Recent content by Ocis

Well obviously not 1? I have never differentiated pi before...? Or it is a constant and it should therefore cancel out the first part of the product rule because it equals 0 U'V leaving just {\Pi}{\frac{12}{27}}h^{2} ??

Ok I get it all now very silly mistakes indeed. So using the product rule is it anything close to {\frac{4}{27}} h^{3} + {\Pi}{\frac{12}{27}}h^{2}

Ok, so am I wrong to think that constants such as Pi get canceled out and then everything else just individually divided by three (2/3h^2 / 3) (* h/3 )' which should = 4/27h ?? Sorry about this...

Now I am even more confused... How do I differentiate both sides of the volume equation with respect to t ?

Regrettably I am confused, again.. If I differentiate that volume I am getting {\frac{dv}{dh}} ?? = {\frac{4}{9}} h then {\frac{dv}{dt}} = ? {\frac{dv}{dh}} x {\frac{dh}{dt}} where {\frac{dh}{dt}} = ( 3 x 10^{-2} ms^{-1} ) ??

Ok so how is this? Volume = {\frac{\Pi X ({\frac{2}{3}}h)^{2} X h}{3}} ?? Then...?

Could I just use trig to work out the new radius tan \Theta = {\frac{3}{2}} = 56.31 {\frac{2}{tan 56.31}} = 1.34 (r) Vol = {\frac{\Pi X 1.34^{2} X 2}{3}} = 3.72 m^{3} But then how do I involve the flow rate which at that height 2m is 3 X 10^{-2} m s^{-1}

[SOLVED] Volume of liquid flowing into a tank A tank in the shape of an inverted cone has liquid flowing into it at constant rate. If the rate at which the water rises in the tank is 3x10^-2 m s^-1 when the liquid is at a depth of 2 m, calculate the volume of liquid flowing into the tank at...

Yeah you are right, I do - many thanks. Is the second derivative right that i wrote earlier?

I have in the past but am not overly keen on them. Would it be something along the lines of -t log e = 6/12 and then 6 /12 log e = -t well that looks wrong, what else..?

but then to differentiate it a second time it becomes what - 12 e^-t ? how do I work out what the time is then for part 3. i would have e^-t = 6/12 ? I am stuck where to go from there how do I do it?

constant = 0 and derivative of an exponential would be - e ^-t so its just 12 e ^-t ?

[SOLVED] Differentiation of displacement and time to get velocity A moving body is related to a fixed point by the displacement equation: s = 12( 1 - e^-t). Assuming the body is moving in a straight line on a flat plain. 1. obtain an expression for the velocity after time t seconds. 2...

Oh yeah of course it is, thanks to you all. Panic over!

Is the solution anywhere close to this? \frac{\left(25x ^{3}\right)}{6} - \frac{\left(10x ^{4}\right)}{4} + \frac{\left(x ^{5}\right)}{5} Thanks,