# Ex 6.5,26

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan –1 13 Let 𝑟 , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & θ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = π 𝑟2+𝜋𝑟𝑙 S = π 𝑟2+𝜋𝑟𝑙 S – π 𝑟2=𝜋𝑟𝑙 𝑆 − 𝜋 𝑟2 𝜋𝑟=𝑙 𝑙 = 𝑆 − 𝜋 𝑟2 𝜋𝑟 We need to find minimize volume of a cone & show that semi vertical angle is sin −13 i.e. θ =𝑠𝑖𝑛 −13 sin θ = 13 We know that sin θ = 𝑟𝑙 Volume of a cone = 13𝜋 𝑟2ℎ V = 13𝜋 𝑟2 𝑙2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟2𝜋𝑟2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 𝜋2 𝑟2− 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 − 𝜋 𝑟2 𝑟2 𝜋2 𝑟2 V = 13𝜋 𝑟2 𝑠 − 𝜋 𝑟22 − 𝜋2 𝑟4 𝜋2 𝑟2𝑟 V = 𝜋 𝑟23𝜋𝑟 𝑠−𝜋 𝑟22− 𝜋2 𝑟4 V = 𝑟3 𝑠2+ 𝜋 𝑟22−2𝑆 𝜋𝑟2− 𝜋2 𝑟4 V = 𝑟3 𝑠2+ 𝜋2 𝑟4−2𝑆𝜋 𝑟2− 𝜋2 𝑟4 V = 𝑟3 𝑠2−2 𝑆𝜋 𝑟2 V = 13 𝑟2 𝑠2−2 𝑠 𝜋 𝑟2 V = 13 𝑟2 𝑠2−2 𝑠𝜋 𝑟4 Diff w.r.t 𝑟 𝑑𝑉𝑑𝑟= 13 𝑑𝑑𝑟 𝑠2 𝑟2−2 𝑠𝜋 𝑟2 𝑑𝑣𝑑𝑟= 13× 12 𝑠2 𝑟2−2 𝑠𝜋 𝑟2× 2 𝑠2𝑟−2𝑠𝜋 𝑟3 𝑑𝑣𝑑𝑟= 1 𝑠2𝑟 − 4𝑠𝜋 𝑟33 𝑠2 𝑟2 − 2 𝑠𝜋 𝑟2 Putting 𝑑𝑣𝑑𝑟=0 13 𝑠2 𝑟2 − 2𝑠𝜋 𝑟4× 𝑠2𝑟−4𝜋𝑠 𝑟3=0 Hence, S = 4𝜋 𝑟2 Now, Surface area of cone = 𝜋 𝑟2+𝜋𝑟𝑙 𝑆=𝜋 𝑟2+𝜋𝑟𝑙 Putting S = 4𝜋 𝑟2 4𝜋 𝑟2=𝜋 𝑟2+𝜋𝑟𝑙 𝜋 𝑟2+𝜋𝑟𝑙=4𝜋 𝑟2 Dividing both sides by 𝜋𝑟 𝜋 𝑟2+ 𝜋𝑟𝑙𝜋𝑟= 4𝜋 𝑟2𝜋𝑟 𝑟+𝑙=4𝑟 𝑙=4𝑟−𝑟 𝑙=3𝑟 𝑙𝑟=3 𝑟𝑙= 13 But we know that sin θ = 𝑟𝑙 Putting value of 𝑟𝑙 sin θ = 13 θ = 𝒔𝒊𝒏−𝟏 𝟏𝟑 Hence proved

Ex 6.3,5
Important

Ex 6.3,7 Important

Ex 6.3,12 Important

Ex 6.3,15 Important

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

Ex 6.5,18 Important

Ex 6.5,20 Important

Ex 6.5,23 Important

Ex 6.5,26 Important You are here

Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

Misc 11 Important

Misc 13 Important

Misc 17 Important

Misc 22 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.