A water trough is 8 cm wie at the top, 50 cm wie at the bottom and 40 cm deep and 3 m long. Water is poured in at 1500cm ^3/s how fast is the water level rising when itis 20 cm deep?

To find the rate at which the water level is rising, we need to apply the concept of related rates.

Let's denote the depth of the water as "h" and consider its rate of change with respect to time, dh/dt. We are given that the water trough is 8 cm wide at the top, 50 cm wide at the bottom, and 40 cm deep. It is also mentioned that the trough is 3 m long.

Given that the water is poured in at a rate of 1500 cm^3/s, we need to find dh/dt when the depth of the water is 20 cm.

To solve this problem, we can use two similar triangles: one formed by the top surface of the water and the top surface of the trough, and the other formed by the current water level and the bottom of the trough.

Let's denote the width of the water at a depth of "h" as "w(h)." Using similar triangles, we can determine the relationship between the width and the depth:

w / (40 cm) = (w(h) - 8 cm) / h

Rearranging the equation, we get:

w = (40 * (w(h) - 8)) / h

Since the width at the bottom is 50 cm, we have:

50 = (40 * (w(h) - 8)) / h

Simplifying further, we get:

h * (w(h) - 8) = (2/5) * w(h) * 40

h * w(h) - 8h = (8/5) * w(h)

Rearranging the equation, we get:

h * w(h) - (8/5) * w(h) = 8h

Combining like terms, we get:

(5h - 8) * w(h) = 8h

Now, we can differentiate both sides of the equation with respect to time "t" to find the rate at which the width is changing with time, dw(h)/dt:

(5h - 8) * dw(h)/dt + w(h) * dh/dt = 8 * dh/dt

Rearranging the equation to solve for dw(h)/dt, we get:

(5h - 8) * dw(h)/dt = 8 * dh/dt - w(h) * dh/dt

Simplifying further, we have:

dw(h)/dt = (8 * dh/dt - w(h) * dh/dt) / (5h - 8)

Next, we need to find a relationship between the volume of water and its depth. The volume V(h) at a depth "h" can be calculated by multiplying the cross-sectional area with the length:

V(h) = (w(h) + 50) * h * (3 m)

Now, we can differentiate the volume equation with respect to time "t" to find the rate at which the volume is changing with time, dV(h)/dt:

dV(h)/dt = ((dw(h)/dt) * h * (3 m)) + (w(h) + 50) * dh/dt

Since the water is poured into the trough at a rate of 1500 cm^3/s, we can substitute dV(h)/dt with 1500 cm^3/s:

1500 cm^3/s = ((dw(h)/dt) * h * (3 m)) + (w(h) + 50) * dh/dt

Now we need to find the values of w(h), dh/dt, and h when the depth of the water is 20 cm.

From the geometry of the trough, we know that w(h) = 8 + (42/4) * h, where h is the depth of the water.

dh/dt represents the rate at which the water level is rising, which is what we want to find.

Substituting the values into the equation, we have:

1500 cm^3/s = ((dw(h)/dt) * (20 cm) * (3 m)) + ((8 + (42/4) * (20 cm)) + 50) * dh/dt

Simplifying the equation, we get:

1500 = (60 * dw(h)/dt + (770 * dh/dt)

Now we have an equation with dw(h)/dt and dh/dt. We can solve for dh/dt by isolating it on one side of the equation:

1500 - 60 * dw(h)/dt = 770 * dh/dt

dh/dt = (1500 - 60 * dw(h)/dt) / 770

Finally, we need to find the value of dw(h)/dt when the depth of the water is 20 cm. Substituting h = 20 cm into the equation w(h) = 8 + (42/4) * h, we get w(h) = 8 + (42/4) * 20 cm.

By evaluating the value of dw(h)/dt when w(h) is 8 + (42/4) * 20 cm, we can substitute that value into the equation to find dh/dt.