Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 19, 2021 by Teachoo

Transcript

Ex 6.1, 9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.Since Balloon is spherical
Let r be the radius of balloon .
& V be the volume of balloon.
We need to find rate at which balloon volume is increasing when radius is 10cm
i.e. We need to find change of volume w.r.t radius when r = 10
i.e. we need to find 𝒅𝑽/𝒅𝒓 when r = 10 cm
We know that
Volume of sphere = V = 4/3 πr3
Now,
𝑑𝑉/𝑑𝑟 = (𝑑 (4/3 𝜋𝑟3))/𝑑𝑟
𝑑𝑉/𝑑𝑟 = 4/3 π 𝑑(𝑟3)/𝑑𝑟
𝑑𝑉/𝑑𝑟 = 4/3 π 3𝑟^2
𝑑𝑉/𝑑𝑟 = 4𝜋𝑟^2
When r = 10
𝑑𝑉/𝑑𝑟 = 4 × π × (10)2
𝑑𝑉/𝑑𝑟 = 400π
Since volume is in cm3 & Radius is in cm
So, 𝑑𝑉/𝑑𝑟 = 400π cm3/cm
Hence, volume is increasing at the rate of 400 π cm3/cm when r = 10 cm

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!