If a fluid enters a hemispherical vat with a radius of 10meter . at a rate of 0.5cubic meter per minute, how fast will the fluid be rising when the depth is 5meter?
Trick:
at depth h
the volume increase with delta h
= the surface area * delta h
draw that and remember it.
surface area = pi r^2
what is r of surface at h = 5 ?
that takes some geometry
triangle from center
5^2 + r^2 = 10^2
r^2 = 75
r = 5 sqrt 3
so
dV = pi (75) dh
dV/dt = 75 pi dh/dt
so
dh/dt = .5 ft^3/min / (75 pi ft^2
To find the rate at which the fluid is rising, we can use the concept of related rates from calculus.
First, let's define some variables:
- r: the radius of the hemispherical vat
- V: the volume of the fluid in the vat
- h: the depth of the fluid in the vat
We are given:
- r = 10 meters
- dV/dt = 0.5 cubic meters per minute (rate at which the volume is changing)
We want to find:
- dh/dt: the rate at which the depth is changing when the depth is 5 meters
The volume of a hemisphere can be calculated using the formula:
V = (2/3) * π * r^3
To relate the variables V and h, we need to find an expression for V in terms of h. From the definition of a hemisphere, we know that at any given depth, the volume of water above that depth is half the volume of the hemisphere. So, we have:
V = (1/2) * (2/3) * π * r^3
V = (1/3) * π * r^3
Now, let's differentiate both sides of the equation with respect to time t:
dV/dt = (1/3) * π * (3r^2 * dr/dt)
Since r is constant (given as 10 meters), the term (3r^2 * dr/dt) becomes 0. So, we have:
dV/dt = 0
Now, let's differentiate V = (1/3) * π * r^3 with respect to time t:
dV/dt = (1/3) * π * (3r^2 * dr/dt)
Since we want to find dh/dt, we need to find dr/dt. We can use the Pythagorean theorem to relate the variables r and h:
r^2 = 10^2 - h^2
Differentiating both sides with respect to time t:
2r * dr/dt = - 2h * dh/dt
We can simplify this to:
dr/dt = (-h/r) * (dh/dt)
Now, we have all the expressions we need to find dh/dt when h = 5 meters. Given that r = 10 meters and dV/dt = 0.5 cubic meters per minute:
0.5 = (1/3) * π * (3(10^2) * (-5/10) * (dh/dt))
Simplifying:
0.5 = (1/3) * π * (3) * (-5/10) * (dh/dt)
0.5 = -0.5 * π * (dh/dt)
Now, solve for dh/dt:
dh/dt = 0.5 / (-0.5 * π)
dh/dt = -1 / π
Therefore, the fluid will be rising at a rate of -1/π meters per minute when the depth is 5 meters. Note that the negative sign indicates that the fluid is decreasing in depth.