## 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) 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)

Illustration post, 5 metres high. The rate at which the length of his shadow increases is (A) 3 m/min SOLUTION: (B) FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) A man 2 metres high, walks at a uniform speed of 6 metres per minute away from a lamp (C) 6 m/min (B) 4 m/min (D) 8 m/min Let AB be the lamp-post. Let at any ti man CD be at a distance x metres from the lamp- post and y metres be the length ofhis shadow CE Then, dt =6metres/minute dt [Given] (i) Clearly, triangles ABE and CDE are similar AB AE CD CE x + y = dy d dt dt 3-2(6) [using ( ) Thus, the shadow increases at the rate of 4 metres/ mmute. 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)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration hole at the vertex in the bottom. When the slant height of the water is 3 cm, the rate of decrease of the slant height of the water-cone. (Given that the vertical angle of the funnel is 120) Water is dripping out from a conical funnel at a uniform rate of 4 cmIsec through a tiny 32 (C) 93 32 32 cm/sec (B) Cm/ sec 27 (D) None of these SOLUTION: (B) Let at any time t, Vbe the volume of the water in the cone i.e., the volume ofthe water-cone VA'B', and let l be the slant height. Then, I sin 60 -_ and VO-| cos 60 = O' 60 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) V=- dt dV We are given that 4 cm dt When I-3, we have -4=3 12 dl d -32 -32 cm/sec dt 3 (3)2 27 dV Putting-#-4 in (i) Thus, the slant height of the water-cone is decreas- dl32 32 27T ing at the rate ofcm/sec. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration times the rate of change of the opposite angle, then that angle is (A) /6 A variable triangle is inscribed in a circle of radius R. If the rate of change of a side is R (B) /4 (C) /3 (D) /2 SOLUTION (C) Let the side be BC-a and A be the opposite angle. R=-a- 2 sin A Then, a 2R sin A 2 R cos A dt dA dt da dA dt dt dA R2R cos A as R cos A- 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!!