Application of Derivatives

The surface area of sphere is S=4πr2,r is radius
Now, differentiate w.r.t x, we get
‌

ds

dt

=4π⋅2r⋅‌

dr

dt

=8πr‌

dr

dt

Also, given that ‌

ds

dt

=4cm2∕sec and r=8cm
‌∴‌‌‌

ds

dt

=πr⋅‌

dr

dt

⇒4=8π(8)⋅‌

dr

dt

‌⇒‌‌‌

dr

dt

=‌

1

16π

cm∕sec
Volume of Sphere is : v=‌

4

3

πr3
Differentiate both sides w.r.t r, we get
‌

dv

dt

‌=‌

4

3

π⋅3r2⋅‌

dr

dt

=4π⋅(8)2⋅‌

1

16π

‌ [using Eq. ‌
‌=16cm3∕sec
∴ Rate of change in the volume of the bubble =16cm3∕sec