A container is in the shape of a right circular cone (inverted) with both height and diameter 2 meters. It is being filled with water at a rate of (pi)m^3 per minute. Fine the rate of change of height h of water when the container is 1/8th full(by volume).
(Volume of a right circular cone of radius r and height h is 1/3(pi)r^3h)
Step by step working please.
Thanks and God bless :)
make a sketch to see that by ratios,
r = h/2 , like bobpursely noted
The sneaky part of the question is that when the cone is 1/8 full , the water is NOT 1/8 of the way up
Full volume = (1/3) π (1^2)(2) = 2π/3 m^3
we want r and h when volume = (1/8)(2π/3) or π/12
(Again, see bobpursely above)
Volume = V = (1/3)π(r^2)(h)
= (1/3)π(h^2/4)(h) = (π/12) h^3
so when cone is 1/8 full,
(π/12) h^3 = π/12
h^3 = 1
h = 1 , and r = 1/2
Now back to actual Calculus,
V = (π/12 h^3
dV/dh = (π/4) h^2 dh/dt
plug in the given dV/dt = π, and h = 1
π = (π/4) (1^2) dh/dt
dh/dt = 1/4
So when the cone is 1/8 full, the height is changing at
1/4 m/minute
We could have done the 1/8 part in our heads by realizing that ..
The volume of two similar solids is proportional to the cube of their sides, and since
(1/2)^3 = 1/8 ......
the height must have been 1/2 of the 2 m of the cone, or 1 m
for any height h, volume is 1/3 PI r^2 h
Notice your formula is wrong.
but for r for any height h is r=h/2. Think on that.
so volume=1/3 PI *(h/2)^2*h
V=1/12 * PI * h^3
and h= cuberoot (12V/PI) solve for h when V=1/8*1/12*PI*2 (max volume, r=1,h=2)
Now for the calculus work.
dV/dh=3/12 PI h^2
but dV/dh*dh/dt=dV/dt
or dh/dt=dV/dt / dV/dh
you area given dV/dt as PI m^3/minute
and you found dV/dh=3/12 PI h^2
so figure dh/dt
I understand the rest but not the 1/8th of the volume part. Can you give me a further explanation on that?
Thanks :)
Answer is dh/dt= m/min
The volume when it's 1/8th full is Pi/12
I don't understand how they got that....
To find the rate of change of the height of the water, we can use the concept of related rates.
Step 1: Determine the known and unknown variables
Known variables:
- Height of the container (h) = 2 meters
- Diameter of the container (d) = 2 meters
- Rate of change of volume (V') = π m³/minute
Unknown variables:
- Rate of change of height of water (h') when the container is 1/8th full
Step 2: Find the volume of water when the container is 1/8th full
To find the volume of water when the container is 1/8th full, we need to find the volume of the whole container and then multiply it by 1/8.
The volume of the whole container is given by the formula for the volume of a right circular cone:
V = (1/3)πr²h
Since the diameter of the container is equal to 2 meters, the radius (r) is equal to half the diameter:
r = d/2 = 2/2 = 1 meter
Substituting the values into the formula, we get:
V = (1/3)π(1²)(2) = (2/3)π m³
To find the volume of water when the container is 1/8th full, we multiply the volume of the whole container by 1/8:
V(1/8th full) = (1/8)(2/3)π = (1/12)π m³
Step 3: Apply the concept of related rates
To find the rate of change of the height of water (h'), we need to relate it to the rate of change of volume (V').
The volume of a right circular cone is given by the formula:
V = (1/3)πr²h
To differentiate both sides of the equation with respect to time (t), we get:
V' = (1/3)π[2r(r')h + r²(h')]
Since the radius (r) is constant and the rate of change of radius (r') is zero because the diameter is not changing, the equation simplifies to:
V' = (1/3)πr²(h')
Step 4: Substitute known values and solve for the unknown variable
We know that the rate of change of volume (V') is given as π m³/minute, and the radius (r) is 1 meter. Substituting these values into the equation, we get:
π = (1/3)π(1²)(h')
Simplifying the equation, we find:
1 = (1/3)(h')
To isolate the rate of change of the height of water (h'), we multiply both sides of the equation by 3:
3 = h'
Therefore, when the container is 1/8th full, the rate of change of the height of the water is 3 meters per minute.