How do you  work out the divergence of (u x v) in a vector field?
353111
I think you mean a div (u x v), where "x" is a cross product.\n\n(a,b,c) x (d,e,f) = (bf-ce, cd-af, ae-bd)\n\nu = (y^2,-z,3)*(i,j,k), where * is dot product\nv = (z+1,xz,x(1+y))*(i,j,k)\nu x v = (-xz(1+y)-3xz, 3(z+1)-y^2*x(1+y), xzy^2+z(z+1)) =\n = (-4xz+xyz, 3z+3-xy^2-xy^3, xzy^2+z^2+z)\n\nNext, div F = (F_1,F_2,F_3) = dF_1/dx + dF_2/dy + dF_3/dz\n\ndiv (u x v) = div (-4xz+xyz, 3z+3-xy^2-xy^3, xzy^2+z^2+z) =\n = (-4z+yz) + (-2xy-3xy^2) + (xy^2+2z+1) =\n = -2xy - 2xy^2 + yz - 2z + 1\n\nSo, the answer is div (u x v) = -2xy - 2xy^2 + yz - 2z + 1
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
86765
Plus infinity (It appears!)\n\nWe must compute the Surface Integral of the vector field F overthe surface S of the Sphere.\n\nBy divergence theorem of Gauss it is equal to the \ntriple integral of div F d(x,y,z).\n\nIntroducing spherical coordinates\nx=rCos[u]Sin[v]\ny=rSin[u]Sin[v]\nz=rCos[v]\n\nSubstition of F=(z,y,x) in the formula for divF in spherical coordinates \nand computing the triple integral inside the sphere\n\nIntegral ( -Pi/2 < v < Pi/2 )\n  Integral ( 0 < u < 2Pi )\n    Integral ( 0 < r < 1 )\n      (1/r)(2Cos[v]+cos[u]+2Cos[u]Cos[v])\n    dr\n  du\ndv\n\nThe integral (0<r<1) of (1/r)dr is infinity!\nJust ignore the other parts.
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
242020
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
749461
u = 3i - 5j\nv = i - 2j\n\nso in order to find vector 2u - 3v, u have to find out 2u and 3v first, then subtract them.\n\nso, 2u = 2(3i - 5j) = 6i - 10j\nand 3v = 3(i-2j) = 3i - 6j\n\nnow, 2u - 3v = 6i - 10j - (3i - 6j) \n\n2u - 3v = 6i - 10j - 3i + 6j\n2u - 3v = 3i - 4j\n\nso your answer is: 3i - 4j\nhope that helped!
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
49129
To find the direction of maximum ascent, find the directional derivative (DD), and find the direction where it has its max.\nNote: This is the equation of a plane, so the directional derivative is independent of the point where it is taken.\n\nIf v is a unit vector, then the DD in the direction of v, is the vector dot product of v with the vector  formed from the two partial derivatives\nPartial Z with respect to x is 1.5\nPartial Z with respedt to y is -.8\n\nDD = (1.5, -.8) dot v\n\nSince v is  unit vector, you can represent it\n\n(sin t, cos t)\n\ngiving us DD = (1.5)sint - (.8)cost \n\nTo find direction the max, take the derivative with respect to t, and set to 0.\n\n0 = (1.5)cost + (.8)sint\n-(1.5)cost = .8sint\n-1.5/.8 = tan t\n\nThe solutions to that say that v has the same direction as either of the (non-unit) vectors\n(1.5, -.8) or (-1.5, .08). The first is maximum ascent, the second maximum descent.
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
416901
A-a??\n\nA gradient is a vector field consisting of derivative vectors of a source of data (like a moving ocean)\nYou could use a gradient to show how fast the water is moving across the ocean surface. The derivitaive vectors that make up the gradient are instantaneous calculations of the rate of flow at several points on the surface of the ocean)
728779
Sounds like youve got a nasty integral. Your going to have to integrate a polar vector in cartesian coordinates. Write the B field in cartesian and then integrate one side of the square.\n\nB.ds= miu (0) * I is true. I forget which theorem but I think its stokes theorem from vector calculus. It doesnt matter which shape you choose. The symmetry of the circle allows a simple integral because there is no dot product for the B.s since they are always parrallel. Choosing a square just makes your life horrible. If you correctly write the B field and then use the vector identity:\n\ncos theta = a.b/(|a||b|)\n\nto find the angle for the dot product you will get the same answer. It's just hard to set up the problem.\n\nIm sorry dude, good luck.
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!
846591
go for pentominoes.. search this word in the web and you have something related with maths to do. i did it in my class 10. it was nice. all you ll need is some thermocoal. you can also go for the explanation of gradient and its use. you can do this with the help of some models. or you can also opt for the vector related problems, like in the forensic science they calculate the speed of a car involved in accident based on the impact. it deals with vectors. you can do this with the help of something to measure impact. if you need help mail me i ll help you in whichever way i can.
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted