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Miscellaneous Mathematics (open thread)

opaltiger

opaltiger

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Because maths is awesome! Plus I want to practice a little. And I have a nice problem!

Let two vertices of a rectangle lie on the x-axis and the other two on the curve y = e^(-x^2). a) Find an expression for the area of the rectangle. b) Find the maximum area.

So we know that the curve will be symmetrical over the y-axis, because the function is even (since -x^2 = x^2). Thus we can determine that the vertices on the curve will have the coordinates (x,y) and (-x,y). This means that one side (the base) of the rectangle will be 2x (x - (-x)). The other side (the height) is y, or e^(-x^2). Thus we can express the area as

A = 2x * e^(-x^2)

If we then differentiate:

dy/dx = 2x * e^(-x^2) * (-2x) + e^(-x^2) * 2
dy/dx = e^(-x^2)(-4x^2 + 2)

equal to zero:

e^(-x^2)(-4x^2 + 2) = 0
-4x^2 + 2 = 0 (because e^n where n is any real number is always positive)
2 = 4x^2
1/2 = x^2
x = +/- 1/sqrt(2)

substitute back into the area expression (taking the positive result, because area cannot be negative):

A(max) = 2x * e^(-x^2)
A(max) = 2/sqrt(2) * e^(-1/sqrt(2))^2
A(max) = sqrt(2) * e^(1/2)

And this is the kind of maths I really like, both because it is such a pleasure to work your way through a longer problem and get the right answer at the end and because you can actually see what this kind of thing might be used for.

So. This is an open thread; feel free to ask for help, post your own interesting problems, or whatever.
 
i'll contribute to this thread if we have a need for some advanced algebra stuff as i'm of course studying to be an engineer

and i've done diff. eqs and all that
 
lol maybe you can explain shit right when my calculus class cannot.

god I love math but what can I do if I am not given the challenges I need.
 
I'm taking high school-level Physics and Algebra 2, so I doubt I could be of any help here...
 
I'm only taking Geometry, so you kinda got me a little lost there. But I'd like to help with anything else!
 
Walk me through sum derivative shit, yo.

They're asking to find dy/dx of y=(x^2 + 3)/x with a. quotient rule and b. I keep coming up with different things and gah.
 
Well, with the quotient rule, we would set f(x) as the numerator, x^2 + 3, and g(x) as the denominator, x. Then:

y' = (f'(x)g(x) - f(x)g'(x))/(g(x))^2 = (2x*x - (x^2 + 3)*1)/x^2

y' = (2x^2 - x^2 - 3)/x^2

y' = (x^2 - 3)/x^2 = 1 - 3/x^2

Though I'm not sure exactly what you mean; are you supposed to be finding it by some other means as well? If you're supposed to do it by any other method you feel like, the most straightforward one would be to just rewrite it to y = x^2/x + 3/x = x + 3/x and then differentiate each part separately to get y' = 1 + (-3/x^2) = 1 - 3/x^2.
 
That's what I thought, ok, thanks.

Oh sorry I obviously forgot to type out what b was. They want me to divide out first and then differentiate? Cept like ... the only way to really divide it out is just far too complicated, bah. I can't imagine why you would.
 
Cool, a math thread. 8D Maybe I can ask for help with my homework whenever I'm in a rut....

I got a two basic questions I want to ask because I see them being used here which has got me curious, and plus I would like to learn them because I have no knowledge of them... yet:

1) What is "dy/dx"? (This is my first time seeing this just so you know).

2) What is the quotient rule? (Also my first time).

EDIT: Actually, I got another one: "3) What is sum derivative?" (I'm pretty sure I'm gonna learn this next class after the test).
As you can see, I'm really curious to know what these are. :3
 
dy/dx is Leibniz's notation for a derivative, which is basically a function describing a base function's rate of change. The simplest way to explain it is that if you for instance plot a graph with time on the x axis and location on the y axis, you can create a function describing how your location changes with time as you, say, drive in a straight line. The derivative of this function will describe how your speed changes during this journey. (Then you could differentiate (i.e. find the derivative of) that, and you'd get a function describing how your acceleration changes.)

The quotient rule is a rule of differentiation which states that the derivative of a fraction, if f(x) is the numerator, g(x) is the denominator and f'(x) and g'(x) are their respective derivatives, is (f'(x)*g(x) - f(x)* g'(x))/(g(x))^2.
 
In other words, for the quotient rule (this is how I think of it, it's the same but is just much less messy):

if you wanna take the derivative of u/v, where u and v are two functions.

(v*du - u*dv)/v^2

I just happen to hate function notations. So many parentheses and blah.
 
Thanks for explaining what those two are guys. ^w^ I would like to clarify one thing though: dy/dx is the same as f'(x) right?
 
Yep. There's like twenty bagrillion different ways to write out notation for derivatives.
 
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