- A trough in the cross sectional shape of an inverted equilateral triangle is being filled at a rate of 355 cm^3/min. The trough is 2 m long and has a side length of the triangle of 25 cm. How fast is the water level rising when the water is 12 cm deep?
Would this be the way to do it?
Draw a figure of the trough.
Cross section inverted equilateral triangle
with side 25 cm and length = 200 cm
Height h --> area A = 1/2(h* 2h/√3)
A = h²/√3
V(h) = A*200 = 200h²/√3
dV/dh = 400h/√3
dV/dt = dV/dh * dh/dt , dV/dt = 355 and h = 12
355 = 400*12/√3 * dh/dt
dh/dt = 355/(400*12/√3)
dh/dt = 0.1281 ≈ 0.1 cm/min
@zxcvbnm1216 looks right; I don’t immediately see anything incorrect…
A small tip: I like to keep careful track of the units to avoid any mistakes or misunderstandings, so 355 would be replaced with 355 cm^3/min, and dV/dh would have the units cm^2 (or cm^3/cm).
ok thanks @MITer94