I want nothing more than to take calculus

This is a discussion on I want nothing more than to take calculus within the A Brief History of Cprogramming.com forums, part of the Community Boards category; You don't actually need conics until you've done a lot of calc... have no fear....

  1. #16
    Pursuing knowledge confuted's Avatar
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    You don't actually need conics until you've done a lot of calc... have no fear.
    Away.

  2. #17
    Toaster Zach L.'s Avatar
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    Originally posted by MrWizard
    Trig sure, but vector operations? That doesn't come up until Calculus III usually. I'm pretty sure he's just talking about Cal I in which case you just need a strong trig background and like alg. II or something.
    Yeah... There isn't too much involving vectors until Calc III, but it is possible that you'll deal with them somewhat while going over kinematics (yes, physics) type material, so in my opinion, its good to at least know a bit about them.
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  3. #18
    Registered User major_small's Avatar
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    if you can figure out:

    (56x^6+48x^5-14x^4+468x^3+45x^2+189x-189)/(2x^6+18x^5-48x^4+165x^3-59x^2-486x+48)

    assuming x==489488436498899878648897898 then you're good for calculus...

    btw... the answer is around 28... (limits)
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  4. #19
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    > My math teacher said having a solid understanding of trig
    > identities is key, which is believable, but he didn't say what
    > else I would have to get into, and honestly I'm good at trig and
    > math in general.

    If he won't tell you, can't you ask other students at your school what is needed?

  5. #20
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    Shiro, I've been bugging the ........ out of two of my friends that are taking trig/precalc, and they confirmed that they are doing complex trig identities but otherwise it is similar to my math class.
    What's the last thing you learned in math class?
    My math class is kind of a joke and not representative of what I actually understand. We're defining what a function is versus a relation i.e the tan at 90 degrees isnt' a function because it produces an undefined slope and polar coordinate graphs are relations not functions because there can be more than one y value for every x.

    I originally had a plan to do trig/precalc over the summer so I could take calculus next year, but the teacher doesn't want to do it anymore with me because I've been slipping up in this math class (not handing stuff in on time). Actually I've been slipping up in school altogether.

    But yeah, anyway, my plan for right now is to just read what I can before college. I'm starting from the beginning, and I'm reading about limits and how to find the slope at a point using lines that are secant to the curve. I plan on asking questions sometime (I haven't really delved into it yet).

  6. #21
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    would the reason you cant take the class be because you dont have the required pre-requisit?

    if that is the case i cant blame your teacher...

  7. #22
    Just because ygfperson's Avatar
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    Calculus is easy. For some people. Probably for you, too, a self-proclaimed nerd.

    A limit is what a function is almost when the independant variable is almost. In clearer terms:
    f(x) = x / x (don't simplify this, i'm proving a point)

    f(0) is undefined. but f(0.01) is 1 and so it f(0.001)...
    So f(x) as x approaches 0 is 1. There's a strong mathematical reasoning behind that, too, but I'll leave it for you to explore.


    A derivative is the slope of a curve at some value 'x'.

    Slope is rise over run. For a line, no matter how many steps forward, the number of steps up is proportionally the same. But for a curve, the slope changes based on the part of the curve.

    Here you can use the limit I just described to make an expression showing the slope for any point x. Since slope is rise over run, let the run approach zero, and the rise approach zero. That's your instantaneous slope. You will learn many derivative formulas which will help you do this quickly and keep it interesting.


    An integral is the accumulation of values.
    Imagine that f(x) = 3. If you integrate over 3 steps somewhere, you will get an accumulation of 9. (3 times 3). It's much harder to do when f(x) is more complicated, so you accumulate rectangles under the curve, each one of the same width (which is approaching zero to gain an exact area.) In clearer terms:
    The integral is the
    width * (sum of height of each rectangle over some distance)
    The integral is the opposite of the derivative.

    It's easier to think it out physically. Your car's speed isn't always constant. The integral of its speed over time is the distance traveled. And the derivative of the position of the car is its current speed.

    I know its a lot to handle, but I'm just providing some thinking material here. Calculus provides a lot of formulas you probably use right now. It lets your calculator calculate logs. (It's the accumulation of rectangles over the function 1/x). It provides area and volume formulas for the same reasons. It gives flexibility to any formula by letting it change with time.
    Last edited by ygfperson; 05-31-2003 at 10:27 PM.

  8. #23
    Back after 2 years Panopticon's Avatar
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    This may be a bit off topic but since you're all "calculus dudes" I have a query. How do you find dy/dx of something with xy as a term?

    Ok the actual question (not homework, I made this up for myself when I was bored) is find dy/dx if:

    x^2 + y^2 -2xy -6x -6y +3 = 0
    (This is actually a parabola with directrix as y=-x-1 and focal point of (1,1))

    I had a brief read of implicit differentiation and got to the -2xy part and went blank.
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  9. #24
    Back after 2 years Panopticon's Avatar
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    Originally posted by major_small
    if you can figure out:

    (56x^6+48x^5-14x^4+468x^3+45x^2+189x-189)/(2x^6+18x^5-48x^4+165x^3-59x^2-486x+48)

    assuming x==489488436498899878648897898 then you're good for calculus...

    btw... the answer is around 28... (limits)
    umm, what's that got to do with calculus?
    I AM WINNER!!!1!111oneoneomne

  10. #25
    Registered User rahaydenuk's Avatar
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    Originally posted by Panopticon
    This may be a bit off topic but since you're all "calculus dudes" I have a query. How do you find dy/dx of something with xy as a term?

    Ok the actual question (not homework, I made this up for myself when I was bored) is find dy/dx if:

    x^2 + y^2 -2xy -6x -6y +3 = 0
    (This is actually a parabola with directrix as y=-x-1 and focal point of (1,1))

    I had a brief read of implicit differentiation and got to the -2xy part and went blank.
    Well if you want your dy/dx in terms of just x, you can complete the square on the above equation to get it as y = f(x) and then explicitly find dy/dx as follows:

    x^2 + y^2 - 2xy - 6x - 6y + 3 = 0
    y^2 - (2x + 6)y + (x^2 - 6x + 3) = 0
    (y - (x + 3))^2 - (x + 3)^2 + x^2 - 6x + 3 = 0
    (y - (x + 3))^2 = 12x + 6
    y = (12x + 6)^(1/2) + x + 3

    Then dy/dx = 6*(12x + 6)^(-1/2) + 1

    Or, if you don't mind having y terms in your expression for dy/dx, you could differentiate implicitly. To do this, just differentiate normally with respect to x, but remember that y is a function of x, so by the chain rule if you were to differentiate y^2 with respect to x, you would get 2y(dy/dx). When you get to the 2xy term, remember you must use the product rule. Differentiating your equation implicitly yields:

    x^2 + y^2 - 2xy - 6x - 6y + 3 = 0
    2x + 2y(dy/dx) - 2y - 2x(dy/dx) - 6 - 6(dy/dx) = 0
    2y(dy/dx) - 2x(dy/dx) - 6(dy/dx) = 2y - 2x + 6
    dy/dx = (2y - 2x + 6)/(2y - 2x - 6)

    Hope that helps,

    Regards,
    Richard Hayden. (rahaydenuk@yahoo.co.uk)
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  11. #26
    Registered User rahaydenuk's Avatar
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    Originally posted by Panopticon
    umm, what's that got to do with calculus?
    I think the original poster expected someone to use L'Hopital's rule to find the limit of that expression as x tends towards infinity (which is 28) and then use that value as an approximation for when x = 489488436498899878648897898.

    Regards,
    Last edited by rahaydenuk; 06-01-2003 at 08:05 AM.
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  12. #27
    Back after 2 years Panopticon's Avatar
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    Ohhhh I get it now!
    so if i wanna dy/dx implicitly, from -2xy am i allowed to express it as -2(dxy/dx) (taking 2 out of the term, then applying d*/dx to the inside)? I wasnt sure before whether i was allowed to excluse the coefficient -2 from the d*/dx.

    Damn... calculus is hard, you all seem so adept, college material and all. I'm only grade 11, and math seems pretty gay right now.
    I AM WINNER!!!1!111oneoneomne

  13. #28
    Registered User rahaydenuk's Avatar
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    Originally posted by Panopticon
    Ohhhh I get it now!
    so if i wanna dy/dx implicitly, from -2xy am i allowed to express it as -2(dxy/dx) (taking 2 out of the term, then applying d*/dx to the inside)? I wasnt sure before whether i was allowed to excluse the coefficient -2 from the d*/dx.
    Yes, you can take constant coefficients out of the differential operator and put them on the 'outside'. For example, if a is constant with respect to x, then d(a*f(x))/dx = a*d(f(x))/dx. The same is true of integration.

    Regards,
    Richard Hayden. (rahaydenuk@yahoo.co.uk)
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  14. #29
    Just because ygfperson's Avatar
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    Originally posted by rahaydenuk
    Yes, you can take constant coefficients out of the differential operator and put them on the 'outside'. For example, if a is constant with respect to x, then d(a*f(x))/dx = a*d(f(x))/dx. The same is true of integration.

    Regards,
    And just because you can doesn't mean you have to. (See my expression manipulator program for proof... it does not seperate out constants.)

    Say you have u * c.
    d(u * c) = u * dc + c * du because of product rule
    since c is a constant, dc = 0, and you only have c * du, which is what you wanted in the first place.

    Not that that way's easier or preferable... it just shows the universality of calculus ideas.

  15. #30
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    ygf I think I understand everything you've said in at least the broadest terms. Are integrals used to find the area/volume of any weird forumla (even graphed in 3d?). It seems that integrals are only an approximation.

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