Simple Related Rates Question (implicit differentiation)?
921273
Let x be the distance of the man from spotlight on ground. Let his shadow on wall be s. Then we have, x/10 = 2/s, by similarity of triangles. or  s = -x/20 if we see that as x increases s decreases. \n\nThis gives ds/dt = -(1/20).(dx/dt) = -.06 m/s independent of x. so the Shadow decreases at this rate even when the man is 3m from wall or x = 7 m
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
193440
Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
826981
It looks to me like it's .5d which is not too far from your answer.\n\nFor simplicity, use a unit of length in which d = 1 and a unit of time in which the speed of each of the two bugs is also 1. Let r be the distance between them and let h be the angle between the y direction and the vector from S to R. In other words h starts out pi/2 and approaches 0.\n\nThe time derivatives are:\n\ndh/dt = -sin(h)/r , dr/dt = -1 + cos(h)\n--->\ndr/dh = r(1-cos(h))/sin(h)\n\nSeparating the variables r and h to opposite sides of the equation and integrating both sides with the initial condition (r=1 at h=PI/2) gives:\n\nr = tan(h/2)/sin(h)\n\nYou want the distance between the two bugs as t-->oo or equivalently as h --> 0 which is r=1/2 (use L'hopital's rule for example), or d/2 in terms of the initial distance d.\n\nOf course S never actually reaches the y axis but approaches it asymptotically.\n\n(Maybe I should add, I first tried to formulate the problem in terms of the Cartesian coordinates and was able to write a differential equation for the curve but it looked very difficult to solve. Finally tried using the above coordinates (r,h) and it became much easier)
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
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
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
899768
Its like instantaneous speed. It is the speed or wateva you are trying to find between two really reallly reallllly close points. Example if you are trying to find the derivitive of 2 then plug in 1.999999 or 2.000001 into the equation, doesn't matter. \n\nOtherwise, it is the slope of the tangent of a point.
434625
You are trying to find d/dt (0.51) ...\n\n0.51 -> alpha [ which inturn points "CONSTANT" ]\n\nDerivative of a constant is zero\n\nso d/dt (alpha) = 0
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
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
4004
lambda =velocity/freq\n       =360/10\n       =36 m.\n\nb)1/10 secs.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2
825491
x=ut+(1/2)*a*t^2\n\nso, x=(0*16)+0.5*9.81*16^2=1255.68m\nwhere x is distance, u is initial velocity, t=time , v=velocity, a=acceleration(in this case due to gravity, 9.81m/s^2)\nthats a pretty deep well....... Are you sure its 16 seconds and not 1.6?\n\n:-)