Recent content by sunnyceej

Homework Statement If a and b are both quadratic residues/nonresidues mod p & q where p and q are distinct odd primes and a and b are not divisible by p or q, Then x2 = ab (mod pq) Homework Equations Legendre symbols: (a/p) = (b/p) and (a/q) = (b/q) quadratic residue means x2 = a (mod p) The...

prove that for any graph G, kappa (G) ≤delta (G).

I reworked the problem and got a negative exponent as well. Thanks again for all your help!

Thanks for all your help SammyS!

Thanks. That is a lot easier to integrate. After integrating, I get v2=\frac{mg}{k}-ce^{\frac{2kx}{m}} with initial value V(0)=0, I get c=\frac{mg}{k} so v=\sqrt{\frac{mg}{k}\left(1-e^{\frac{2kx}{m}}\right)} x would have to represent the height fallen, and it will have to be...

With the initial condition: v(0)=0, c=1 and e^{ln\sqrt{1+\sqrt{\frac{k}{mg}v}}}e^{-ln\sqrt{1+\sqrt{\frac{k}{mg}v}}}=e^{\frac{t\sqrt{kmg}}{m}} This gives me velocity if I know time, but I know height instead. How do I figure that into my problem?

Ok. I see why. I also now see why I should have 1/\sqrt{kmg} in front instead of what I had. So I now have \frac{1}{\sqrt{kmg}}\left(ln\sqrt{1+u}-ln\sqrt{1-u}\right)=\frac{t}{m}+c \left(ln\sqrt{1+u}-ln\sqrt{1-u}\right)=\frac{1}{\sqrt{kmg}}\left(\frac{t}{m}+c\right)...

So this gives me \int\frac{dv}{mg-kv^{2}} = \int\frac{dt}{m} with u substitution: \sqrt{\frac{k}{mg}v}(\frac{1}{2}\int\frac{du}{1+u} + \frac{1}{2}\int\frac{du}{1-u}) = \int\frac{dt}{m} becomes \sqrt{\frac{k}{mg}v}(\frac{1}{2}ln\left|1+u\right| + \frac{1}{2}ln\left|1-u\right|) =...

After thinking more, differentiating with respect to height would not change my equation (except from t to h), but my initial value would be different for each height. So I would have a different specific solution for each height. I'm still curious as to whether I solved the equation...

If mg, and k =1, then I know I would get arctanh v. It's been a while since calc, so I'm not sure what to do with the constants. My other thought is to go the natural log route: ln(mg-kv2)=t/m+c then raising both sides to e: mg-kv2=cet/m using v(0)=0 as my initial condition, c=180...

Homework Statement Find the velocity of a 180 lb object falling from a state of rest from 5 ft, 15 ft, 25 ft, and 45 ft Homework Equations using the formula f=mg-Kv2, where g is the acceleration due to gravity, m the mass of the body, v its velocity. The terminal velocity is 120mph...

Thanks!

How would you solve differently if you had m(dv/dt)= mg- kv2

Thanks! I got y=x^2/(c-x) Every time I use the x2 button at the top I get SUP/SUP I thought I also had to plug it back into the original to verify, but I found out solving for y is good! :)

Homework Statement Consider the equation (y^2+2xy)dx-x^2dy=0 (a) Show that this equation is not exact. b) Show that multiplying both sides of the equation by y^-2 yields a new equation that is exact. C) use the solution of the resulting exact equation to solve the original equation. d) Were...