0
$\begingroup$

A large vase has a square base of side length $6 \text{ cm}$, and flat sides slopingoutwards at an angle of $120^{\circ}$ with the base. Water is flowing in at $12 \text{ cm}^3/\text{s}$. Find, to three significant figures, the rate at which the height of water is rising when the water has been flowing in for $3$ seconds.

Spent around an hour trying to do it, but I keep getting the answer wrong. I think I'm not getting the right volume function.

$\endgroup$
3
  • $\begingroup$ What formula are you using for the volume of the vase? It should be a function of the height. $\endgroup$
    – Jeff
    Apr 28, 2014 at 7:25
  • $\begingroup$ @Jeff yes i expressed the volume in terms of h, the height, but i think my expression is wrong. $\endgroup$
    – Rishi
    Apr 28, 2014 at 7:36
  • $\begingroup$ Welcome to math.stackexchange! Please remember to vote up helpful answers, and to mark an answer as accepted if it answers your question. This will increase your reputation on the site and give you more privileges. $\endgroup$
    – Jeff
    Apr 28, 2014 at 16:07

1 Answer 1

0
$\begingroup$

The volume of the vase is $\frac{1}{3}\left(6^2 + (6+\frac{2h}{\sqrt{3}})6 + (6+\frac{2h}{\sqrt{3}})^2 \right)h$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .