what is the answer to (dP/dt)=4P^2 +t*P^2?
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
976124
dy/dx = (dy/dt)/(dx/dt)\ndy/dt = 1/(2√t)\ndx/dt = 1\n\ndy/dx = 1/(2√t)\nx=t\ny=√t\n\nSubstituting t=1/4:\n\ndy/dx = 1\nx = 1/4\ny = 1/2\n\nUsing point-slope form:\n\ny-1/2 = 1(x-1/4)\n\nSimplifying:\n\ny=x+1/4
330850
S A D + M A D \n A D = A D\n     = MS
994673
I am not sure of the final answer you are looking for.  If you are looking for the number of molecules transferred after a certain time t, the answer is simply n=-DAt(dc/dz).  This of course assumes that dc/dz remains relatively constant.  If not you will need a dc/dz as a function of t.
829349
Take the derivitive of your equation with respect to t\nd(9x^2+16y^2)/dt = 18x dx/dt + 32y dy/dt = 0\n\nThey give you dx/dt = 4y.\nSubstitute this in the above equation and then solve for dy/dt.\n\nThe maximum dy/dt will be pretty easy to find from the equation you get.
878312
V=4/3pier^3\nA=4pier^2
791163
2r*2r*(p/4)\n=4r^2(p/4)\n=pr^2
982484
i think the volume is constant, so\nV= pi r^2 h, so\nh = V/(pi r^2)\nthen\n dh/dt = V/pi * (-2)/r^3 * dr/dt =\n- 128*2/(1,8)^3 * 0.05 which i think is a) \nBtw it's pi not pie....lol
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
1051597
start with   d = m / v        multiply both side by v; \nyou get     v * d = m        divide both sides by d; \nyou get     v = m / d
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
461412
1) assuming it's inversely proportional then you have multiplied by m+2/m so it's dm/(m+2)\n2)(m+2)d/m
150092
t=to\n_____\n√1-v2/c2
432536
Oops………..!\n\ndac2chari\n\n\n u = 1/(ln t) Then du = t dt ………………..?  How that come …!\n\nIf u = 1/(ln t) then  u = [ ln t ] ^-1 \n\nAnd  du = - [ln t ] ^-2 [ 1/t ] dt = dt / [-t (ln t ) ^2] … please be careful.\n------------------------------------------------------------------------------------\n\nOK… this integral can’t carry out using analytical method; however, the flowing one can be solve\n\n∫ dt / [t ln(t)] \n\nu = ln (t)\ndu =  dt / t  \n\nGives\n\n∫ dt / [t ln(t)]  = ∫ du/ u  ln(u)\n\nSubstitute for u\n\n∫ dt / [t ln(t)]  = ln [ ln (t) ]
845493
we shall use chain rule\n\nd/dx(sec(pi*x)^2\n= - sec(pi*x) csc(pi*x)d/dx(pi*x))^2\n= -2pi^2x sec(pi*x) cosec(pi*x)
889055
Power:\nP=dE/dt = dW/dt\n\nForce:\nf=ma\n\nEnergy:\ne=mc^2
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)