Addmaths Nov 13 P 23 Q 10(4037)

[musiclover gurl]

Can anyone explain to me how this number is worked? I don't know the technique to work such questions. Plzz. I've tried to work it out but in vain!

(ii) Find the rate at which h is increasing when h = 50 [3]
(iii) Find the rate at which the circular area of the water’s surface is increasing when h = 50 [4]

 

[Rony020298]

Can anyone explain to me how this number is worked? I don't know the technique to work such questions. Plzz. I've tried to work it out but in vain!
View attachment 46126
(ii) Find the rate at which h is increasing when h = 50 [3]
(iii) Find the rate at which the circular area of the water’s surface is increasing when h = 50 [4]

(i) If the Radius is equal to 30cm when the height is 120cm,
Then,
120cm=30cm
1cm = 30/120
h cm = 1/4* h = h/4 <------ represents the radius

Volume= 1/3 pi r^2 h
1/3 * pi * (h/4)^2 * h = (pi * h^3) / 48 <---- shown

(ii) differentiation of part 1, you obtain,
dV/dh= (pi*h^2)/16
when h= 50
dV/dh= 156.25*pi (ZZ)

accordingly, rate of change is gvien by
dh/dt = dh/dV * dV/dt

dh/dV = 1/ZZ

dV/dt = 20*pi (given in question)

dh/dt = (1/156.25 *pi) * (20*pi)

Ans : 0128 cm/s

(iii) Use dA/dt
use formula
dA/dt = dA/dh * dh/dt
Area is obtained by differentating again (50*pi/8)
same then applies.

 

[musiclover gurl]

(i) If the Radius is equal to 30cm when the height is 120cm,
Then,
120cm=30cm
1cm = 30/120
h cm = 1/4* h = h/4 <------ represents the radius

Volume= 1/3 pi r^2 h
1/3 * pi * (h/4)^2 * h = (pi * h^3) / 48 <---- shown

(ii) differentiation of part 1, you obtain,
dV/dh= (pi*h^2)/16
when h= 50
dV/dh= 156.25*pi (ZZ)

accordingly, rate of change is gvien by
dh/dt = dh/dV * dV/dt

dh/dV = 1/ZZ

dV/dt = 20*pi (given in question)

dh/dt = (1/156.25 *pi) * (20*pi)

Ans : 0128 cm/s

(iii) Use dA/dt
use formula
dA/dt = dA/dh * dh/dt
Area is obtained by differentating again (50*pi/8)
same then applies.

Thanks a lot!