Question
A tank in shape of inverted prism is 10ft higher, 20ft wide and 20ft long water is initially 4ft deel and a hole in the bottom releases water at the rate of square root the depth. Find the relationship between depth and time ? And when the tank will be empty
Answers
how is an inverted prism different from a regular prism?
Do you mean a triangular prism? If so, then using similar triangles, you see that the width of the water surface is twice the depth. So, with the height h of the water
v = 20*(h)(2h)/2 = 20h^2
dv/dt = 40h dh/dt = -√h
dh/dt = -1/(40√h)
√h dh = -1/40 dt
2/3 h^(3/2) = -1/40 t + c
h^(3/2) = -3/80 t + C
we have h(0) = 4, so C = 8
h^(3/2) = 8 - 3/80 t
when h=0, t = 640/3
you didn't specify the units of time.
Do you mean a triangular prism? If so, then using similar triangles, you see that the width of the water surface is twice the depth. So, with the height h of the water
v = 20*(h)(2h)/2 = 20h^2
dv/dt = 40h dh/dt = -√h
dh/dt = -1/(40√h)
√h dh = -1/40 dt
2/3 h^(3/2) = -1/40 t + c
h^(3/2) = -3/80 t + C
we have h(0) = 4, so C = 8
h^(3/2) = 8 - 3/80 t
when h=0, t = 640/3
you didn't specify the units of time.