which method?

[sdracula]

below jpg

 

[sdracula]

for this, cross section or disk method or washer method? when should we use it (difference)?

 

[Deleted member 4993]

for this, cross section or disk method or washer method? when should we use it (difference)?

As far as I know - there are no hard-and-fast rule. You can use either in most of the problems I have met.

It is useful to look at the "physics" of the problem - to help you choose the method. In the given problem, the level is rising disk by disk. So disk method is probably most expedient!

 

[Jagulba]

I don't know what you mean by washer method but I went to disk method naturally.

 

[MarkFL]

I would first find a function expressing the radius of the mug as a function of the height above the bottom:

[MATH]r(h)=\frac{1}{20}h+2=\frac{h+40}{20}[/MATH]
Now, considering the volume of the mug as a stack of disks, we may write:

[MATH]dV=\pi r^2\,dh=\pi\left(\frac{h+40}{20}\right)^2\,dh=\frac{\pi}{400}(h+40)^2\,dh[/MATH]
Or equivalently:

[MATH]\frac{dV}{dt}=\frac{\pi}{400}(h+40)^2\,\frac{dh}{dt}[/MATH]
Hence:

[MATH]\frac{dh}{dt}=\frac{400}{\pi(h+40)^2}\,\frac{dV}{dt}[/MATH]
Can you proceed?

 

[sdracula]

problem: (how fast...)
I think dh/dt is required.
dh/dt = dh/dV * dV/dt
we replace the given value - 50

I'm trying to understand the difference between part a) and part b).
Is it possible ''surface area of revolution''

and how we achieved the formula of frustum height from similarity.

 

[Jagulba]

I'm trying to understand the difference between part a) and part b).
Is it possible ''surface area of revolution''

the part "a" is at height 10cm.
the part "b" is at height x where the volume is half.... you see sense the top part has bigger radius it will contain more volume. so if you stay in the middle of mug you will have bigger volume on top compare to the bottom. so the height when we have two equal volume its closer to the surface.