how to compute the PDE of "dv/dt=g-(Dv^2)/m"? Given when t=o,v=0; when t=t, v=v?
815728
first this is not a PDE but ODE\n\nsimple integration would do
776431
Integrate the acceleration equation with respect to time, twice.  Use, Xo = 500 and Vo = 10...solve for t.
98111
Code:\nx'' - 2nd derivative of x with respect to time (t) (acceleration in x)\n\nx' - 1st derivative of x wrt time (velocity in x)\n\nAnd the same system for the y coordinate.\n\nx''=1.2 then by integration wrt time gives x'=1.2t+constant. Boundary conditions are x'=4750 m/s at t=0, so the constant = 4750 m/s. Equation is now Vx = x' = 1.2t + 4750. Substitute in t=743s, answer is Vx=5641 m/s. \n\nSame method for y'' with acceleration of 8.5 m/s^2, and boundary conditions of y'=0 at t=0 gives Vy=6315.5 m/s.\n\nFinal answers:\nVx=5641.6 m/s\nVy=6315.5 m/s
24383
here are some simple mechanics formulas dealing with velocity:\n\nv=d/t\na=v/t\nvf^2=vi^2+2ad\nd=vit + 1/2at^2\n\nwhere v is velocity, vi is initial velocity, vf is final velocity, a is acceleration, and t is time.  also the i and the f after the v are just subscripts, and ^2 just means sqared.
705605
formula\n\nv^2 = u^2 + 2gh\n\nv = final velocity\nu = initial velocity\n\nso v^2 = 0 + 2*9.81*0.041 = 0.80442 meter\n\nthus v = 0.8968 meter / second\n\nhere mass and angle is not required.. unless some conditions are given ... which is none here ... so 0.8968 meter/second is the answer
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
978230
Let u = t\ndu = dt\n\nLet dv = sin(2t)dt\nv = -1/2 cos(2t) \n\nintegral (u dv) = uv - integral( v du)\n\n= -(1/2)*t*cos(2t) - integral[-1/2 cos(2t) dt]\n\n= -(1/2)*t*cos(2t) + 1/2*1/2*sin(2t)\n\n= -(t/2)cos(2t) + (1/4)sin(2t)\n\nIt looks like you did the integration by parts correctly, but you didn't evaluate the second integral.\n\nKeep at it!
150092
t=to\n_____\n√1-v2/c2
795982
∫_0_2 g(t) dt = 5\n\n(a)\ny= ∫_0_4 g(t/2) dt\nlet t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2\nso, y= ∫_0_2 g(x) 2 dx = 2  ∫_0_2 g(x) dx = 2 * 5 = 10\n\n(b)\ny= ∫_0_2 g(2-t) dt\nlet 2-t=x; so dt = -dx; range becomes x:2 to 0\ny= - - ∫_0_2 g(x) dx = 5
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
899944
first, I would put it in an easier to express format g(t)=[t^(1/2)](1-t)  then take the derivative g'(t)=[(1/2)t^(-1/2)](1-t)-t^(1/2)   now just solve for g'(t)=0, so 0=[(1/2)t^(-1/2)](1-t)-t^(1/2)      now we distribute to get 0=[(1/2)t^(-1/2)]-(3/2)t^(1/2)     we add (3/2)t^(1/2) to both sides to get (3/2)t^(1/2)=(1/2)t^(-1/2)   simplfy this to [3sqrt(t)/2]=[1/2sqrt(t)]   we multiply both sides by 2sqrt(t) to get 6t/2=1 or 3t=1, which means that t=1/3\n\nAnd by the way, this isn't the chain rule, this is the Product Rule
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
731860
x² - y² = 16\nSo 2x - 2ydy/dx = 0\nie dy/dx = x/y\n\nd²y/dx² = (y - xdy/dx)/y² (NOTE: x² - y² = 16  → y² = x² - 16)\n= (y - x.x/y)/(x² - 16)\n= (y² - x²)/(x²y - 16y)\n= -16/(x²y - 16y) (As x² - y² = 16)
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
732220
dx/dt=1/2\ndy/dt=2t\ndy/dx=dy/dt/dx/dt\n=2t/1/2=4t\nat t=-1 slope=-4\nat t=1 slope=4