A water trough is 10m long and a cross-section has the shape of anisosceles trapazoid that is 40cm wide at the bottom, 75cm wide atthe top, and has a height of 55cm. If the trough is being filledwith water at the rate of 0.7m^3 /min, how fast is the water levelrising when the water is 45cm deep? (Round the result to thenearest hundreth)

I get the answer of 8.2cm/min when I work this out, its supposed tobe over 10 cm/min though. If someone could work this out please I'dappreciate it.

dv/dt = .7 m^3/min

h= .45

Height = .4 + h

total area of filled .5(.8 + h)h => (h^2 +.8h)/2

volume = 5(h^2 + .8h)

dv/dt = 5(2h + .8) dh/dt

.7 = 5(2 * .45 + .8) dh/dt

dh/dt = .7 / 8.5

dh/dt= .8235m^3/min

dh/dt= 8.24cm^3/min

Have no idea where I went wrong. Thank you for the help.

I get the answer of 8.2cm/min when I work this out, its supposed tobe over 10 cm/min though. If someone could work this out please I'dappreciate it.

dv/dt = .7 m^3/min

h= .45

Height = .4 + h

total area of filled .5(.8 + h)h => (h^2 +.8h)/2

volume = 5(h^2 + .8h)

dv/dt = 5(2h + .8) dh/dt

.7 = 5(2 * .45 + .8) dh/dt

dh/dt = .7 / 8.5

dh/dt= .8235m^3/min

dh/dt= 8.24cm^3/min

Have no idea where I went wrong. Thank you for the help.

## Answers ( 1 )

dV/dt = 2dh/dt +3.5h/.55 dh/dt

dh/dt = .7 /(2 + (3.5/.55)(.45)) = .1439m/min

=14.39cm/min