The volume of a cube is increasing at the rate of 8 cm^{3}/S. How fast is the
surface area increasing when the length of its edge is 12 cm?

 

 

 

 
 
 
 
 

Answers (1)

Let x be the strength of side, v be the volume & s be the surface area of cube
Then v= x^{3}\: and\: s= 6x^{2},where x is a function of
line t
Now \frac{dv}{dt}= 8\frac{cm^{3}}{s} \left [ given \right ]
\therefore 8= \frac{dv}{dt}= \frac{d}{dx}\left ( x^{3} \right )= 3x^{2}\frac{dx}{dt}\left ( chain \: rule \right )
8= 3x^{2}\frac{dx}{dt}
\frac{8}{3x^{2}}= \frac{dx}{dt}-(i)
Now, \frac{ds}{dt}= 12x\frac{8}{3x^{2}}\left [ using(i) \right ]\Rightarrow \frac{32}{x}
Hence when x= 12cm
then, \frac{ds}{dt}= \frac{32}{12}= \frac{8}{3}\: cm^{2}/S
 

Most Viewed Questions

Related Chapters

Preparation Products

Knockout CUET (Physics, Chemistry and Mathematics)

Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout CUET (Physics, Chemistry and Biology)

Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan.

₹ 7999/- ₹ 4999/-
Buy Now
Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Knockout NEET (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions