Rates and Changes help needed (1 Viewer)

cssftw · Sep 3, 2011

Hi guys need a bit of help on this question:

Q. Water is poured into an inverted cone at a rate of 10cm^3 /second. The radius of the cone is 5cm and its height is 10cm.

(i) At what rate is the level rising after 2 seconds?
(ii) At what rate is the level rising when the level is 8cm?

----------

My solution (please someone figure out what is wrong with it):

dV/dt = 10 cm^3 / second

We are trying to find dx/dt (where x is the length from the tip of cone to the level of water --> remember the cone is INVERTED

dx/dt = dV/dt * dx/dV

V=(1/3)(pi)(r^2)(h)

x=h, r=r, therefore:

V=(1/3)(pi)(r^2)x

By using similar triangles:

x/10 = r/5 (sorry I don't have a diagram)

Therefore r = x/2, subbing back into V:

V = (1/3)(pi)((x^2)/4)(x)
V = ((pi)(x^3))/12

Therefore:

dV/dx = 3(x^2)(pi)/12
dV/dx = (x^2)(pi)/4

THEREFORE: dx/dV = 4/(x^2)(pi)

Substituting this back into the chain rule expression:

dx/dt = dx/dV * dV/dt

dx/dt = 40/((x^2)(pi)) m/s

So that's what I got for my expression of the rate of change of water level -- however it doesn't look very right, does it? I mean how do I find the rate of change at t=2 seconds?

If anyone could find the proper expression for dx/dt, and how to find the answer to the questions it would be immensely appreciated!

rolpsy · Sep 3, 2011

$\begin{align*} \frac{\mathrm{d} V}{\mathrm{d} t}&=10\\ V&=10t+c\\ \textup{when }t=0&, \ V=0 \ \ \therefore c=0\\ V&=10t\\ \textup{but } V&=\frac{\pi x^{3}}{12}\\\\ \therefore 10t&=\frac{\pi x^{3}}{12}\\ x^{3}&=\frac{120t}{\pi}\\ \textup{when } t=2&, \\ x^{3}&=\frac{240}{\pi}\\ x^{2}&=\left (\frac{240}{\pi} \right )^{\frac{2}{3}} \end{align*}$

your expression for dx/dt is correct

$\begin{align*} \frac{\mathrm{d} x}{\mathrm{d} t} &= \frac{40}{\pi x^{2}}\\ &=\frac{40}{\pi \left (\frac{240}{\pi} \right )^{\frac{2}{3}}}\\ &=\frac{40(240)^{-\frac{2}{3}}} {\pi^{\left (1-\frac{2}{3} \right )}}\\ &=\frac{\sqrt[3]{\frac{10}{9}}}{\sqrt[3]{\pi}}\\ &=\sqrt[3]{\frac{10}{9\pi}} \textup{ cm/second} \end{align*}$

and

$\begin{align*} \textup{when } x=8&,\\ \frac{\mathrm{d} x}{\mathrm{d} t}&=\frac{40}{\pi (8)^{2}}\\ &=\frac{5}{8\pi}\textup{ cm/second} \end{align*}$

cssftw · Sep 3, 2011

rolpsy said:
$\begin{align*} \frac{\mathrm{d} V}{\mathrm{d} t}&=10\\ V&=10t+c\\ \textup{when }t=0&, \ V=0 \ \ \therefore c=0\\ V&=10t\\ \textup{but } V&=\frac{\pi x^{3}}{12}\\\\ \therefore 10t&=\frac{\pi x^{3}}{12}\\ x^{3}&=\frac{120t}{\pi}\\ \textup{when } t=2&, \\ x^{3}&=\frac{240}{\pi}\\ x^{2}&=\left (\frac{240}{\pi} \right )^{\frac{2}{3}} \end{align*}$

your expression for dx/dt is correct

$\begin{align*} \frac{\mathrm{d} x}{\mathrm{d} t} &= \frac{40}{\pi x^{2}}\\ &=\frac{40}{\pi \left (\frac{240}{\pi} \right )^{\frac{2}{3}}}\\ &=\frac{40(240)^{-\frac{2}{3}}} {\pi^{\left (1-\frac{2}{3} \right )}}\\ &=\frac{\sqrt[3]{\frac{10}{9}}}{\sqrt[3]{\pi}}\\ &=\sqrt[3]{\frac{10}{9\pi}} \textup{ cm/second} \end{align*}$

and

$\begin{align*} \textup{when } x=8&,\\ \frac{\mathrm{d} x}{\mathrm{d} t}&=\frac{40}{\pi (8)^{2}}\\ &=\frac{5}{8\pi}\textup{ cm/second} \end{align*}$

^repped, thanks a million mate

Rates and Changes help needed (1 Viewer)

cssftw

Member

rolpsy

Member

cssftw

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)