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

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
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.

Jawab arr sendiri

Payah aa nak jawab soalan ni

Kelakar lah korang ni..

To find the rate at which the surface of the liquid is rising, we need to use the concept of related rates. We can break down the problem into two parts: the cylinder and the cone.

a) To find how fast the surface is rising at h = 1/2m, we need to focus on the cylinder since h = 1/2m is within the height of the cylinder.

First, let's calculate the volume of the cylinder. The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height.

Given that the radius of the tank is 1m and the height of the cylinder is 3m:
V_cylinder = π(1^2)(3) = 3π m³

Since the liquid is flowing into the tank at a rate of 1m³/4min, the rate of change of the volume is dV_cylinder/dt = 1/4 m³/min.

To find how fast the surface is rising at h = 1/2m, we need to find dh/dt when h = 1/2m. We can use the formula V = πr^2h to relate the volume and height:

3π = π(1^2)h
h = 3

Differentiating both sides with respect to time, we get:
dh/dt = (1/r^2)(dV_cylinder/dt)

Substituting the given values of r (1m) and dV_cylinder/dt (1/4 m³/min), we can calculate dh/dt:
dh/dt = (1/1^2)(1/4) = 1/4 m/min

Therefore, the surface is rising at a rate of 1/4 m/min when h = 1/2m.

b) To find how fast the surface is rising at h = 2.5m, we need to focus on the cone since h = 2.5m is within the height of the cone.

The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius and h is the height.

Given that the radius of the tank is 1m and the height of the cone is 2m:
V_cone = (1/3)π(1^2)(2) = 2/3π m³

Since the liquid is flowing into the tank at a rate of 1m³/4min, the rate of change of the volume is dV_cone/dt = 1/4 m³/min.

To find how fast the surface is rising at h = 2.5m, we need to find dh/dt when h = 2.5m. We can use the formula V = (1/3)πr^2h to relate the volume and height:

2/3π = (1/3)π(1^2)h
h = 2/3

Differentiating both sides with respect to time, we get:
dh/dt = (1/r^2)(dV_cone/dt)

Substituting the given values of r (1m) and dV_cone/dt (1/4 m³/min), we can calculate dh/dt:
dh/dt = (1/1^2)(1/4) = 1/4 m/min

Therefore, the surface is rising at a rate of 1/4 m/min when h = 2.5m.