Learn instantaneous rate of change of volume of sphere with respect to radius, practice problem of surface area and volume, cylinder volume formula.

- Volume of sphere = \(\dfrac {4}{3}\pi r^3\) where 'r' is the radius of sphere.
- Surface area of sphere = \(4\,\pi\, r^2\), where 'r' is the radius of sphere.
- All balloons and bubbles are of spherical shape.

- For a sphere

Volume = V = \(\dfrac {4}{3}\pi r^3\) where 'r' is the radius of sphere.

- The average rate of change of volume of sphere when radius changes from r
_{1}to r_{2}is

\(\dfrac {\dfrac {4}{3}\pi r_2^3-\dfrac {4}{3}\pi r_1^3}{r_2-r_1}\)

\(=\dfrac {4}{3}\pi\;\dfrac {(r_2^3- r_1^3)}{r_2-r_1}\)

A 54.12 cm2

B 117.23 cm2

C 5022 cm2

D 1.1 cm2

- \(V=\dfrac {4}{3}\;\pi\,r^3\;\)

- \(\Rightarrow\dfrac {dV}{dr}=4\,\pi\,r^2\)
- Observe that instantaneous rate of change of volume w.r.t. r is equal to the surface area of the sphere.

A \(2\,\pi\) cm2

B \(196\,\pi\) cm2

C \(78\,\pi\) cm2

D \(121\,\pi\) cm2

V = volume of cube = \(\ell^3\), where '\(\ell\)' is length of its sides.

\(\dfrac {dV}{d\ell}=\)rate of change of volume w.r.t. side length = \(\dfrac {d}{d\ell}(\,\ell^3)=3\,\ell^2\)

A 75 cm2

B 15 cm2

C 1.3 cm2

D 550 cm2

Area of rectangle = \(\ell\;×b\)

where \(\ell\;=\) length, \(b=\) breadth

\(\Rightarrow A=\ell\;b\)

- If the rate of change of length and breadth of rectangle are known, then we can find the rate of change of area at some instance when the length and breadth of rectangle is known.

\(\dfrac {dA}{dt}=\) Rate of change of area.

\(\dfrac {db}{dt}=\) Rate of change of breadth.

\(\dfrac {d\ell}{dt}=\) Rate of change of length.

\(\Rightarrow\dfrac {dA}{dt}=\ell\dfrac {db}{dt}+b\dfrac {d\ell}{dt}\) (Product Rule)

A \(2\, cm^2/min.\)

B \(13\, cm^2/min.\)

C \(-5\, cm^2/min.\)

D \(191\, cm^2/min.\)

Consider the volume of cylinder V, whose radius is 'r' and height 'h'

\(V = \pi\,r^2\,h\)

If we fill some liquid in the tank, then

Rate of change of volume \(= \dfrac {dV}{dt}=\pi\,r^2\,\dfrac {dh}{dt}\) (Because 'r' is constant and 'h' increases as V increases)

where, \(\dfrac {dh}{dt}\) is the rate of change or increase of height of liquid in the tank.

A \(\dfrac {36}{\pi}\,cm/min\)

B \(\dfrac {1}{18\pi}\,cm/min\)

C \(18\pi\,cm/min\)

D \(2\pi\,cm/min\)

Consider sand being poured on the ground such that it always has a conical shape.

Volume of Cone = \(V=\dfrac {1}{3}\;\pi r^2h\)

where, V = Volume of cone,

r= radius of base,

h = height of cone

- If 'r' and 'h' are related in some way, we can use that relation and then differentiate.
- This will give a relation between \(\dfrac {dV}{dt}\) and \(\dfrac {dh}{dt}\) i.e. the rate of change of volume (pouring of cone) in term of rate of change of height of the pile that forms on the ground.
- Let \(V=\dfrac {1}{3}\;\pi r^2h\) such that \(r=\dfrac {h}{2}\) (always)

\(\therefore V=\dfrac {1}{3}\;\pi \left (\dfrac {h}{2}\right)^2h\)

\(\Rightarrow V=\dfrac {1}{3}\;\pi \left (\dfrac {h^3}{4}\right)=\dfrac {\pi\,h^3}{12}\)

\(\therefore \dfrac {dV}{dt}=\dfrac {3\,\pi\,h^2}{12}\;\dfrac {dh}{dt}\)

\(\Rightarrow\; \dfrac {dV}{dt}=\dfrac {\,\pi\,h^2}{4}\;\dfrac {dh}{dt}\)

Steps:

(1) \(V=\dfrac {1}{3}\;\pi r^2h\) where, V = Volume of cone, r = its radius, h = height)

(2) Use the relation between 'r' and 'h' (say \(r=g\,(h)\)), then

\(V=\dfrac {1}{3}\pi\, (g(h))^2h\)

(3) Differentiate both sides with respect to 't', this will relate \(\dfrac {dV}{dt}\) and \(\dfrac {dh}{dt}\)

A \(\dfrac {1}{2\,\pi}\,cm/min\)

B \(\dfrac {3}{\pi}\,cm/min\)

C \(2\,\pi\,cm/min\)

D \(4\,\pi\,cm/min\)