Question

A vessel is in the shape of a cone. The radius of the top
is 8 cm and the height is 40 cm. Water is poured into the
vessel at a rate of 20 cm/s.
Calculate the rate at which the water level is rising when
(i) the water is 12 cm from the vertex,
(ii) the vessel is 1/4 full​

Answer 1

Answer:Let r be radius of cross section at height h.

h=9cm and r=5cm....given

⇒  

5

r

​  

=  

9

h

​  

⇒r=  

9

5h

​  

 

Given,  

dt

dh

​  

=  

A

π

​  

 

⇒  

dt

dh

​  

=  

πr  

2

 

π

​  

 

⇒  

dt

dh

​  

=  

25h  

2

 

81

​  

 

⇒  

81

25

​  

∫  

0

9

​  

h  

2

dh=∫  

0

9

​  

 

3

h  

3

 

​  

dt

⇒  

81

25

​  

×  

3

729

​  

=t

⇒t=75 seconds.

Step-by-step explanation:

Answer 2

Given:

Radius of the cone $=8cm$

Height of the cone $=40cm$

Water filling rate $=20cm/s$

To find:

Rate at which the water level is rising

Solution:

Step 1

We have been given that in the given cone of radius $R=8cm$ and height $H=40cm$ is being filled at the rate of $20cm/s$.

Hence,      $\frac{dH}{dt}=20cm/s$

We can see that the exact value of height is 5 times the radius of the cone. Hence, we can write the relation between height and radius as

$H=5R$  $;$ $or$

$R=\frac{H}{5}$

Now, since the water is filling at the rate of $20cm/s$ and we know the volume of cone is given by,

$V=\frac{1}{3} \pi R^{2} H$

Hence, the rate at which the volume of the cone is increasing is

$\frac{dV}{dt}=\frac{d(\pi R^{2} H)}{3dt}$

Substituting $R=\frac{H}{5}$

$\frac{dV}{dt} = \frac{d\pi( H)^{2} H)}{3(25)dt}$

$\frac{dV}{dt}=\frac{\pi }{75} \frac{dH^{3} }{dt}$

$\frac{dV}{dt}=\frac{\pi }{75} (3H^{2} )\frac{dH}{dt}$      

Step 2

Now, to find the rate when the water is 12 cm from the vertex, we need to

Substitute the values of $H=12cm$ and $\frac{dH}{dt}=20cm/s$ , we get

$\frac{dV}{dt} =\frac{\pi }{75} 3(12)^{2} (20)$

$\frac{dV}{dt}=\frac{22*3*12*12*20}{7*75}$

$\frac{dV}{dt}=361.8cm^{3} /s$

Hence, at the rate of 361.8 cm³/s, the water level is rising when the water level is 12 cm from the vertex.

Step 3

To find the rate when the vessel is $\frac{1}{4}$ full.

We need to find the volume of vessel when it is $\frac{1}{4}th$ full. Hence, the volume

$V'=\frac{1}{4}$ × $\frac{1}{3}\pi R^{2} H$

Rate can be written as

$\frac{dV'}{dt}=\frac{d}{dt}( \frac{1}{12}\pi R^{2} H)$

$\frac{dV'}{dt}=\frac{d}{dt} (\frac{1}{12}\pi\frac{ H^{3} }{25})$

$\frac{dV'}{dt}=\frac{\pi }{300} ( 3H^{2}) \frac{dH}{dt}$

Substituting $H=40cm$ and $\frac{dH}{dt}=20cm/s$

$\frac{dV'}{dt}=\frac{3.14}{300} 3(40)^{2} (20)$

$\frac{dV'}{dt}=1004.8cm^{3} /s$

Final answer:

Hence,

1. The water level in the beaker is rising at the rate of 361.8 cm³/s when the water is 12 cm from the vertex.
2. The water level in the beaker is rising at the rate of 1004 cm³/s when the vessel is one-fourth full.