Slide26.JPG Slide27.JPG Slide28.JPG

You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.


Transcript

Ex 9.3, 20 In a bank, principal increases continuously at the rate of ๐‘Ÿ% per year. Find the value of r if Rs 100 double itself in 10 years (logโกใ€–2=0.6931ใ€— )Let Principal = p Given, principal increases ar rate r % per year โˆด ๐’…๐’‘/๐’…๐’• = ๐’“ % ร— P โˆด ๐‘‘๐‘/๐‘‘๐‘ก = ๐‘Ÿ/100 ร— p ๐’…๐’‘/๐’‘ = ๐’“/๐Ÿ๐ŸŽ๐ŸŽ dt Integrating both sides โˆซ1โ–’๐‘‘๐‘/๐‘ = ๐‘Ÿ/100 โˆซ1โ–’๐‘‘๐‘ก log p = ๐’“๐’•/๐Ÿ๐ŸŽ๐ŸŽ + log c log p โˆ’ log c = ๐‘Ÿ๐‘ก/100 log ๐‘/๐‘ = ๐‘Ÿ๐‘ก/100 ๐’‘/๐’„ = ๐’†^(๐’“๐’•/๐Ÿ๐ŸŽ๐ŸŽ) As we have put Rs 100 initially Putting t = 0 and p = 100 in (1) 100/๐‘ = ๐‘’^((๐‘Ÿ ร— 0)/100) 100/๐‘ = e^0 100/๐‘ = 1 c = 100 Putting value of C in equation (1) ๐’‘/(๐Ÿ๐ŸŽ๐ŸŽ ) = ๐’†^(๐’“๐’•/๐Ÿ๐ŸŽ๐ŸŽ) Also, given that Rs 100 will double itself in 10 years โˆด Putting t = 10, p = 200 in the equation 200/(100 ) = e^(10๐‘Ÿ/100) 2 = e^(๐‘Ÿ/10) log 2 = ๐‘Ÿ/10 0.6931 = ๐‘Ÿ/10 r = 6.931 โˆด Rate of interest = r = 6.931 %

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo