Recent content by jumboopizza

multivariable limit of (x,y)--->(1,0) of ln(1+y^2/x^2+xy)) Homework Statement limit of (x,y)--->(1,0) of ln(1+y^2/x^2+xy)) Find the limit, if it exists, or show that the limit does not exist. Homework Equations The Attempt at a Solution so i have: lim(x,y)---->(1,0)...

Homework Statement the only part of the question I am unsure about is finding the force of reaction as a function of time Homework Equations F=Ma F=-Kx The Attempt at a Solution F=Ma F=M(Dv/Dt) F Dt= M Dv i integrate force with respect to time from final time=t and...

V(X final)=1/2Kx^2 V(X initial)=V(0)=0 T(final)=0 T(initial)=1/2Ka^2

I think you are getting a bit tangled there: lol that's what i meant to say,im just not good at typing everything here... ok xi=initial position=0 xf=final position and what about the integration of Dv from xf-xi? i thought the contant terms would cancel each other out like this...

ok the problem states that the mass is projected from x=0 so x0 has to be 0 A) but i forgot that x and x0 were supposed to represent lol...i just remembered that they represent the beginning and end points of movements so by that we can say that T(o)+V(xo)=E=1/2 Ka^2 since the potential is...

Homework Statement a particle of mass m is contrained to lie on along a frictionless,horizontal plane subject to a force given by the expression F(x)=-kx. It is projected from x=0 to the right along the positive x direction with initial kinetic energy T(o)=1/2kA^2, k and A are positive...

multivariable calculus 1...the way you did it looks really complicated lol...would it be the same if i used this for length? (sqrt((dx/dtheta)^2+(dy/dtheta)^2)dtheta)? i get an integral of sqrt(25^x(1+(ln5)^2) however when i try to compute the exact integral, the results i get from...

Homework Statement r=5^theta theta goes from 0 to 2Pi Homework Equations Length= integral between a and b of sqrt(r^2+(dr/dtheta)^2)dtheta The Attempt at a Solution r^2=25^theta or 5^(2theta) dr/dtheta=5^theta (ln 5) (dr/dtheta)^2=25^theta+10^theta (ln 5)+...

or is it better to make V dot V= [V]^2 sqrt [V]^2= sqrt u^2 ---------> [v]= u d[v]/dt [v]= du/dt u = 0 --------> d[v]/dt = [a] magnitude of the acceleration

ok that's what i thought dv/dt= A but its not necessarily 0, if they( vector V and Vecotor A) are perpendicular to eachoter the dot product would be zero... ok so V dot A=0 so when i first derived V dot V= u^2 and got to dv/dt V+V dv/dt= u^2 du/dt=0 du/dt is the rate of change of speed...

Homework Statement the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2... V=velocity A=acceleration u=speed t=time Homework Equations i suppose...

Using the approximation, explain why the second derivative works. Give three exam- ples for each scenario of the second derivative test. isnt that what the approximation is? the f(x+delta x,y +delta y)? its asking about finding local mins,local max and saddle points...now i can show...

[b]1. Homework Statement [/ Using the approximation, explain why the second derivative test works approximation=f(x0+delta x, y0+delta y) delta x and delta y are small... Homework Equations f(x0+delta x,y0+delta y) The Attempt at a Solution ok so i know the first derivative...