Water trough being filled with water, related rates?

I guess that the ends of the tank are triangles with their bases on the top...

If the tank has a level of x (in ft), its filled volume is the volume of a triangular prism whos base has a height of x. As the triangles are equilateral, x and B satisfy:

x/B = sin(60°) = (√3)/2.

Then the area of the triangle (the section under water) is

A(x) = (1/2) B x = (1/2) (2x/(√3)) x = x² / √3.

The volume of water is then

V(x) = A(x) L = 5 x² / √3.

This volume is increasing at a rate of 3 (cubic ft by minute), then

dV/dt = dV/dx dx/dt = (10/√3) x dx/dt = 3

=>

dx/dt = (3√3/10)/x

and when x = 6 in = 0.5 ft

dx/dt (x=0.5) = 3√3/5 = 1.039 ft/min

There could desire to be an overabundance of water troughs contained in the calculus international. it extremely is the third such question i've got responded in as many day. I be conscious that the numbers contained in the question do not experience the numbers on your "answer". right here is answer utilizing numbers in question: First, we could desire to apply comparable measurements (m) for all dimensions: Water trough is 10 m long circulate section is shape of isosceles trapezoid 20 cm = 0.2 m huge at backside 80 cm = 0.8 m huge at precise has height 60 cm = 0.6 m component of trapezoid = h * (b? + b?)/2 quantity of trapezoidal trough = h * (b? + b?)/2 * L ---------- the subject with looking quantity of in part crammed trough is that b? variations with height. while h = 0, b? = b? = 20cm = 0.2 m while h = 0.6, b? = 80 cm = 0.8 m when you consider that distinction between bases is 0.8m - 0.2m = 0.6m and h=0.6m, then width of precise base will enhance by utilizing 0.1m for each 0.1m enhance in height. so we are able to specific b? = 0.2 + h Now we are able to rewrite equation for quantity as function of h: V = h * (b? + b?)/2 * L the place b? = 0.2, b? = 0.2 + h, L = 10 V = h * (0.2 + 0.2 + h)/2 * 10 V = 10h * (h + 0.4) / 2 V = 5h² + 2h ---------- Water enters at fee of 0.one million m³/min dV/dt = 0.one million Water is 50 cm = 0.5 m deep h = 0.5 how briskly is water point increasing? locate dh/dt V = 5h² + 2h dV/dt = (10h + 2) dh/dt ----> as much as date: I had 5h+2 earlier dh/dt = (dV/dt) / (10h + 2) dh/dt = 0.one million / (10*0.5 + 2) dh/dt = 0.one million / 7 dh/dt = one million/70 = 0.0143 So water point is increasing at fee of 0.0143 m/min or one million.40 3 cm/min ==================== ==================== i might have achieved calculations utilizing numbers on your "answer", yet i'm undecided if H = .4m refers back to the peak of the trough or the peak of the water. to unravel, i want the two heights.