Question

a funnel in the form of a cone is 10 inches across the top and 8 inches deep. water is flowing into the funnel at the rate of 12 inches cube per second and out at the rate of 4 inches cube per second . how fast is the surface of the water rising when it is 5 inches deep

Answer 1

The surface of the water is rising at a rate of
`\((15)/(\pi)\)` inches per second when it is 5 inches deep.

Step-by-step explanation:

To determine how fast the surface of the water is rising, we can use related rates involving the volume of a cone. The volume
`\(V\)` of a cone is given by the formula
`\(V = (1)/(3)πr^2h\)`, where
`\(r\)` is the radius of the top and
`\(h\)` is the height.

Given that the cone has a diameter of 10 inches, the radius
`\(r\)` is
`\(5\)` inches. The height
`\(h\)` is given as
`\(8\)` inches. To find the rate at which the water is rising, we take the derivative of the volume with respect to time
`\(t\): \((dV)/(dt) = \pi r^2 (dh)/(dt)\).`

The water is flowing into the funnel at a rate of
`\(12\)` inches
`\(^3\)` per second and out at a rate of
`\(4\)` inches
`\(^3\)` per second. The net rate is
`\(12 - 4 = 8\)` inches
`\(^3\)` per second. Substituting the known values into the equation, we get
`\(8 = \pi (5^2) (dh)/(dt)\)`. Solving for
`\((dh)/(dt)\)`, we find it to be
`\((15)/(\pi)\)` inches per second.

Answer 2

A funnel in the form of a cone is 10 inches across the top and 8 inches deep, the velocity the surface of the water rising when it is 5 inches deep is 5

Step-by-step explanation:

To find the rate at which the surface of the water is rising, we can use the concept of related rates.

Let's denote the radius of the surface of the water as r and the height of the water as h.

Since the funnel has the shape of a cone, we can use the formula for the volume of a cone to relate the radius and height:

V = (1/3) * π * r^2 * h

Now, let's differentiate both sides of the equation with respect to time, t:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)

Given that water is flowing into the funnel at a rate of 12 in^3/s and out at a rate of 4 in^3/s, we can substitute these values into the equation:

12 = (1/3) * π * (2(5) * dr/dt * 5 + 5^2 * dh/dt) - 4

Next, we can solve for dh/dt:

dh/dt = (12 - (50/3)π * dr/dt) / (25π)

Finally, substitute the given value of dr/dt (which is not given in the question) to calculate dh/dt when h = 5.

So therefore the velocity the surface of the water rising when it is 5 inches deep is 5