I haven't looked at physics in over 4 years but I think that the forces are equal, and the reason the bug is squished and the car isn't deterred is because the momentum of the system must be conserved.

Hey guys, can you guys tell me if what I'm doing here is correct?

I need to solve this equation:

max subject to:



So, I know I can solve this using Lagrange multipliers. However, I'm really lazy and don't want to solve a 5 variable quadratic system of equations. So here is what I have:



(from the first constraint)


So now, can I just use single variable calculus to solve for the max? I think I need to find the local maxima of this function, since the constraints imply that
,
from squaring the first constraint, and subtracting away the second constraint. So, certainly needs to hold, but I can probably get the bound tighter, since something like violates the first constraint.

Is my way a correct way to go about it, or should I just man up and solve with Lagrange mutipliers?

Evaluate the following limits:

lim x->1
[(sqrt(x) - 1)] / [cbrt(x)-1]

lim x->6
[sqrt(x-2)-2] / [x-6]

how are these done without just subbing in x?

-----

also
lim x->infinite
sqrt(x^2 -x) - x = sqrt(x^2 (1+ 1/x)) - x
= sqrt(infinite^2 (1+0)) - infinite
= sqrt (infinite^2) - infinite
= infinite
is this right?

seems like it should be 0 if this is mathematically sound (because sqrt(x^2) = x) but i believe using infinity as a number is a faux-pas. my limit skills suck though, so it might be ok.

for the first one you need to somehow rewrite the functions because they are not defined in the limit. i'm trying these exercises myself below in the spoiler, anyone want to comment on it? EDIT: i have 2 solved.



(2) lim x->6 [sqrt(x-2)-2] / [x-6] = lim x>6 [sqrt(x-2)-2] / [x-6] * ([sqrt(x-2)+2] / [sqrt(x-2)+2]) = lim x>6 [(x-2) - 4] / [(x-6)*sqrt(x-2)+2] = [x-6] / [(x-6)*(sqrt(x-2)+2)] = lim x>6 [1] / [sqrt(x-2)+2] = lim x>6 [1] / lim x>6 [sqrt(x-2)+2] = 1 / [sqrt(6-2)+2] = 1/[2+2] = 1/4 = 0,25

edit: what i did was basically multiply the limit by 1, but doing it in such a way to simplify one part of the function and stop the other part from being undefined in the limit so we can use the continuity principle Lim x>a F(x) = F(a)

[img]http://mathstat.uohyd.ernet.in/equationeditor/rawequation.php?\[
\lim_{x \to \6} \frac{sqrt{x-2}-2}{x-6} = lim_{x \to \6} \frac{sqrt{x-2}-2}{x-6} *\frac{sqrt{x-2}+2}{sqrt{x-2}+2} = \frac{(x-2)-4}{(x-6)(sqrt{x-2}+2)} = \frac{lim_{x \to \6} 1}{lim_{x \to \6}sqrt{x-2}+2} = \frac{1}{sqrt{6-2}+2} = \frac {1}{4}\][/img]

LOL why is that LaTeX sticky up if it does not even work :-(

You don't have to use the $ signs or [] for latex on the site; assume it's already been done.

Also, I think I figured out the solution to my problem, so disregard, unless you have a hankering for algebra I or something.

Not really sure what this question is asking. Amino acids don't have an "inner membrane" because any membrane is going to be some "gross" structure made up of individual molecules (i.e. cell membrane made up of many many phospholipids). This question, however, makes a lot more sense if it refers to a protein which is a polymer of amino acids. So let me rephrase the question, and you tell me if this is what it actually is:

What best describes the amino acids on the inner membrane portion of a protein?

In this case the answer would be hydrophobic. Polar and hydrophilic are one in the same because polar molecules seek out water to diffuse their charge (it is this charge that makes them polar molecules). Acid and base aren't really the best answers here because there are cases where they work and where they don't, but hydrophobic molecules (and only hydrophobic molecules) will be found in the fatty acid tails of the inner portion of the plasma membrane.

Hey guys, long time reader on the forums, but never really got around to making an account. Now in my last year of studies, the math is catching up to me. I am in a abstract algebra course to complete my math minor and my professor constantly references previous courses that he assumes I have taken, so I am often lost, but I do my best. Here is a question that I found easy to answer, but the proof I have no idea:


Find all positive integers n for which there exists non-negative integers x, y such that 6x+7y=n. Provide a proof for your claim.

I understand that for all of the possible combinations of x and y, there will be a point at which you can make any natural number. I found this point to be when n=30. After 30, all numbers are attainable through variations of x and y. Below n=30 the first n is obviously 6, and starting from 0 all possible values of n are attained as follows:

Skip 5 numbers (0-5) then 2 values of n (6,7)
Skip 4 numbers (8-11) then 3 values of n (12-14)
Skip 3 numbers (15-17) then 4 values of n (18-21)
Skip 2 numbers (22,23) then 5 values (24-28)
Skip 1 number 29 then inf values n>29

Now I notice the obvious inverse trend of 54321 skipped n values to 23456( with 6 then overlapping into the next iteration hence creating the ability to obtain all natural numbers >29) values of n.

My problem is that I have no clue how to "Provide a proof" of my claim. I do not even know what I am proving. The only thing I know is my professor dislikes proof by example.

Thanks in advance

First off, a bit of a disclaimer: I hate proofs, so while I think what I'm going to write is fairly rigorous and correct, you will need to tart it up with mathematical language, and probably check to make sure I haven't done anything ridiculously stupid. So basically, no guarantees that the proof is perfect, but I think the methodology works, but don't expect it to be mathematically coherent.

Ok, to start off with, the proof for n>=30 is pretty simple, and since I'm not sure if you're saying you're ok with seeing it, or ok with proving it, I'm going to run through how I'd construct the proof for that bit first.

Let x,y,a,b be arbitrary non-negative integers, and let n be an integer
6x+7y = n
6a+7b = n+1

So then
1=6(a-x)+7(b-y)
This can only happen when
a-x=-1
b-y=1

a=x-1
b=y+1

Then

6(x-1)+7(y+1)=n+1
Where x-1>=0
Therefore x>=1

You can run through this (or just mention that it's pretty damn obvious) to show that, above n>e (where e is some number)

6(x-a)+7(y+a)=n+a=n'

This holds when x-a>=0
And x>=a

You can then show that this equation holds for all n>=30, just by showing after that x-a>=0 for all 5>=a>=0, so that it can cycle through any value, so basically this equation holds for any x>=5.

From the equation 6(x-a)+7(y+a)=n+a, you can use it to show the pattern of gaps between n'=n+a values (as in, that 54321 skip).

First off, trivially, the minimum value is when x=0,y=0-->n=0

Then let x=1, then 1-a>=0, 1>=a>=0, so you can get n'=6 and n'=7.
Let x=2, then 2-a>=0, 2>=a>=0, so n'=12,13,14

More generally
x-a>=0, x>=a>=0, so n'=6*x+a, which is the formula which shows all the integers 6x+7y=n, where x and y are arbitrary non-negative integers and n is a non-negative integer.

I suggest the following:

First solve this problem without the restriction x,y >= 0 !

Its not hard:
The equation 6x + 7y = n certainly has the solution x = -n, y = n
The difference of two solution is a solution of the homogenous equation:
6x + 7y = 0
The solutions of the homogenous equation are x = 7k, y = -6k for some integer k.

So all of the solutions of 6x + 7y = n can be written as:
x = -n + 7k, y = n - 6k

Now we require the solutions to be nonnegative, so n has to satisfy:
k*6 <= n <= k*7 for some integer k
The n-s that satisfy this for some k are:
6,7
12,13,14
18,19,20,21
24,25,26,27,28
and anything greater or equal then 30

And we are done

Ok the last part is totally "illegal". I mean you cant just start subbing in infinity whenever you see "x" and end up with things like infinity-infinity
or infinity/infinity that does not tell you anything about the limit of the function.

All of these can be solved using L'Hopital's rule.
For details see: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
Or Calculus textbook of your choice.

For example look at
[sqrt(x-2)-2] / [x-6]
both the numerator and the denominator goes to 0 as x goes to 6.
but if you derive you get
0.5*(x-2)^(-0.5) / 1, which you evaluate at 6 and get that the limit is 1/4 (i think, i suck with the part where you actually plug in numbers )

Try doing the other 2 using this L'Hopital's rule if it doesnt work out poast and i'll help you.

The last one you could rewrite as
[sqrt(1 - 1/x) - 1]/(1/x) and apply the rule for that, altough it seems a bit awkvard, possibly there is a quicker method.

Skillgannon and Artagas: thank you both, both of you helped me understand much better.

Suppose the coefficient of static friction between the road and the tires on a car is.60 and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve 30.5m in radius.

No problem, happy to hear it helped you.

Force pushing the car out as it corners is F=mv^2/r (mass, velocity, radius)
Frictional force is F=uN (coefficient of static friction, Normal force from the road)
N=m*g as the car has no negative lift (and presumably, no lift)

Equate the forces (as at the point of sliding, the frictional force is equal to the force pushing the car out, as the force poushing the car out increases from this point, it's no longer balanced with the frictional force, and so you get a net force away from the center of the circle - a slide), sub in values, and you're done

mgu=mv^2/r
gu=v^2/r
v^2=g*u*r
v=sqrt(g*u*r)

ok thx guys

I have a probability questions amigos....

A buried treasure has equal probability of being on island A, B, or C. If the buried treasure is actually on the island, the chance of it being found is .2, .3, and .4, respectively. What is the probability that the treasure is on island A, given that the treasure was looked for on island A and not found?

so let...

A = Treasure is on island A
B = Treasure was looked for on island A and not found

So we want P(A | B). Using Bayes' Theorem, P(A | B) = [P(B|A) * P(A)] / P(B)

P(B|A), or the probability that the treasure was looked for on the island and not found, given the treasure was on island A, is 1 - .2 = .8

P(A), the probability the treasure is on the island, is 1/3.

What is P(B)?

Thanks

well 2/3 of the time you dont find the treasure because it is not there the reamining 1/3 of the time you have 1/5 chanche of finding it.
So P(B) = 2/3 + 1/3*4/5 = 14/15

Ok guys, I got another fun one that I am stumped on:

K any integer >= 2, n is any natural number. Define g(n)=GCD (n,k)
GCD being the greatest common divisor
Show g(n) is multiplicative.

Multiplicative meaning that if the GCD (a,b)=1 and f is multiplicative then f(ab)=f(a)f(b)

So far I have made the statement that n=ab and GCD(a,b)=1

so g(n)=g(ab)=GCD (ab,k)= GCD (a,k) * GCD(b,k)

I just cant figure out where to go from here. Because a and b are relatively prime, I'm sure that there is some link that I am missing. Any direction would be great.

You're done. GCD(a,k) * GCD (b,k) = g(a)g(b).

You might want elaborate on why GCD(ab, k) = GCD(a,k) * GCD(b, k) is true, as that is the 'main' step.

right, I gathered that. All i can think is that because a and b are relatively prime, then multipying them together will not yield a different GCD respectively; however, theyre GCDs are still independent, so multiplying their GCDs together would yield the new GCD. This rambling makes sense to me, but I do not know how to say it in math terms.

you can argue like this:
gcd(a,k)*gcd(b,k) divides k since they are rel. prime and both divides k.
It also divides a*b (for any a,b)

so gcd(a,k)*gcd(b,k) divides gcd(ab,k)
other implication is obvious, so they are equal.

Ok, Differntial Equations.

Have an exam tomorrow and have been studying doing the practice problems..whatever, sucks that WCOOP was there cuase i didnt get to do practice problems as they were assigned(prob could of lol but lazy).

Anyways, cant seem to get the solution to this problem:

Mass= .15kg
Thrown upward
Velocity when thrown = 20 m/s
Start Position = 30m off ground
Air resistance: |v|/30

a) Maximum Height ball reaches
b) Time it takes to hit ground

FWIW, this IS NOT a physics clas, unfortuantely I can just plu in values into easy formulas.

Stuff I have:

m(dv/dt) = -mg - vk

divide by m

(dv/dt) = -g - (k/m)v

Mu(x) = exp((k/m)t)


exp((k/m)t)*V = -g(m/k)exp((k/m)t) + C

V = -g(m/k) + c/[exp((k/m)t)]

maybe im doing it wrong but for other first order equations i solve for solutions fine but cant sem to get it from this:

Solutions:



Any help?

What do you have for m,g,k, and s(t)?

Basically, write down what you think the end results for v(t) and s(t) are.

I didn't get those values either, but my ability to do arithmetic, especially before caffeinated, is greatly diminished.