Eminem4Life1006 wrote:Any tips on how to improve in Calculus/better understand it? I am completely lost now..it's so fast paced I can't grasp it that easily..any tips would be helpful
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Do you understand algebra? Thats the first step.
Do you understand the concept of limits? Thats the second step.
Without a full grasp of algebra and limits, you're going to struggle with calculus. If you understand both of those, this is how you take the leap:
Taking the derivative of a function means converting your function to a different function. Your original function tells you the location (in y) of every point (in x). For example:
y = 2x
This tells you that for every point (x), the value (y) is twice x. If x is 3, y is 6. if x is 5, y is 10.
Now, the converted function will tell you the slope (in y', read y prime) of every point (in x). Eventually you will learn shortcuts to do the conversion from the original function (y) to the slope function (y prime). I can easily look at the equation and tell you that the slope function is:
y' = 2
This means that at every point the slope is 2. Since x isn't in the y' equation, the slope is independent of y. Heres another example... Original equation:
y = x^2 (x squared)
Converting this function to the slope function (more commonly you would say: taking the derivative of y) will result in:
y' = 2x
OK, this tells you that for every point (x) the slope is 2 times x. At x = 2, the slope is 4. At x = 0, the slope is 0. At x = -4, the slope is -8.
Now the important part. To take those two derivatives I used 'shortcuts', but that doesn't really tell you where the concept came from. That comes from something called the 'limit definition of a derivative'. I will give you a brief overview of that derivative, then give you some links with graphical descriptions.
As I said before, the original function tells you the position (in y) for every point (x). The derivative changes this to a slope function, telling you the slope (in y') for every point (x). From algebra, you know that slope is defined as rise over run (change in y divided by change in x). Another way to write that is:
(y2 - y1) / (x2 - X1)
This equation gives you the slope for a line. It is only correct for a line, or it will result in the 'average slope' between the points x1 and x2. To find the slope of something non-linear, you need to find the 'point slope'. Point slope just means the slope at a point, and a point is just an infinitely small line segment. To find the point slope you need to use the equation above (y2-y1/x2-x1), where the line segment (x2-x1) approaches zero. Basically, you are taking using the concept of limits to change your 'line segment slope' equation into a 'point slope equation'. This is the most important, and difficult, step to understanding calculus (at least to understand derivatives), so make sure you get it (even if it takes some time).
Heres something that will explain it using graphs and pictures:
http://en.wikipedia.org/wiki/DerivativeDefinition via difference quotients (read this section / the section before)
http://tutorial.math.lamar.edu/Classes/ ... ative.aspx (Note: These notes are GREAT for all higher levels of math)
I am assuming you are only doing derivatives right now. Eventually you will get to integrals, which is basically the anti-function to the derivative. Integrals will tell you the area under the curve, from one point (x) to a second (x). Integrals and derivatives are related by the fundamental theorem of calculus. After understanding the limit definition of a derivative, the fundamental theorem of calculus is the next big topic to grasp.
After you fully understand derivatives and integrals, you will basically spend calculus learning ways to apply them to word problems. It is very similar to algebra, you only really learn a handful of mathematical concepts, but spend a year learning how to apply those concepts (techniques, strategies, etc.) If you understand algebra, and get a hang of derivatives and integrals, it will all become a lot easier as you go on.
- The Bomber