RI revision programme calculus (hard)

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision 1 (Zeta) Topic(s): Calculus I (Differentiation, Applications, Maclaurin and Graphing (Solutions) 1 HCI Prelim 9758/2023/01/Q11 [The volume of a right circular cone of radius r and height h is 21 π3 rh .] A student makes a printer nozzle for his self -built 3D printer. The printer nozzle consists of a hollow inverted right circular cone of negligible thickness with radius r mm where 02 r , height h mm and slanted edge 2 mm joined to a hollow cylinder of negligible thickness with radius r mm and height r mm. A height of 1 4 h mm is cut off from the vertex of the cone for the nozzle opening (see diagram). The volume of the printer nozzle is V mm3 . (a) Show that 3 2 2 21π π 4 64V r r r= + − . [3] The student wants V to be a maximum. (b) It is given that 1rr= gives the maximum value of V . Show that 1r satisfies the equation 424537 18736 3136 0rr− + = . [4] (c) Show that one of the positive roots of the equation in part (b) does not give a stationary value of V . Suggest a reason why this value is a solution to the equation in part (b) even though it does not give a stationary value of V . [3] (d) Given that 1r is the largest positive root of the equation in part (b), state the value of 1r and find the corresponding value of h. (You need not show that your answer gives a maximum.) [1] (e) With reference to the value of 1r found in part (d), comment on the practicality of having a maximum volume for the printer nozzle. [1] r mm mm h mm 2 mm mm

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ______________________________________ Y6 H2 Math Term 3 Revision Lecture 1: Calculus I Page 2 of 16 Solution: (a) By Pythagoras’ Theorem, 2 2 2 22 2 4 (reject 4 since 0) rh h r h r h +=  = − =− −  Let 0r be radius of cut off cone. Using similar s , 0 0 1 44 r rrr =  = 2 2 2 0 3 2 2 2 2 3 2 2 1 1 1π ( ) π π ( )3 3 4 11π π 4 π 43 192 21π π 4 (shown)64 V r r r h r h r r r r r r r r  = + − = + − − − = + − (b) 2 2 2 2 2 22 2 2 2 2 d 21 13π π 2 4 ( )( 2 )d 64 24 213π π 2 464 4 21 8 33ππ 64 4 V r r r r rr r rr r r r rrr r = + − + −  − = + − −  − −=+ − At maximum value of V, d 0d V r = , since 02 r 2 2 2 2 2 2 2 2 2 2 2 21 8 33π π 0 64 4 7 8 33π0 64 4 7 8 3 0 6 (since 0 2)64 4 7 3 8 64 4 64 4 21 5 rrr r rrr r rr r r r rr r rr − + = −   −+=  −  −+ =   −  −=  − − −  = 2 2 2 24096 (4 ) (21 56)r r r− = − 2 4 4 2 42 16384 4096 441 2352 3136 4537 18736 3136 0 (shown) r r r r rr − = − +  − + =

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________