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Rate Of Change Of Volume And Surface

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

Rate of change of Volume & Surface Area of Sphere

  • 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 r1 to r2 is

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

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

Illustration Questions

Find the average rate of change of volume of a spherical ball with respect to radius (r) where r changes from 2 cm to 4 cm.

A 54.12 cm2

B 117.23 cm2

C 5022 cm2

D 1.1 cm2

×

Average rate of change of \(V=\dfrac {V(r_2)-V(r_1)} {r_2-r_1}\)

 

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

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

\(=\dfrac {4}{3}\pi\;(r_1^2+r_2^2+r_1r_2)\)

\(=\dfrac {4}{3}\pi\;(2^2+4^2+2×4)\)  (r1 = 2, r2 = 4)

\(=\dfrac {4}{3}\pi\;×28\) cm2

\(=\dfrac {112}{3}\pi\) cm2

= 117.23  cm2

Find the average rate of change of volume of a spherical ball with respect to radius (r) where r changes from 2 cm to 4 cm.

A

54.12 cm2

.

B

117.23 cm2

C

5022 cm2

D

1.1 cm2

Option B is Correct

Instantaneous Rate of Change of Volume of Sphere with Respect to Radius

  • \(V=\dfrac {4}{3}\;\pi\,r^3\;\)

 

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

Illustration Questions

A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with respect to its radius when radius is 7 cm.

A \(2\,\pi\) cm2

B \(196\,\pi\) cm2

C \(78\,\pi\) cm2

D \(121\,\pi\) cm2

×

\(V=\dfrac {4}{3}\;\pi\,r^3=\) Volume of a sphere of radius r.

\(\dfrac {dV}{dr}=\) rate of change of volume w.r.t. radius \(4\;\pi\,r^2\)

\(\dfrac {dV}{dr}\Bigg|_{r=7}=4\,\pi×7^2\)

 = \(4\,\pi×\)49

= 196\(\pi\) cm2

A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with respect to its radius when radius is 7 cm.

A

\(2\,\pi\) cm2

.

B

\(196\,\pi\) cm2

C

\(78\,\pi\) cm2

D

\(121\,\pi\) cm2

Option B is Correct

Instantaneous Rate of Change of Volume of Cube with respect to its Side Length

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 = \(\dfrac {d}{d\ell}(\,\ell^3)=3\,\ell^2\)

Illustration Questions

A cube has a variable edge length. Find the rate at which its volume is increasing with respect to its edge length when edge length is 5 cm.

A 75 cm2

B 15 cm2

C 1.3 cm2

D 550 cm2

×

\(V=\ell^3\) (Volume of cube whose edge length is \(\ell\))

\(\dfrac {dV}{d\ell}=3\,\ell^2=\)rate of change of volume with respect to edge length

\(\dfrac {dV}{d\ell}\Bigg|_{\ell=5}\) = 3 × 52

= 75 cm2

A cube has a variable edge length. Find the rate at which its volume is increasing with respect to its edge length when edge length is 5 cm.

A

75 cm2

.

B

15 cm2

C

1.3 cm2

D

550 cm2

Option A is Correct

Rectangle Problem

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)

Illustration Questions

The length \('\ell\,'\) of a rectangle is decreasing at the rate of 3 cm/min and the width \('w'\) is increasing at the rate of 2 cm/min. Find the rate of change of area of rectangle when \(\ell=10\,cm\) and \(w=6\,cm\).

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

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

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

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

×

Area = length × width 

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

 

 

Differentiate both sides w.r.t. 't'

\(\Rightarrow\dfrac {dA}{dt}=\ell\dfrac {dw}{dt}+w\dfrac {d\ell}{dt}\)    (Product Rule)

Put \(\ell=10\,cm\)\(w=6\,cm\)\(\dfrac {d\ell}{dt}=-3\,cm/min.\), \(\dfrac {dw}{dt}=2\,cm/min.\) 

\(\Rightarrow\dfrac {dA}{dt}=10×2+6×(-3)\)

\(=20-18\)

\(=2\,cm^2/min\)

The length \('\ell\,'\) of a rectangle is decreasing at the rate of 3 cm/min and the width \('w'\) is increasing at the rate of 2 cm/min. Find the rate of change of area of rectangle when \(\ell=10\,cm\) and \(w=6\,cm\).

A

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

.

B

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

C

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

D

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

Option A is Correct

                 The Cylinder Problem

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.

 

Illustration Questions

A cylindrical tank of radius 6 cm, is filled with water at the rate of 2 cm3/min. How fast is its height increasing?

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

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

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

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

×

Volume of cylinder =  \(V = \pi\,r^2\,h\)

 

Differentiate both sides w.r.t. 't' 

\(\Rightarrow \dfrac {dV}{dt}=\pi\,r^2\,\dfrac {dh}{dt}\)

Here,

\( r = 6\, cm\)\(\dfrac {dV}{dt}=2\,cm^3/min\)

\(\therefore 2=\pi×36×\dfrac {dh}{dt}\)

\(\dfrac {dh}{dt}=\dfrac {1}{18\,\pi}\,cm/min\)

A cylindrical tank of radius 6 cm, is filled with water at the rate of 2 cm3/min. How fast is its height increasing?

A

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

.

B

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

C

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

D

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

Option B is Correct

The Cone Problem

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 \(\dfrac {dh}{dt}\)

Illustration Questions

Sand  is being poured onto a conical pile at constant rate of \(50\, cm^3/min\) such that height of cone is always one half of radius of its base. How fast is the height of pile increasing when sand is \(5\, cm\) deep?

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

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

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

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

×

\(V=\dfrac {1}{3}\;\pi r^2h\)   for a cone

 

 

Given, that 

\(h=\dfrac {1}{2}\;r\)

\(\Rightarrow V=\dfrac {1}{3}\;\pi\,(2h)^2×h\)

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

Differentiate both sides w.r.t. 't'

\(\Rightarrow \dfrac {dV}{dt}=\dfrac {4}{3}\;\pi\,3h^2\,\dfrac {dh}{dt}\)

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

Here, \(h=5,\;\Rightarrow \dfrac {dV}{dt}=50\,cm^3/min\)

\(\therefore 50=4\,\pi×5×5×\dfrac {dh}{dt}\)

\(\Rightarrow \dfrac {dh}{dt}=\dfrac {1}{2\,\pi}\,cm/min\)

Sand  is being poured onto a conical pile at constant rate of \(50\, cm^3/min\) such that height of cone is always one half of radius of its base. How fast is the height of pile increasing when sand is \(5\, cm\) deep?

A

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

.

B

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

C

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

D

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

Option A is Correct

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