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the volume of the spherical balloonis increasing at the rate of 25cm³/sec. find the rate of change of surface area at the instance when radius is 5 cm

$G i v e n, t h e v o l u m e o f t h e s p h e r i c a l b a l l o o n s i n c r e a \sin g a t t h e r a t e o f 25 {c m}^{3} / \min . S o, r a t e o f c h a n g e o f v o l u m e = \frac{ⅆ V}{ⅆ t} = 25 {c m}^{3} / \min ....... (1) W e k n o w t h a t, V o l u m e o f s p h e r e, V = \frac{4}{3} π r^{3} D i f f e r e n t i a t i n g w . r . t . t, w e h a v e, \frac{ⅆ V}{ⅆ t} = \frac{4}{3} π (3 r^{2} \frac{d r}{d t}) \Rightarrow \frac{ⅆ V}{ⅆ t} = 4 π r^{2} \frac{d r}{d t} N o w s u b s t i t u t e v a l u e o f \frac{ⅆ V}{ⅆ t} f r o m e q u a t i o n (1), w e o b t a i n, 25 = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{25}{4 π r^{2}} ...... (2) N o w, s u r f a c e a r e a o f s p h e r e, S = 4 π r^{2} D i f f e r e n t i a t i n g w . r . t t, w e o b t a i n, \frac{ⅆ S}{ⅆ t} = 4 π (2 r \frac{ⅆ r}{ⅆ t}) \Rightarrow \frac{ⅆ S}{ⅆ t} = 8 π r \frac{ⅆ r}{ⅆ t} s u b s t i t u t e v a l u e o f \frac{ⅆ r}{ⅆ t} f r o m e q u a t i o n (2), w e o b t a i n, \frac{ⅆ S}{ⅆ t} = 8 π r (\frac{25}{4 π r^{2}}) \Rightarrow \frac{ⅆ S}{ⅆ t} = \frac{50}{r} \Rightarrow {(\frac{ⅆ S}{ⅆ t})}_{r = 5} = \frac{50}{5} = 10 {c m}^{2} / \min H e n c e, r a t e o f c h a n g e o f s u r f a c e a r e a i s 10 {c m}^{2} / \min$

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