A large vase has a square base of side length 6 cm, and flat sides sloping outwards at an angle of 120◦ with the base. How to find the rate when height is rising?

A large vase has a square base of side length 6 cm, and flat sides sloping outwards at an angle of 120◦ with the base. Water is flowing in at 12cm3/s.

How to 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.

1 Answer
Jul 1, 2018

#dh/dt=0.246 cms^-1#

Explanation:

enter image source here From the sketch it can be seen that an element of volume #dv# = #[2y]^2dx# We need to integrate #[2y^2]dx# from #0# to #h#., so we need to find #y# in terms of #x#.

It can be seen that the straight line along the face of the pyramid has equation #y=sqrt3/3x+3#..........#[1]# [ since tan# pi/6#,i.e, tan 30 degrees =#sqrt3/3# and the #y# intercept is #[0,3]#

#[2y]^2# = #[2[sqrt3/3x +3]]^2# = #[ 4x^2/3+ 8sqrtx+36]#.........#[2]#

Thus volume of the frustrum of the pyramid #V# =# int_0^h[[4x^2]/3 + 8sqrt3 + 36]dx# = #[[4h^3]/9 + [4sqrt3]h^2 +36h]......#[3]##

From the question #dv/dt = 12cm^3s^-1# so after #3# seconds the volume #V# = #36 cm^3#

Therefore, #36 = [4h^3]/9 +[4sqrt3]h^2 + 36h#. Solving this cubic for #h# yields a real value of #h# = #0.85249# [ online cubic equation solver].

Differentiating.......#[3]# wrt to t.

#[dv]/dt= 4/3h^2[dh/dt] + 8sqrt3hdh/dt+36dh/dt#.

#[dv]/dt= [dh]/dt[ 4/3h^2+8sqrt3h+36]# But #dv/dt =12# [from the question]

Therefore, #[dh]/dt=12/[ 4/3h^2 + 8sqrt3h+36]#, and evaluated when #h# = #0.85249# will give the answer above. Hope this was helpful, and I believe it to be correct. I will ask it to be checked.