calculus rate question?
931725
it goes an additional 7 1/3 (or 7.33333333333) ft. All you needed to do was divide 44 by 6. You were probably just over complicating it.
951198
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.
750296
-2x + h\nIn about 5 minutes you will be taken the limit of the expression as h ---> 0.  Then, you will be told that then it will be the derivative.\n\nAfter that you take 2 - x^2 do the derivative and get -2x.
965285
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
540612
give me ten points if u need further help just mail to me.\n\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/2413.asp\n\nhttp://www.calculus.org/\n\nhttp://www.math.ucdavis.edu/~mikew/free_calc.pdf\n\nhttp://www.math.gatech.edu/~cain/notes/calculus.html\n\nhttp://www.exambot.com/cgi/topic/list.cgi/math/difc?class=hlink
502732
Let me guess you're taking the AP Calc test tomorrow!  Me too!  :)\n\nWell first take the seperate derivatives of x and y\nto get dx/dt and dy/dt\n\nthen put you dy answer over the dx answer to have dy/dx\n\nthen take the derivative of dy/dx and put that over your dx/dt\n\n\nexample problem\nx=5t^2                 y=4t^3\n\nsolution:\n\ndx/dt= 10t       dy/dt= 12t^2\n\ndy/dx (first derivative) = (12t^2)/(10t)\n\nwhich equals (6/5)t\n\nthen take d(dy/dx)/dt (derivative of (dy/dx))\n\nso that equals (6/5)\n\nthen put that over dx/dt which is 10t\n\nso (6/5)/(10t) which equals 6/(5*10*t)\n\nso answer equals 6/(50t)\n\n\nHope it helps and if you are taking the AP Test good luck!!!
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
679000
time= 2.303/(rate constant) log( initial conc / final conc)\n\nthis is for a first order reaction...
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
158912
1.) If you graph a "linear function" all you have is a line.  The rate of change is the slope.  The change in y is the "rise" and the change in x is the "run", and the slope is "rise/run", so in this case, the rate of change is 3/9, or 1/3\n\n2.) slope is, again, the change in y divided by the change in x.  Given 2 coordinates, the formula for slope is\n\n(y2 - y1)/(x2 - x1)\n\nSo in this case it's (-7 - (-4))/(3-2) which equals -3/1, or -3\n\n3.) If this means that the slope = 1, then the equation of a line is y = mx + b, and you need to solve for the y-intercept, b.  You know that the slope, m, equals 1, so\n\ny = 1x + b\n\nYou also have a coordinate, (3,8), so plug it into the equation\n\n8 = 1(3) + b\n8 = 3 + b\nb = 5\n\nSo the equation would be y = x + 5\n\n4.) The slope of a perpindicular line is the negative reciprocal of the original line.\n\nx - 3y = 4  solve for y\nx - 4 = 3y\ny = (1/3)x - 4/3\n\nSo the slope for this line is 1/3, therefore the slope for the perpindicular line is - 3\n\n5.) solve the equation for y to get it in the form of y = mx + b\n\n3x - 4y = 4\n3x - 4 = 4y\ny = (3/4)x - 4/3\ny = mx + b\n\nSlope is "m" and y-intercept is b, so slope is 3/4 and y-intercept is -4/3
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
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.
735157
let x be the distance AP and let Y be the angle <PLA.\n\nThen x=3tanY then differentiate implicitly to time\n\ndx/dt= 3sec^2Y dY/dt\n\nNow dY/dt= 4 rev/min * 2pi radians/rev= 8pi radians/min\n\nWhen x=1, we have tanY= 1/3 sec^2Y=1+tan^2Y= 10/9\n\nThus dx/dt=3*10/9*8pi= 80/3 pi which approximately equals\n\n83.8\n\nThe spot of light is moving along the shoreline at a rate of about \n\n83.8km/min when it is1km from p.\n\n*It is essential to convert revolutions to radians
626855
This is your garden variety related rates problem.\n\nAt any instant when the monkey is on the wall, you have yourself a right triangle, with the taut rope being the hypotenuse, the distance from fastening point to the wall being the horizontal leg, and the distance the monkey climbs vertically being the vertical leg. The one equation which relates all three of those quantities together is the Pythagorean Theorem: x^2 + y^2 = z^2 (x and y being the legs and z being the hypotenuse).\n\nSince this is a related rates problem, you need to differentiate this equation with respect to time (t). The result is 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt). All of the d*/dt terms represent the rate * is changing over time.\n\nSo, now it's a matter of plugging in the appropriate values given in the word problem. According to the word problem, x=8 ft, dy/dt=4 ft/s, y=6 ft (at that instant). From this info, one can derive that z=10 ft (via Pythagorean Theorem) and dx/dt=0 ft/s (because the distance between the fasten point and the wall is fixed). The problem is asking for dz/dt. Now, plug in:\n2(8)(0) + 2(6)(4) = 2(10)(dz/dt)\nSolving for dz/dt, your answer should be 2.4 ft/s (if the math is correct).