related rates question (calculus)?
970749
500sin30 =  x  m/h altitude gained
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
316045
kg m^2\n------- = Joule\ns^2\n\nb = kg m/s
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
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
200117
distance= rate*time\ntime= distance/rate
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
1298
derivative of 5(sec(x)^2) with respect to sec(x) = 10(sec(x))\n\nderivative of 5(sec(x)^2) with respect to x = 10 * derivative of sec(x) with respect to x
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
950679
(8.5m/s)/.008s = 1062.5 m/s^2 = a\nm = 12\nF=ma=12750N. <----- b)\nImpulse is Fdt, so 12750*.008 = 102 N/sec
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:-)
867482
Let t be the time, the same time for both then because distance is rate times time the equation is\n\n90t + 80t = 425\n170 t = 425\nso t = 2.5 hours
860522
v = 8(10 - r²)\n\na) Area ≈ 2πr.δr (= length of circumference at radius r x width)\n\nb) flowrate = Cross-sectional area x speed\n≈ 2πr.δr * 8(10 - r²)\n= 16πr(10 - r²)δr\n\nc) Flow rate = ∫(r = 0 to 3)16πr(10 - r²)dr\n= 16π∫(r = 0 to 3) (10r - r³)dr\n= 16π[5r² - ¼r^4](r = 0 to 3)\n= 16π{(45 - 81/4) - 0]\n= 16π x 99/4\n= 396π cu in / s