I do not see your geometry arrangement.
However in general
change in volume = surface area of liquid * change in surface height
dV = Area * dh
dV = pi r^2 dh
so
dV/dt = pi r^2 dh/dt
(1/4) = pi r^2 dh/dt
so
dh/dt= 1/( 4 pi r^2)
You will have to find r from your geometry at those two values of h
Liquid is flowing into a tank consist of vertical circular cylinder and vertical circular cone at the rate of 1m³/4min.The height of the cylinder is 3m while the height of cone is 2m.If the radius of the tank is 1m,
a)how fast is the surface rising at h=1/2m
b)how fast is the surface rising at 2.5m
6 answers
Liquid is flowing into a tank consist of vertical circular cylinder and vertical circular cone at the rate of 1m³/4min.The height of the cylinder is 3m while the height of cone is 2m.If the radius of the tank is 1m,
a)how fast is the surface rising at h=1/2m
b)how fast is the surface rising at 2.5m
Math - Damon, Friday, September 2, 2016 at 10:01am
I do not see your geometry arrangement.
However in general
change in volume = surface area of liquid * change in surface height
dV = Area * dh
dV = pi r^2 dh
so
dV/dt = pi r^2 dh/dt
(1/4) = pi r^2 dh/dt
so
dh/dt= 1/( 4 pi r^2)
You will have to find r from your geometry at those two values of h
-Mr Damon you said that to find r from my geometry at those two values-means that i have to substitute the value of h in 1/(4 pi r^2)..is it to be like that.Sorry.I'm a little bit not understand about that.Please answer my question
a)how fast is the surface rising at h=1/2m
b)how fast is the surface rising at 2.5m
Math - Damon, Friday, September 2, 2016 at 10:01am
I do not see your geometry arrangement.
However in general
change in volume = surface area of liquid * change in surface height
dV = Area * dh
dV = pi r^2 dh
so
dV/dt = pi r^2 dh/dt
(1/4) = pi r^2 dh/dt
so
dh/dt= 1/( 4 pi r^2)
You will have to find r from your geometry at those two values of h
-Mr Damon you said that to find r from my geometry at those two values-means that i have to substitute the value of h in 1/(4 pi r^2)..is it to be like that.Sorry.I'm a little bit not understand about that.Please answer my question
Sketch the graph of a continuous function f that satisfies all the stated conditions.
f(0)=0; f(2)=-2 ; f(4)=f(10)=-1;f(6)=0 f’(2)=f'(6)=0
f'(x)<0 throughout (-infinity,2) and (6,+infinity)
f'(x)>0 throughout 2<x<6
f''(x)<0 throughout (4,+infinity)
f''(x)>0 throughout (-infinity,4)
Hence,state the maximum,minimum and inflection point(s),if any.
f(0)=0; f(2)=-2 ; f(4)=f(10)=-1;f(6)=0 f’(2)=f'(6)=0
f'(x)<0 throughout (-infinity,2) and (6,+infinity)
f'(x)>0 throughout 2<x<6
f''(x)<0 throughout (4,+infinity)
f''(x)>0 throughout (-infinity,4)
Hence,state the maximum,minimum and inflection point(s),if any.
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