This is a realted promblem question. When you solve include to formula and put the answer ty?

Question

The cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25 com deep and water is flowing in at a rate of 0.25m^3, how fast is the water level rising when the water is 10cm deep at the deepest point

Answer 1

let v be the volume of water filled.

So dv/dt = 0.25

We don't want v, we want water height, h. v is related to h by being length on an equilateral triangle.

v = 0.5*base*h*length

Length is 5m, and we can get the base in terms of h if we use trigonometry.

base = h/sin60

so v = 2.5*h^2/sin60

We want dh/dt.

dh/dt = (dv/dt)/(dv/dh) (from the chain rule)

dv/dh = 5h/sin60, and dv/dt = 0.25

so dh/dt = 0.25*sin60/5h

So when h = 0.1 metres

dh/dt = 0.25*sin60/0.5 = 0.43301270189221932338186158537647 m/s

Answer 2

Hehe - might p!ss your prof off, but you don't need calculus to solve this. You can simply divide the rate of flow by the area of the water surface when the water level is 10cm. I get a width of 10*3^(1/2) (<--please check my trig here) times a length of 5m = an area of 3^(1/2)/2 m^2. Dividing the rate of 0.25m^3/sec (or is it per minute? You didn't include that) into the area yields

1/(2*3^(1/2))m per second or minute, whichever

Do please check my trig and math on this, and only submit it if you want to amaze/piss off your prof. Sorry, no time for the calculus solution, maybe later, ok?