## 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 JEE MAIN AND ADVANCED 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) Derivatives as the rate of change: If a variable quantity y is some function of time ti.e. y-f(t), then small change in time At have a corresponding change y in y Thus, the average rate of change- When limit At-0 is applied, the rate of change becomes instantaneous and Ay At we get the rate of change with respect to at the instant x. i.e., Hence, it is clear that the rateof variable is derivative of first variable with respect to other variable. MATHEMATICS FOR IIT-JEE lim _ = dy change of any variable with respect to some other ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example 1 Find the rate of change of the area of a circle per second with respect to its radius r when r 5 cm Solution The area A of a circle with radius r is given by A- of change of the area A with respect to its radius r is given by a. Therefore, the rate dA d, dr dr dA When r5 cm = 10 . Thus, the area of the circle is changing at the rate of dr 10 cm3/s. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing? Solution The area A of a circle with radius r is given by A of change of area A with respect to time t is Therefore, the rate dA d, = (Tr2 )= (Tr2 ).ar-m r (By Chain Rule) dt dt dr dt dA dt It is given that = 4cm/s Therefore, when r= 10 cm. -2n (10) (4) 80 Thus, the enclosed area is increasing at the rate of 80 cm2/s, when r= 10 cm MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is (1) a constant Ex Sol.(1) Let v be the velocity of the particle when the distance covered is s. Then voc given = constant - .. The acceleration is constant MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. For what values of x is the rate of increase of x35x25x8 is twice the rate of increase of x? (3) 3, (4) 3, 3 3 Sol. (4) Let y = x3-5x2+ 5x + 8, then, dy dt dy dx dt when 2, we have dt dx dx (3x2- 10x 5)2 dt dt 3 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The distance moved by the particle in time t is given by x = t -12t2+ 6t (1) 42 8. At the instant when its acceleration is zero, the velocity is- (2)-42 (3) 48 (4)-48 Sol. (2) we have, x = f-12E+ 6t + 8 -3t2 24t6 and6t dt dt Now, Acceleration 0 2 d x 06t 24 0t 4 dt Att=4, we have dx Velocity - dt )-4 - 3 42 24 4642 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex Side of an equilateral triangle expands at the area of 2 cm/s The rate of increase of its area when each side is 10 cm is (1) 10 v2 cm2/sec 210 V2 cm2/sec (3) 10 cm2/sec (4)5 cm/sec Sol. (2) Let x be the side and A be the area of equilateral triangle at time t. Then 3 4 3 dx 2 dt dA dA dt 2 dx dt 10 2 103 x-10 and 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)