At any instant t, let
r be the radius,
V be the volume and
S be the surface area of the balloon.
Given:
dS∕dt=2cm2∕sec As we know that, surface area of sphere is given by:
4πr2 where
r is the radius of the sphere. i.e
S=4πr2 Now by differentiating S with respect to t we get,
⇒=.------(By chain rule)
⇒==8πr By substituting the value of dS/dr in dS/dt we get
⇒=8πr. By substituting dS/dt = cm
2/sec in the above equation we get
⇒2=8πr.⇒= As we know that volume of sphere is given by:
πr3 where
r is the radius of the sphere.
i.e
V=πr3 Now differentiating V with respect to t we get
⇒=. -----(By chain Rule)
As,
V=πr3 ⇒=(πr3)=π.3r2=4πr2 By substituting dV/dr in dV/dt we get
⇒=4πr2. Now substitute dr/dt = 1/4πr in the above equation we get
⇒=4πr2.=r By substituting r = 6 cm in the above equation we get,
⇒[]r=6=6cm3∕sec