(adsbygoogle = window.adsbygoogle || []).push({}); a) 1. The problem statement, all variables and given/known data

A water trough is 10m long, and a cross section has the shape of an isosceles triangle that is 1m across at the top and 50cm high. The trough is being filled with water at a rate of 0.4m^3/min. How fast is the water level rising when the water is 40cm deep?

b) As a volcano erupts, pouring lava over its slope, it maintains the shape of a cone, with height twice as large as the radius of the base. If the height is increasing at a rate of 0.5 m/s, and all the lava stays on the slopes, at what rate is the lava pouring out of the volcano when the volcano is 50m high?

2. Relevant equations

h=height

w=width

3. The attempt at a solution

a)

dV/dt = 0.4m^3/min

V=(1/2)hw(10)

=5hw

w/1=h/0.5

w=2h

V=5hw = 5h(2h) = 10h^2

dV/dt=20hh'

0.4 = 20(0.4)h'

0.05=h'

I am almost sure it is correct but I am just looking for a confirmation. I will add units of course later.

b)

h' = 0.5m

r=(h/2)

h=50

v'=?

I think we are searching for the rate the volume decreases..so it would be 981.25m^3/s.

**Physics Forums - The Fusion of Science and Community**

# Water Trough (Related Rates)

Have something to add?

- Similar discussions for: Water Trough (Related Rates)

Loading...

**Physics Forums - The Fusion of Science and Community**