Calculus Related Rate Problem?
930510
Express volume as a function of depth by using the fact that the volume of the frustum is the difference between the volume of two cones and the fact that the ratio between the depth plus 6 inches and the radius at the surface is 12/2:\nV(x)=1/3*(x+6)*pi*((x+6)/6)^2-1/3*6*pi*1^2=pi/108*(x+6)^3-2pi.\nNow differentiate with respect to time, assuming that x is a function of time:\ndV/dt=pi/36*(x+6)^3*dx/dt.\nNow, remember that dV/dt=-3 and x=4, so -3=pi/36*1000*dx/dt, so dx/dt=-27/(250pi).
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
918147
You have a right triangle with legs 106 and 88, a & b respectively. By the Pythagorean Theorem, the hypotenuse is square root of 18980. Now, use related rates. (dc/dt)2c=(da/dt)2a+(db/dt)2b. That becomes (dc/dt)*2*sq. rt of 18980=19(212)+22(176). dc/dt=28.67 knots per hour.
1016838
Use the quotient rule for differentiation.  Taking the case where the firetruck is approaching:\n\nThe numerator of the derivative has the denominator of the original function times the derivative of the numerator minus the numerator of the original function times the derivative of the denominator.  \n\nThis is pretty easy, since the numerator of the original function is a constant.  The derivative of a constant is 0.  When approaching, you only have to worry about the original numerator times the derivative of (331 - v).  The 331 is a constant with a derivative of 0 leaving you the derivative of -v.  That's -1.  \n\nTherefore, the numerator of your derivative is -132400 * -1, or 132400 when approaching.\n\nThe denominator of the derivative is just the square of the original denominator, or (331 - v)^2 when approaching.  When approaching, that would give you 301^2 (301 squared), or 90601. \n\nWhen approaching, the rate of change would be 132,400/(301^2), or about +1.46.\n\nFinding the rate of change when moving away works the same, but with a +v instead of a -v.
958697
10.4 m/sec\n\nV(t) = (30+a*t) m/sec\n\nwhere a = - 9.8m/sec^2\n\nV(2) = (30 - 09.8*2) m/sec\n\nV(2) = 30-19.6 = 10.4 m/sec
497479
When calculating with error, remember to MAXIMIZE error wherever possible. For example, since V1 and V2 are in the denominator of their respective terms, error should be subtracted to maximize the overall terms. Check me on this, but I found an average velocity of (19.5 ± 0.5) m/s.
776431
Integrate the acceleration equation with respect to time, twice.  Use, Xo = 500 and Vo = 10...solve for t.
864537
r1 * A1 = r2 * A2\n0.21*(pi*1.2^2) = r2 * (pi*0.23^2)\nr2 = 5.72 m/s
748998
Assume smaller pipe fills the tank in x hours and the larger one in (x-2.5) hours. Taking the given inf that both together can fill the tank in 8.4 hours, develop a quadratic equation in x, and solve it. \n\nLarger pipe fills the tank in 15.65 hours whereas the smaller one takes 18.15 hours to fill it.
794278
The distance between them is calculated by:\n\nD = √((50t)^2 + (60t)^2) = √(6100t^2) = 10√(61)t\n\nD' = 10√(61) (a constant rate of change)\n\nSo I agree with you!
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
194533
Let the second man's rate be x.  We are given the first man's rate is x+0.5.  Finally, let y be the distance the first man can walk in 5 hours.  We know that this is the same distance the second man can walk in 6 hours.\n\nSo you get two equations, and two unknows.  We know the first man takes 5 hours to walk the distance y:\n\ny/(x+0.5) = 5\n\nWe know the second man takes 6 hours:\n\ny/x=6\n\nSo solve for x, and that gives you the second man's rate.  The first man's rate will be x+0.5
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.
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.
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.
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.
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page