A trough is 14 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is filled with water at a rate of 13 ft3/min, how fast (in ft/min) is the water level rising when the water is 5 inches deep?

Answer 1

Did you make a sketch?
I made a sketch of an isosceles triangle with width of 4 ft and a height of 1 ft

At some time of t min, let the width of the water level be 2r and h
then for the water in the trough at that moment,
V = (1/2)(2r)(h)(14) = 14 hr

we have similar triangles: so r/h = 2/1 ===> r = 2h
V = 14(2h)(h) = 28 h^2
dV/dt = 56 h dh/dt
so for my given data ...
13 = 56(5/12)dh/dt
dh/dt = 156/280 ft/min = 39/70 ft/min
= appr .557 ft/min

better check my arithmetic

Answer 2

ratio of width to depth of water = 4/1 = 4
if depth is 5 in = 5/12 ft then w = 4 * (5/12) = 5/3 ft
area of surface = w L = (5/3)*14 = 23.33 ft^2
dh * w L = d volume
dh / dt * wL = d volume / dt = 13 fT^3/min
dh/dt = 13 ft^3/min / 23.33ft^2 = 0.557 ft/min

Answer 3

LOL Your arithmetic checks :)