how fast is the water level rising when the water is 4 inches deep?

Question

A trough is 12 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is filled with water at a rate of how fast is the water level rising when the water is 4 inches deep?

Answer 1

If your filled with water at a rate of 14 ft^3/ min

then you can draw a triangles first
look at the triangle, you will find out
5/1=b/h by similar triangle
so b=5h

v=1/2bh(12) = 6(5h)h=30h^2

dv/dt = 60h dh/dt
since the rate is 14
so 14=60h dh/dt
dh/dt = 7/30h

when h = 1/3 then jsut plug in the number to dh/dt
21/30 ft/min is the final answer
hope it help
since I don't know what's your rate. you can just look at urs and change the number

hope it help!

Answer 2

Let x = the water depth in feet Let V = the volume of water in the trough Let t = time V = (10)½(5x)(x) = 25x² By the chain rule, dV/dt = (dV/dx)(dx/xt) Solving for dx/dt, we get: dx/dt = (dV/dt) / (dV/dx) Since dV/dt is given, we know that dx/dt = 14 / (dV/dx) dV/dx = d(25x²)/dx = 50x thus, dx/dt = 14 / (50x) = 7/(25x) At x = 1/3, dx/dt = 21/25, so: the water level is rising at the rate of 0.84 ft/min.

Answer 3

Critical data missing...fill rate.