What is the second derivative of V=D/t?
109422
This is pretty close to an question that was asked previously.\n\nI presume you mean Velocity= Distance/time, and you want the derivatives with respect to time t (for velocity in a 1-dimensional space; velocity strictly speaking is a vector and needs magnitude and direction). \n\nThe first derivative of velocity with respect to time is called acceleration. The second derivative of velocity with respect to time is called the boost or jerk. \n\nYou have not stated if distance D is a function of time or not. Assume that D is a constant. \n\nThen dV/dt=-D/t^2 is the acceleration\n\nAnd d/dt(dV/dt)=2D/t^3 is the boost.
1186433
Well, I think that for the first oneyou can get the acceleration by deriving it twice, then you get the integral from 1 to 5, let's see...\n\nds/dt= 6t^2+3=v\ndv/dt= 12t=a\n\nfor a\n\nI think you should do it this way\n\nintegral of vdt/ integral of dt from 1 to 5\n\nso you will get\n\n(2t^3+3t)/t from 1 to 5\n\nso it will be\n\n2t^2+3 from 1 to 5\n\nevaluating you'll get\n\n2*(25)+3-2-3\n\n=50-2\n\naverage speed = 48\n\nfor b\n\nyou just substitute t=4 to the first derivative\n\nV=6*(4)^2+3=99ft/min
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
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.
858220
Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
46701
v=u+at\nv^2-u^2=2as\ns=ut+1/2at^2\ns=disance traveled\nu=initial vel.\nv=final vel.\na=accelaration\nt=time
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
99360
In kinematics, the 3rd time derivative of position x(t) (x'''(t)) gives the rate of change of acceleration with respect to time (da(t)/dt). It is also known as jerk. This quantity is relevant in the context of problems involving rockets that have a mass that changes with time.
1107625
I assume you are able to take partial derivatives. So take the partial derivatives that appear in the heat equation (you did look up the heat equation, right?) and plug them in to see if you get equality.\n\nSo, the partial derivative with respect to t is\n-lambda^2 k*u(x,t). This is the same as k times the second partial derivative with respect to x. Looks right to me!
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
904537
d = v*t + 0.5*a*t^2 = 9.7sin(x*t)\ntake the derivative\nv + a*t = 9.7x*cos(x*t)\ntake the derivative\na = 9.7*x^2*-sin(x*t)\n\nso:\n10g = 9.7*x^2\nx^2 = 10*9.8/9.7\nx = 10.1\n\n2pi = 10.1*T\nT = 0.62 sec\n\nfreq = 1/T\nfreq = 1/0.62\nfreq = 1.61 Hz
3246
So for the first derivative:\n\n4(x-1)\n-----------\n3x^(2/3)\n\nThe second derivative:\n\n4(x+2)\n-----------\n9x^(5/3)
660602
d/dx[y/x^2] gives y'/x^2 - 2y/x^3 on expansion
813109
this is a simple dervitive problem....we recall that given a position the first derivitive would give a velocity at any time t, and a second derivitive would give an accelaration at any time t\n\nso we need to take the derivitive of h(t)\n\nso h(t) = -6.5t^2 +40t+6\nh'(t)= -13 t +40\n\nnow you find when h'(t) is zero...this will tell you when the slope of a tangent line of position is zero...in otherwords...a max or min\n\nso you find h'(t)=0=-13t+40\nso t= 40/13\n\nnow we need the second derivitive to  apply the second derivitive test (recall this tells us that if the h''(t) is negative at a specific t value we have a max, and if its positive at that t value its a min)\n\nso h''(t)=-13, so for all values of t h''(t) is negative so at t=40 h'(t) is zero and h''(t) is negative so there is a max height at t= 40/13\n\nnow to find the height you plug 40/13 into h(t)\n\nso h(40/13) i will leave to you cause this is teh easy part\n\nhope this helps\n\nmatttlocke
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
252215
2nd Law of motion:  the net force = the time derivative of the momentum vector.
821342
THe VdW equation is P= nRT/(V-nb) - a(n/V)^2, just plug and chug.  But dont forget to switch your T into Kalvin