Question

A water trough is 15 m long and has a cross-sectional shape of an equilateral triangle with a side length of 2 m. It is being filled at a rate of 0.6 m3/min. How fast is the water level rising when the water is 0.2 m deep?

Answer 1

regardless of side length; you need to find away to equate a change in height with a change in volume i believe

Answer 2

V = basearea*15; gotta determine the area for an equilat i spose

dV/dt = .6

Answer 3

base = s

height = s sin(60)

V = 15 s^2 sqrt(3)/4

dV = 30sqrt(3)/4 *s *ds/dt perhaps?

Answer 4

find dh/dt

height = s sin(60)

s = h/sin(60) = 2h/sqrt(3)

ds/dt = 2/sqrt(3) dh/dt

dh/dt = sqrt(3)/2 * ds/dt

Answer 5

\[\frac{dV}{dt} = \frac{30s\sqrt{3}}{4}\frac{ds}{dt}\]
\[.6 = \frac{30s\sqrt{3}}{4}\frac{2}{\sqrt{3}}\frac{dh}{dt}\]
\[.6 = \frac{30s}{2}\frac{dh}{dt}\]
\[\frac{.6(2)}{30s} =\frac{dh}{dt}\]
\[\frac{.12}{30s} =\frac{dh}{dt}\]

when we know s at time (t) then we know dh/dt

Answer 6

well, we dont need to know t :) but the problem allows us to determine s when the water is at any given height

Answer 7

1-sqrt(3)-2 = b/2-h-s

h = .2
h = sqrt(3)

scalar is .2/sqrt(3)

2(.2)
----- = s
sqrt(3)

\[\frac{dh}{dt}=\frac{.12\sqrt{3}}{2(.2)(30)}\]
\[\frac{dh}{dt}=\frac{.12\sqrt{3}}{.4(30)}\]
\[\frac{dh}{dt}=\frac{12\sqrt{3}}{40(30)}\]
\[\frac{dh}{dt}=\frac{\sqrt{3}}{10(10)}\]
\[\frac{dh}{dt}=\frac{\sqrt{3}}{100}\]
maybe?

Answer 8

by the chain rule:
\[\frac{dh}{dt}=\frac{dh}{ds}\frac{ds}{dV}\frac{dV}{dt}\]

dV = .6
...................................................
V = 15 s^2 sqrt(3)/4

1 = 30s sqrt(3)/4 ds

ds = 4/(30s sqrt(3)) ; s = .4/sqrt(3)
= 4/30(.4)
= 1/30(.1)
= 1/3
...................................................

h = s sin(60)

dh = sin(60)
= sqrt(3)/2

\[\frac{dh}{dt}=\frac{\sqrt{3}*1*6}{2*3*10}\]
\[\frac{dh}{dt}=\frac{\sqrt{3}}{10}\]
...........................................

which may be a better result