Challenging Problem

[maxhk]

[Question] "(a) Water is being poured into a hemispherical bowl of radius 3 inch at the rate of 1 inch^3/s. How fast is the water level rising when the water is 1 inch deep ? (b) In (a), suppose that the bowl contains a lead ball 2 inch in diameter, and find how fast the water level is rising when the ball is half submerged." [Difficulty] Part (b) I can't see what to do Thanks for your help [Thoughts] solved Part (a) (a) V = (pi/3)* h^2*(3R-h) (volume of a segment of height h of a sphere) => dV/dt = pi*h*(2R-h)*dh/dt => dh/dt = 1/(pi*h*(2R-h))*dV/dt R = 3 inch ; h = 1 inch ; dV/dt = 1 inch^3/s so dh/dt = 1/(5*pi) inch/s

(b) ????

 

[Deleted member 4993]

[Question] "(a) Water is being poured into a hemispherical bowl of radius 3 inch at the rate of 1 inch^3/s. How fast is the water level rising when the water is 1 inch deep ? (b) In (a), suppose that the bowl contains a lead ball 2 inch in diameter, and find how fast the water level is rising when the ball is half submerged." [Difficulty] Part (b) I can't see what to do Thanks for your help [Thoughts] solved Part (a) (a) V = (pi/3)* h^2*(3R-h) (volume of a segment of height h of a sphere) => dV/dt = pi*h*(2R-h)*dh/dt => dh/dt = 1/(pi*h*(2R-h))*dV/dt R = 3 inch ; h = 1 inch ; dV/dt = 1 inch^3/s so dh/dt = 1/(5*pi) inch/s

(b) ????

If you can do the first part - 2nd part should not be too difficult.

Just subtract the volume of the "submerged" ball from your previous volume expression (at any height h ≤ 1")

 

[maxhk]

If you can do the first part - 2nd part should not be too difficult.

Just subtract the volume of the "submerged" ball from your previous volume expression (at any height h ≤ 1")

V = (pi/3)* h^2*(3R-h) - (2pi/3) (2pi/3 volume of half sphere of radius 1)

dV/dt = pi*h*(2R-h)*dh/dt

If I do that rate is still the same!
is this ok ?

 

[Deleted member 4993]

V = (pi/3)* h^2*(3R-h) - (2pi/3)

As you are pouring water the lead ball is getting submerged gradually.

You need the volume of the "part of the sphere" that is submerged in "h" depth of water (not the total hemisphere).

(2pi/3 volume of half sphere of radius 1)

dV/dt = pi*h*(2R-h)*dh/dt

If I do that rate is still the same!
is this ok ?

No - see the explanation above

 

[maxhk]

No - see the explanation above

I actually don't understand the problem (part b); I am unable to see what is going on physically.
Can you draw a picture for me ? or explain things better ?

the ball submerged will cover only part of this "h" ?

Thanks a lot.

 

[galactus]

I think what SK is trying to say is do the same thing for the ball as you did for the hemisphere, then subtract the two formulas and differentiate implicitly.

The water will rise a little faster with the ball in there.

 

[burakaltr]

[Question] "(a) Water is being poured into a hemispherical bowl of radius 3 inch at the rate of 1 inch^3/s. How fast is the water level rising when the water is 1 inch deep ? (b) In (a), suppose that the bowl contains a lead ball 2 inch in diameter, and find how fast the water level is rising when the ball is half submerged." [Difficulty] Part (b) I can't see what to do Thanks for your help [Thoughts] solved Part (a) (a) V = (pi/3)* h^2*(3R-h) (volume of a segment of height h of a sphere) => dV/dt = pi*h*(2R-h)*dh/dt => dh/dt = 1/(pi*h*(2R-h))*dV/dt R = 3 inch ; h = 1 inch ; dV/dt = 1 inch^3/s so dh/dt = 1/(5*pi) inch/s

(b) ????

If you draw 2 intersecting lines inside a circle you will have the following relation

Say AB intersects CD in some point O

AO * OB = DO * OC

SO

(x-h)(x+h)=2Rh-h**2


x=sqrt(2R)*(sqrt(h))

dx=sqrt(2R)*0.5 h**(-0.5) dh WHERE dh=2*dx((2R)**-0.5) sqrt(h)

The differential Volume generated is

V=Pi*(x**2)dh

V=Pi*(x**2)dh but we know x**2=2Rh

R=given of the hemisphere

V=integral[ ( 2*3*Pi*h)dh from 0 to some height h(t)

6(h**2)Pi/2 = 3 Pi h**2

v=3Pi*h**2=1t

5Pih(dh/dt)=1


at h=1 unit deep

5Pi ( differential) = 1

dh/dt=0.2(1/Pi) = 0.0636...

 

[burakaltr]

For beginners, PART B also requires

you have a rotated strip around y-axis

where the formula for the Volume generated is always

Pi(x**2)dy=dV

where x**2 + y**2 = 1

 

[burakaltr]

For beginners, PART B also requires

you have a rotated strip around y-axis

where the formula for the Volume generated is always

Pi(x**2)dy=dV

where x**2 + y**2 = 1

answer oughta be 1/(4Pi)


according to cord multiplications

If I am not mistaken

the result is About


( 0.0796) m/s

 

[burakaltr]

answer oughta be 1/(4Pi)


according to cord multiplications

If I am not mistaken

the result is About


( 0.0796) m/s

A greater value than the one found in Part A ( 1/5Pi ) is a Harbinger that 1/(4Pi) might be the answer.

When I am sure , I'll post it here

 

[burakaltr]

It is the matter of putting the smaller sphere on where the Big sphere touches the horizontal

Then, calling u, the distance from the vertical ortogonal to the edge of the big sphere

and

x, from the orth. to the edge of small sphere

then

Relations are found

u**2 =(1+y)(5-y)

and

x**2=(1+y)(1-y)

y is the height

when rotated the strips :


integral[Pi(u**2 - x**2 )] dy=1*dt=dt

substituting yields

Pi(4y+4)dy=dt


dy/dt when the problem starts that is y=0

gives

dy/dt=1/[4Pi(y+1)]

plug y=0

to get

Change in Height over time =

1/(4Pi)