## Vineet Loomba is teaching live on Unacademy Plus

IIT-JEE VIDEO COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) IIT-JEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR Search vineet loomba unacademy" on GOOGLE

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!

APPLICATION OF DERIVATIVES APPROXIMATIONS IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) APPROXIMATIONS Q (r +Ar, y +Ay) S (x +dx, y + dy) dy dy Aj dy dyAy Y' MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Use differential to approximate 36.6 Solution Take y=Vx. Let x = 36 and let Ar = 0.6. Then or 36.6 6+Ay Now dy is approximately equal to Ay and is given by dh dx )-2.06 (0.6) = 0.05 (as y=JX) 2Vx Thus, the approximate value of V36.6 is 6 + 0.05 = 6.05. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Use differential to approximate (25) Solution Let y =x, Let x = 27 and let r =-2. Then or (25). 3 + y Now dy is approximately equal to Ay and is given by dy dx 3x3 -2 27 (-2) =-=-0.074 3((27)3)2 Thus, the approximate value of (25)3 is given by 34 (-0.074) = 2.926 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Exampie Find the approximate value of f(3.02), where f(x) 3x 5x +3. Solution Let x = 3 and Ar= 0.02. Then Note that Ay =f(x + )-f(x). Therefore f(x+Ax) =f(x) + y f (x) f (x) Ar f (3.02) (3x25x 3)(6x 5) Ar (as d-Ar) or = (3(3)2 + 5(3) + 3) + (6(3) + 5) (0.02) - (27 153)(185) (0.02) = 45+0.46=45.46 (as x = 3, Ar= 0.02) Hence, approximate value of f(3.02) is 45.46 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 2%. Solution Note that dV-( Ar.ar) . or = (3x2) (0.02x-0.06x3 m3 (as 2% of x is 0.020 Thus, the approximate change in volume is 0.063m MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume Solution Let r be the radius of the sphere and Ar be the error in measuring the radius. Then r 9 cm and Ar - 0,.03 cm. Now, the volume V of the sphere is given by 4 dV or dr dV Therefore - 4T(9) (0.03) 9.72Tt cm Thus, the approximate error in calculating the volume is 9.72T cm MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates. * Share among your peers as SHARING is CARING!!