Quizing math & physics

[5thhouse]

That is correct. (and your way is so much easier than my way!)
lol my much more challenging approach goes like this.
You set up a differential equation:
dy/dt=kty
dy=kydt
integral (1/y)dy=integral of (k)dt
ln(abs(y))=kt+C
e^ln(abs(y))= ln^kt+ ln^C
abs(y)=Ce^kt
plugging in values for this problem:
1/2=e^5730k
ln(.5)= 5730k
k= about -.000121
y=Ce^-0.000121t
y=10e^(-0.000121 * 2000)
y= 7.851
and 7.851 grams is the answer!

Your way is slightly less accurate but nontheless closs enough to be considered correct and deffinitly much faster and easier.

btw abs=absolute value

[5thhouse]

So no one but me likes to give questions?

f"(x)= x^2+3cosx when f(1)1 and f'(1)=5

[TrekkieMage]

I'm assuming you want f(x)? Here's what I got:

f(x) = (1/12)x^4 - 3cosx + (14/3)x - 3/4

[5thhouse]

That is close but not all the way correct (and yes I was asking so..and that should be F(1)=1)

[TrekkieMage]

That should be right...I just double checked it. What was your answer?

[5thhouse]

I checked with my CAS so I know I'm right

f(x)= (1/12)x^4 - 3cos(x) + (14/3)x - (3sin1)x - 25/6 +3cos1 +3sin1 + C

[TrekkieMage]

But if f(1)=1 and f'(1)=5 you should be able to solve for c.

Where did you get the extra sines and cosines? And 3cos1 = 3 and 3sin1 = 0 so that last part would be 25/6 + 3 + 0 + c

[5thhouse]

haha sorry the C shouldn't be there...reflex

f'(x) = (1/3)x^3 +3sinx + C

since f'(1) = 5

f'(1) = (1/3) 1^3 +3sin1 + C = 5

C= 5- (1/3) - 3sin1 = 14/3 -3sin1

Therefore

f'(x) = (1/3)x^3 + 3sinx + 14/3 - 3sin1

F(x) = (1/12)x^4 - 3cosx + (14/3)x - (3sin1)x + C

f(1) = 1 and so,

f(1) = (1/2) - 3cos1 + (14/3) - 3sin1 + C = 1

C= 1- (1/2) - (14/3) + 3cos1 + 3sin1 = -25/6 + 3cos1 +3sin1

f(x)= (1/12)x^4 - 3cos(x) + (14/3)x - (3sin1)x - 25/6 +3cos1 +3sin1

[TrekkieMage]

Ah, but you made a mistake. 3sin1 is a constant, not a variable, it becomes (some number) multiplied by x.

[lionhead]

Great, now they think they can put 3 in 1.


i'm sorry, inside joke.

[Defiant]

I llike math.

[5thhouse]

[TrekkieMage]

But both 3sin1 and (3sin1)x are 0. So the final equation can certainly be simplified - does the CAS just ignore it? I guess I just don't understand what this CAS is...

[Sevensquare]

if your thinking in radians sin(1)=.8415 and cos(1)=.5403
everything else is perfect

[TrekkieMage]

Just for kicks:

What does {(2^n)/(3^(n+1))} converge to?

[Manitou]

you are saying lim x -> n (2^n/3^n+1) this is 0.
It is the same as ((2/3)^n) * (1/3) Any fraction that is powered goes to 0
I've got a good one for the audience.

[Manitou]

If a chord is created at random on a fixed circle, what is the probability that its length exceeds the radius. (a chord is what happens when a line goes through a circle)

[Manitou]

There are three possible answers as explained in the book. There is also a fourth answer that is also correct. Anyone who wants to try.

[Manitou]

I should be posting some images to provide some idea about the answer. The first solution is the simplest and the other two a little more detailed. The fourth I discovered after some meditation but they all make sense.

[Oliver]

[Oliver]

[Manitou]

The first answer is given by a hexagon inscribed on the circumferance of the circle since each side is equal to r. The line from the mid-point of one side to the center (in relation to r) is sq. rt. of 3 divided by 2. This is also the height of one of the equilateral triangles.

[Oliver]

[Oliver]

sorry, double post

[Manitou]

Yes but if the chord passing through the circle is longer than any of the sides of the equilateral triangle its mid-pt sits somewhere on the height of the triangle, which would be =< then sq. rt. 3 divided by 2. If the chord is less than the radius it's mid-pt would sit somewhere of what's left after sq.rt. 3 divided by two.