \ \ dx = 2[\frac{x^{1+1}}{1+1}] + C = \cancel{2} * \frac{x^2}{\cancel{2}} + C = x^2 + C\)
anonymous
  • anonymous
Parth, don't forget the C
ParthKohli
  • ParthKohli
Yep, because the derivative of any constant is 0, and \(x + 0 = x\).
anonymous
  • anonymous
The "C" means some constant. Why put it on indefinite integrals? Because \(\frac{d}{dx}\) constant = 0
anonymous
  • anonymous
You got it lol :-D
ParthKohli
  • ParthKohli
I am just starting it ;)
ParthKohli
  • ParthKohli
Thank you, all! You are a great help!
ParthKohli
  • ParthKohli
What concept should I start with?
anonymous
  • anonymous
Limits.
ParthKohli
  • ParthKohli
Any resource?
anonymous
  • anonymous
parth,I learn it on MIT , and Khanacedemy
anonymous
  • anonymous
Yes: http://tutorial.math.lamar.edu/sitemap.aspx
anonymous
  • anonymous
That is a phenomenal reference.
anonymous
  • anonymous
also , this awesome book
ParthKohli
  • ParthKohli
Hah! Paul's Notes! I find it a little difficult to start with the basics on that.
ParthKohli
  • ParthKohli
P.S. I learned my logs and trig on there. :)
anonymous
  • anonymous
Try E-book "Calculus for dummies"
anonymous
  • anonymous
http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf
anonymous
  • anonymous
That book help me a lot
anonymous
  • anonymous
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)
anonymous
  • anonymous
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?
ParthKohli
  • ParthKohli
I find these as the answers: 1) \(-1\) 2) \(\infty \) 3) \(\infty\) You use L'Hopital's Rule.
ParthKohli
  • ParthKohli
You may do it visually by taking the numbers closer to the number which it is approaching.
anonymous
  • anonymous
Aww come on draw it! ^_^ Make a sketch
ParthKohli
  • ParthKohli
|dw:1342969380728:dw|
ParthKohli
  • ParthKohli
Hmm. There's a thing that I am missing: CONIC SECTIONS.
ParthKohli
  • ParthKohli
Would I be able to do Calculus without conic sections?
anonymous
  • anonymous
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.
anonymous
  • anonymous
This is why sketching/graphing things visually is so helpful in Calculus :-D
ParthKohli
  • ParthKohli
Hmm—you should have included that I am a beginner too :)
anonymous
  • anonymous
I had to get lunch, sry about the delay
anonymous
  • anonymous
Hey no worries, we all have to learn to fly sometime right? ^_^
ParthKohli
  • ParthKohli
Isn't 'Does NOT Exist' just infinity?
anonymous
  • anonymous
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|
ParthKohli
  • ParthKohli
Oh!
anonymous
  • anonymous
\(\huge \lim_{x \rightarrow 1} | \frac{1}{x-1} | = \infty\) |dw:1342971089127:dw|
anonymous
  • anonymous
^_^ Make more sense now?
ParthKohli
  • ParthKohli
I, for once, believe in visual techniques :)
anonymous
  • anonymous
Amen brotha! ;-P
ParthKohli
  • ParthKohli
Heh! That was so clear :')
anonymous
  • anonymous
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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