If a fluid enters a hemisperical vat with a radius of 10 meter, at a rate of 0.5cubic meter per minute, how fast will the fluid be rising when the depth is 5 meter
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0.0577meter per minute
I get A = pi r^2 = 75 pi
r = 5 sqrt 3
so
dV = pi (75) dh
dV/dt = 75 pi dh/dt
so
dh/dt = .5 m^3/min / (75 pi m^2)
.5/(75 pi) = .00212 m/min
To find the rate at which the fluid is rising, we can use the concept of related rates.
Let's denote the depth of the fluid as "h" and the radius of the hemispherical vat as "r." We are given that the radius is 10 meters, so r = 10 m.
We are also given the rate at which the fluid is entering the vat, which is 0.5 cubic meters per minute. This is the rate at which the volume of the fluid is changing with respect to time (dv/dt).
We want to find the rate at which the depth of the fluid is changing with respect to time (dh/dt) when the depth is 5 meters.
First, we need to relate the volume of the fluid to the depth of the fluid. The volume of the fluid in the hemispherical vat can be calculated using the formula for the volume of a hemisphere:
V = (2/3) * π * r^3 * (1 - cos(θ))
Where θ is the angle between the surface of the fluid and the horizontal plane. Since we are considering the depth of the fluid, θ can be expressed in terms of the depth h as follows:
cos(θ) = (r - h) / r
Substituting this expression for cos(θ) into the volume formula, we get:
V = (2/3) * π * r^3 * (1 - (r - h) / r)
V = (2/3) * π * r^3 * (r + h) / r
Now, let's take the derivative of both sides of this volume equation with respect to time t:
dV/dt = (2/3) * π * r^3 * (dr/dt + dh/dt) / r
We are given that dV/dt = 0.5 cubic meters per minute and r = 10 m. So we have:
0.5 = (2/3) * π * 10^3 * (10 * dh/dt + dr/dt) / 10
0.5 = (2/3) * π * 10^2 * (10 * dh/dt + dr/dt)
Now, we can solve this equation for dh/dt, which represents the rate at which the fluid depth is changing:
dh/dt = (0.5 * 3) / (2 * π * 10^2) - dr/dt / 10
dh/dt = 0.075 / (π * 10^2) - dr/dt / 10
We want to find dh/dt when the depth is 5 meters, so we substitute this value into the equation:
dh/dt = 0.075 / (π * 10^2) - dr/dt / 10
dh/dt = 0.075 / (π * 10^2) - dr/dt / 10
Now we need to find the value of dr/dt, which represents the rate at which the radius is changing. Unfortunately, we don't have enough information to determine this value. We would need to know how the radius changes with respect to time in order to calculate the rate at which it is changing.
So, without knowing the rate at which the radius is changing, we cannot determine the exact rate at which the fluid is rising when the depth is 5 meters.