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[ein]

Verify that the equation is true.
tanθ +   cosθ  = secθ
       1 + sinθ

[Jim Groovester]

Okay.

tanx + cosx/(1 + sinx)

(tanx + sinxtanx + cosx)/(1 + sinx)

(tanx + (sin²x + cos²x)/cosx)/(1 + sinx)

(tanx + secx)/(1 + sinx)

(sinx + 1)/(cosx(1 + sinx))

1/cosx

secx = secx

If a step doesn't make sense just let me know.

[Argembarger]

Argembarger, I worked out the leaky bucket in well problem myself, and I got somewhere ~1e4 ft·lb of energy (I'm not doing your homework for you). I think I have your error. Your integral should be of the form

W = ∫ F(l)·dl

Instead you have something like this

W ≠ ∫ F(l)·(l(l) dl)

This is wrong, wrong, wrong. Once you get the force as a function of height, you are halfway there.

Edit: in particlar, when the bucket is empty, the work done on the bucket should be a linear function of height. Is the antiderivitive of your second integral (where the bucket is empty) linear?

...Oh good lord, you're absolutely right. I'm glad I decided not to e-mail my professor.

And don't worry about "doing my homework for me", as it's already been submitted. I just decided to take the 0 for that portion (Edit: just to clarify, I took the 0 because the assignment autosubmitted at 8:00 this morning, not because I refused to fix it or take a 0 out of shame or anything); and anyway if you had just given me an answer without explanation that wouldn't really help me in the long run at all, like, say, the midterm at the end of the week

Anyway, since we can keep submitting after it's due (simply for practice purposes) I redid the problem, I got it now, thanks a lot!

[ein]

stuff

Awesome, thanks.
I have trouble with math when I can't see the application in things.
Like the chapter on sinusoidal regressions I could do because it was all about meteorology, and the occasional physics problem that gets thrown in here and there.
But I find proofing things and these verifications painful.
Stupid higher-level math.

Oh wow, and then the next problem is sinx cotx = 1.
I needed to make it cosx = 1.
Two steps, man. Two steps...
So easy.

[eerr]

steps? mmmm

Alternate method
cosx/(sinx+1)=secx-tanx
cosx=secx-tanx+sinxsecx-tanxsinx

sinx=o/h
cosx=a/h
tanx=o/a

opposite, hypotenuse, and adjacent sides.

a/h=h/a-o/a+(o/h)(h/a)-(o/h)*(o/a)
a/h=h/a-o/a+o/a-o^2/ha
a/h=h/a-o^2/ha

*ha

a^2=h^2-o^2
o^2+a^2=h^2

h^2=o^2+a^2
pythagorean theorem

(This won't work well in later math classes because you can have more than one θ, but it may save your ass if you don't know a specific formula.)

[Virex]

Can't you use cot(x) = 1/tan(x)?

[ILikePie]

Just a quick question, if I have two triangles, A and B. If each of A's segments equal half B's segments will A's medians equal half of B's medians? (eg, A's dimensions are 2,3,2 and B's are 4,6,4, and A's medians equal, say, 4,5,4, will B's medians be 8,10,8?)

[Jim Groovester]

Yes.

[ILikePie]

That's great then, thanks.

[Ottofar]

I only came to this thread to complain about how our mathematics teacher didn't give us proper explanation of why anything to the power of zero is one (as in 7⁰=1, if my vocab fails yet again).

[KaminaSquirtle]

I only came to this thread to complain about how our mathematics teacher didn't give us proper explanation of why anything to the power of zero is one (as in 7⁰=1, if my vocab fails yet again).

Does this explain it more satisfactorily?
xⁿ/xⁿ=1 because you are dividing a number by itself.
However, because xⁿ/x^m=x^n-m, xⁿ/xⁿ=x^n-n=x⁰
Therefore x⁰=1

[Ottofar]

I only came to this thread to complain about how our mathematics teacher didn't give us proper explanation of why anything to the power of zero is one (as in 7⁰=1, if my vocab fails yet again).

Does this explain it more satisfactorily?
xⁿ/xⁿ=1 because you are dividing a number by itself.
However, because xⁿ/x^m=x^n-m, xⁿ/xⁿ=x^n-n=x⁰
Therefore x⁰=1

Yeah, thanks.

[Vector]

I only came to this thread to complain about how our mathematics teacher didn't give us proper explanation of why anything to the power of zero is one (as in 7⁰=1, if my vocab fails yet again).

Or, alternatively, we often define something to the zeroth power as one for convenience.  Makes all the rest of the stuff work out well.


EDIT: All right, off to make another attempt at proving calculus.  Wish me luck.

[lordnincompoop]

Wait so how did you factor terms like 45a⁵ again. I just suddenly forgot this; 9a⁴ is a factor of 45a⁵, and I'm stuck wondering how they did that. Any help?

[KaminaSquirtle]

Why would you want to factor things like 45a⁵?
As to the second sentence: (9a⁴)(5a)=45a⁵
I don't see the point though...