Thread: Equations

ITT: Nerds pissing on people who obviously haven't taken college level math.

Who would ever want to do these anyway?

A circle is:

Code:

(x - h)2 + (y - k)2 = r2

So, it isn't really a function, but when you graph it, it looks like one.

You'd graph it like this, anyway:

Code:

k + √(r2 - (x - h)2) = y

No you'd need two functions to graph it. There's no such thing as a one-many function.

I know.

You can't have a y²-- So you'd have to have that and a negated version of it to get a circle.

y - x = 7

2y^2 + 14x^2 = 358

Easy, I know, but I couldn't think of anything else.

Oh, also, differentiate: 2 x^4 / 4 x^2

Okay, here's a genuinely interesting question that should be available to high schoolers:

You have the graph y = 2^x. You also have the graph y = kx.

Choose k (a positive constant) so that the graphs only cross once.

Any number less than 0.

Ignoring trivial solutions.

k gotta to be positive.

Code:

f(x) = 2^x -> f'(x) = 2^x ln 2
g(x) = kx -> g'(x) = k
find x such that f(x) = g(x) and f'(x) = g'(x) (functions are tangent)
k = 2^x ln 2
2^x = (2^x ln 2)x
x = log2(x(2^x ln 2)) = log2(x) + log2(2^x) + log2(ln2)
x = log2(x) + x + ln(ln 2)/ln 2
log2(x) = -ln(ln 2)/ln 2
x = 2^(-ln(ln 2)/ln 2)

k = 2^(2^(-ln(ln 2)/ln 2)) ln 2
yeesh o.O
k ≈ 1.884169286

That looks right at first glance on a graph, but I'm not going to put a new equation just to be sure ^^;

Very well done, you managed most of the trick, and got the right decimal. It can be a lot less messy than that though, and you end up with a nice (pure) number. You were very close, you just missed this:

2^x = 2^x*ln 2*x

You got that far.

However you can then find x perfectly legitimately by cancelling the 2^x terms on either side.

Sub in the value you get and see what k is.

Here's another interesting one; it's a bit easier so leave it for someone else it you find it very straightforwards:

Differentiate x^x

You have the graph y = 2^x. You also have the graph y = kx.

Choose k (a positive constant) so that the graphs only cross once.

So there should be only one solution to 2^x = kx.

Code:

log(kx)/log(2) = x
log(kx) = log(2)x
log(k)+log(x) = log(2)x
log(k) = log(2)*x - log(x)
log(k) = log(2^x*x)

I got to go now. Perhaps I'll finish up later.

ive got what im pretty sure is a simple question to high school people.


there are three points - A(1,2,-3) B(4,-2,6) and C(2,4,5)
all i want you to do is find vector AB plus vector BC
you can write the answer in unit vector form because i dont think you can have 3 line long brackets for component form on the computor.

that shouldnt be too hard at all, i just randomly made that up so ill need to figure it out quickly first.

(can't find vector brackets, so { and } will have to do)

{3,-4,9}+{-2,6,-1}={-5,10,-10}

Considering I haven't formally been taught vectors, I think that this works.

hmm, i think you got a bit confused with the negatives there. you found vector AB and BC fine, but your addition was a bit messed up.

Why not just find AC. :l

AB + BC = AC AC= (1,2,8)

f(x)= x^2 - 12x + 27 solve for 2 x values where y=0

What's y?

f(x)=y, and y=0, so f(x)=0

0=(x-9)(x-3)
0=x-9 or 0=x-3
9=x or 3=x

f(x)=y, and y=0, so f(x)=0

Yeah you didn't say that previously.

I was only kidding, the point is that y had nothing to do with the question and could have been anything.

Solve d2y/dx2 - 12dy/dx + 27 = 0.

d2y/dx2-12dy/dx+27=0

y/x-12y/x+27=0

-11y/x=-27

11y/x=27

11y=27x

lawwwwl I would love to write that in an exam...

Figured I'd start this up again.

Let's go easy.

2x=0.

What's x.

I'm confused... x=0?