Learn instantaneous rate of change of volume of sphere with respect to radius, practice problem of surface area and volume, cylinder volume formula.
Volume = V = \(\dfrac {4}{3}\pi r^3\) where 'r' is the radius of sphere.
\(\dfrac {\dfrac {4}{3}\pi r_2^3-\dfrac {4}{3}\pi r_1^3}{r_2-r_1}\)
\(=\dfrac {4}{3}\pi\;\dfrac {(r_2^3- r_1^3)}{r_2-r_1}\)
A 54.12 cm2
B 117.23 cm2
C 5022 cm2
D 1.1 cm2
A \(2\,\pi\) cm2
B \(196\,\pi\) cm2
C \(78\,\pi\) cm2
D \(121\,\pi\) cm2
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
\(\therefore V=\dfrac {1}{3}\;\pi \left (\dfrac {h}{2}\right)^2h\)
\(\Rightarrow V=\dfrac {1}{3}\;\pi \left (\dfrac {h^3}{4}\right)=\dfrac {\pi\,h^3}{12}\)
\(\therefore \dfrac {dV}{dt}=\dfrac {3\,\pi\,h^2}{12}\;\dfrac {dh}{dt}\)
\(\Rightarrow\; \dfrac {dV}{dt}=\dfrac {\,\pi\,h^2}{4}\;\dfrac {dh}{dt}\)
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 \(4\,\pi\,cm/min\)
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
\(\Rightarrow A=\ell\;b\)
\(\dfrac {dA}{dt}=\) 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 \(191\, cm^2/min.\)