math question (1 Viewer)

poptarts12345 · Jul 10, 2020

Sand being poured from a conveyor belt forms a cone with height h and semi vertical angle 60 degrees. Show that the volume of the pile is V = pi h^3, and differentiate with respect to t. At what rate must the sand be poured if its required that the height increase at 8cm/min, when the height is 4 metres?

ANS : 96pi/25 m^3/min

Drongoski · Jul 10, 2020

radius of base of cone r = h x tan 60 = sqrt(3) h

$\therefore V = \frac {1}{3} \times \pi r^2 \times h = \frac {1}{3} \times \pi \times (h\sqrt 3)^2 \times h = \pi h^3 m^3/min\\ \\ \therefore \frac{dV}{dt} = \frac {dV}{dh} \times \frac {dh}{dt} = 3 \pi h^2 \frac {dh}{dt} = 3\pi \times 4^2 \times \frac {8}{100} m^3/min = \frac {96}{25} \pi m^3/min$

CM_Tutor · Jul 19, 2020

Drongoski, I think there are a couple of typos (marked in red) that you might want to fix...

$\begin{align*} \tan 60^o &= \frac{\text{radius of base}}{\text{height of cone}} \\ \sqrt{3} &= \frac{r}{h} \\ \implies r &= h\sqrt{3} \end{align*}$

$\begin{align*} \text{Now, } V_{\text{cone}} &= \frac{1}{3} \times A_{\text{base}} \times h = \frac{1}{3} \times \color{red} \pi \color{black} r^2 \times h = \frac{1}{3} \times \pi \times \big( \color{red} h \color{black} \sqrt{3} \big)^2 \times h = \pi h^3 \\ \implies \frac{dV}{dh} &= 3\pi h^2 \end{align*}$

$\text{We are given that } \frac{dh}{dt} = 8 \text{ cm/min} = \frac{8}{100} \text{ m/min} = \frac{2}{25} \text{ m/min}$

$\begin{align*} \text{Using the Chain Rule, } \frac{dV}{dt} &= \frac{dV}{dh} \times \frac{dh}{dt} = 3 \pi h^2 \times \frac{2}{25} = \frac{6 \pi h^2}{25} \\ \text{When } h = 4 \text{ m: } \frac{dV}{dt} &= \frac{6\pi \times 4^2}{25} = \frac{96\pi}{25} \text{ m}^3\text{/min} \end{align*}$

Drongoski · Jul 19, 2020

Thanks for pointing out the typos in my solution, CM-Tutor. I didn't even notice them.

CM_Tutor · Jul 20, 2020

Drongoski said:
Thanks for pointing out the typos in my solution, CM-Tutor. I didn't even notice them.

No Problem - it is very easy to make typos and other little mistakes, we're all human!

math question (1 Viewer)

poptarts12345

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