Quizing math & physics

[Oliver]

lionhead is correct, I'll post a new question later

[EnsignParis]

For the previous question, is this what we are supposed to be looking at?



And then solving for x?

[Oliver]

I e-mailed you a picture. Perhaps that will clear it up

[EnsignParis]

Sorry, I did get the email. It looks like I deleted it however.

However, I think this is what you told me what it looked like:

[TrekkieMage]

x= 7.5?

I got out rulers and paper and measured. So I guess I cheated...

[EnsignParis]

Hope you don't mind if I post a question. .

You have the function f(x)=x^3. What is the slope of the line tangent to the curve at (1,1)?

[TrekkieMage]

Lemme guess, calculus? I've gotten a bunch of questions like that lately.

You have to find the slope of the curve at (1,1) via derivatives I believe.

[EnsignParis]

Yes, calculus. Related to derivitaves, but you don't need to use them to find this out. You can actually do it via precalculus methods.

However, you do need to understand what a limit is.

I'll give you a hint:

If you take secant lines with the point (1,1), and one other point on the curve, the closer you get to (1,1) with your secant line, the closer the slope of the secant line will be to the slope of the tangent line.

ie: the slope of the tangent line is the limit of the slopes of the secant lines containing the point (1,1)

[webtaz99]

In the real world you could just frikkin measure it.

There's an old (not too funny) joke:

An engineer and a mathematician are put in a room with a beautiful woman. They are told that every five minutes they can move half the distance to her.
The mathematician does his math and wails, "We never really get to her!"
The engineer grins and says, "We'll get close enough."

[EnsignParis]

Alright, here's the answer. I'm going to use calculus to answer it, because well, it's just so much simpler that way.

f(x)=x^3
f'(x)=3x^2
f'(1)=3(1)^2
f'(1)=3

The slope at that point is 3.

If you want the equation of the line, you can just plug it into point slope form, and switch to y=mx+b, if you so wish.

y-1=3(x-1)

y=3x-3+1

y=3x-2

[TrekkieMage]

[EnsignParis]

Yeah, point slope is easier, but it's more clear in slope intercept.

On homework and stuff, I'll just leave it in point slope, unless it specifically says to do otherwise. But for clarity, I like to solve for y.

[Oliver]

What's the derivate of x to the power x?

[webtaz99]

It's been a long time since I took calculus, but....

d/dx of x^n = n*x^(n-1), so

d/dx of x^x = x*x(x-1)

IIRC, multiplying equal variables means adding exponents, so

d/dx of x^x = x^1*x^(x-1) = x^(x-1+1) = x^x

[Sevensquare]

It don't work like that. The calculus students in college have to go through this.
If you're interested in the material---
f(x) = b^x
f' (x) = (b^x)ln b
f(x) = e^x
f' (x) = e^x
There is no derivative for x^x , so the identity
x^x = e^(x ln x)
now things are looking good
Dx e^(x ln x) = e^(x ln x) Dx(x ln x)
= e^(x ln x)(ln x + 1)
= x^x(ln x + 1)
Presto

[EnsignParis]

webtaz,

If you graph x^x, you can clearly see that the derivative isn't going to be x^x.

This however, would work for f(x)=n^x and f' (x)=n^x where n is a constant, but not for f(x)=x^x.

[5thhouse]

oooh this is fun! I've got one....not that hard...

integrate [(2x^3-4x^2-15x+5)/(x^2-2x-8 )] dx

sorry had to fix the smilies with my 8 )

[Sevensquare]

O.K.---
((10x)^4)-(32x^3)-(69x^2)+(134x)+110)/((x^2)-(2x)-^2
Oooooh the math.
If any of those unknowns are exponents your dealing with an impossibly large equation with large numbers.
which nobody does.

[5thhouse]

That is not correct...unless..well I still don't think you would get that and I am lazy to find out.
Since it's been up for so long I'm going to put up the answer if there are no objections?

[5thhouse]

alright, since no objections have been voiced the answer is (and I doubble checked with my CAS so I'm sure it is right ):

- ((Log(x+2))/2) + ((3Log (x-4))/2) + x^2

[5thhouse]

To get that answer you have to long divide x^2 -2x -8 into 2x^3-4x-15x+5. When you do that you get 2x with a remainder of (x+5). So the integral above which I am lazy to retype = the integral 2x dx + integral (x+5)/ (x^2-2x- dx. The first of the two is simple and just gives you x^2. If you factor the denomenator of the second interval you have x^2 + integral (x+5)/((x-4)(x+2)) dx. (x+5)/((x-4)(x+2)) = A/(x+2) + B/(x-4). Solving for A and B you get that (x+5)/((x-4)(x+2)) = -1/2(x+2) + 3/2(x-4). Which means that you now have x^2 + integral (-1/2(x+2))dx + integral (3/2(x-4))dx. When you solve those simple integrals and simplify you get the answer I put above. Of course to be proper you add the integration constant on as well. Well to be really proper you would have been doing C1 etc... with those integrals...but it's not that important here.

[lionhead]

[5thhouse]

sorry I didn't think it was that hard...something around 2nd -semester calculus. Just think if you lived in the federation you would have done that before you were a teenager

[5thhouse]

Okay...given no one liked my last problem maybe something easy?

C14 is a radioactive isotope of Carbon. It decays at a rate proportional to the amount that is present. Its half-life is 5730 years. If 10 grams was origionally present, how much will be present after 2000 years. (hint: if you don't know the formula required, use a differential equation.)

[EnsignParis]