RI Differentiation C6B Lect Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ____________________________________ Chapter 6B: Applications of Differentiation Page 1 of 12 Chapter 6B: Applications of Differentiation SYLLABUS INCLUDES  Problems involving tangents and normal to curves, including cases where the curve is defined implicitly or parametrically  Local maxima and minima problems  Connected rates of change problems CONTENT 1 Tangents and Normals 2 Maximization and Minimization 3 Connected Rates of Change Appendix: Further Applications of Differentiation (For Your Information)

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 2 of 12 1 Tangents and Normals Recall that equation of a (straight) line with gradient m passing through  11, x y is 11 ()yym xx  Let  , f( )Pk k be a point on the graph of f( )yx . In this case,  gradient of the tangent to the curve at P is f' ( )k . So equation of the tangent to the curve at P is  gradient of the normal to the curve at P is 1 f' ( )k , provided f' ( ) 0k  . So equation of the normal to the curve at P is Qn : What are the equations of the tangent and normal to the curve at point P if f' ( ) 0k  ? Example 1 A curve has equation given by 22 22 1 1xx y y  . Find (i) d d y x , (ii) the equations of the tangent and norma l to the curve at the point (1,3), (iii) the coordinates of the points on the curve where the tangent is parallel to the x-axis. Solution (i) 22 22 1 1xx y y  Differentiate with respect to x , dd222 4 0 dd yyxyx y xx   d(4 2 ) 2 2 d yy xx yx  d d2 y x y xy x     f( ) f' ( )yk k x k  1f( ) ( ) f' ( )yk x k k   , provided f' ( ) 0k  Equation of tangent: f( )yk i.e. tangent is horizontal Equation of normal: x k i.e. normal is vertical

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 3 of 12 (ii) At (1,3), d1 3 4 d2 ( 3 ) 1 5 y x   Equation of the tangent to curve at (1,3) is 43 ( 1) 5 4 115yx y x     Equation of the normal to curve at (1,3) is 53 ( 1) 4 5 174yx y x      (iii) When tangent is parallel to the x-axis, d 0( 1 )d2 yx y yxxy x        Sub. (1) into 22 22 1 1xx y y  , 222 1122 1 1 3xxx x    . Coordinates of the points are 11 11 11 11, and ,33 3 3       . Example 2 The parametric equations of a curve are given by cos , sinx ty t . Find the equation of the normal to the curve at the point where 3t  . Solution cos x t d sin d x tt  , sin yt d cos d y tt dd d 1 (cos ) cotdd d s i n yy t ttxt x t       When 3t  , 1cos 32x  , 3sin 32y  , d1 cotd3 3 y x   Gradient of normal at the point where 3t  is 3 . Equation of normal at the point where 3t  is 31 322yx    3 yx

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 4 of 12 Example 3 The parametric equations of a curve are given by (2 ) , 2 ( 1 )xa t t y a t  , where a is a positive constant. Find the equation of the normal to the curve at P when tp . If this normal meets the x axis at G and N is the foot of the perpendicular from P to the x axis, prove that 2NG a . Solution (2 )xa t t 2 2at at d 22 2 ( 1 )d x at a a tt  2( 1 )ya t 22at a d 2d y at dd d 1 dd d 1 yy t xt x t    At P , d1(2 ) , 2 (1 ) , d1 yxa p p y a p xp   Equation of normal to the curve at P is 2 12( 1 ) [ ( 2 ) ] 1 1 (1 ) (2 ) (1 ) 2 (1 ) (1 ) (1 ) ( 22 ) ya p x a p p p yp x a p p p a p yp x a ppp                     At G , 0y ,   2 212 2 (2 2 )1 ap p p xa p p p       At N , (2 )xa p p , since x coordinate of N and x coordinate of P are equal. 2 (2 2 )( 2 ) 2NG a p p ap p a      y x P N G Tangent to the curve at P Normal to the curve at P

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 5 of 12 2 Maximization and Minimization Many real-life situations require that some quantity (e.g. cost of manufacture or fuel consumption) be minimized, i.e. made as small as possible. Othe r situations require that some quantity (e.g. profit on sales or attendance at a concert) be maximized, i. e. made as large as possible. We can use differentiation to solve many of these problems. Recall: To determine whether the required quantity y at x k is minimized or maximized, given that d 0d xk y x   . Method 1 : First Derivative Test Check the signs of d d y x for x k and x k . x kk k kk k d d y x ve 0 ve ve 0 ve Tangent Conclusion y is maximized at x k y is minimized at x k Method 2 : Second Derivative Test Check the sign of 2 2 d d y x at x k . If 2 2 d 0 d xk y x   , then y is maximized at x k If 2 2 d 0 d xk y x   , then y is minimized at x k

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 6 of 12 Example 4 A solid right circular cylinder has a hemispher e hollowed out from each end. Given that the surface area is 264 cm , find the radius of the cylinder in exact form when the volume is a maximum. You need to justify that the volume is a maximum, without the use of the graphing calculator. Solution Let cmh and cmr represent the height and radius of the cylinder respectively. Surface area 2246 4rh r  232 2 rh r  Volume of the cylinder, 23 4 3Vr h r  2 23 332 2 4 10 3233 rrr r r r       2d (32 10 )d V rr  When 2d3 2 1 6 40 d1 0 5 5 V rrr    (since 0r  ) 2 2 d 20d V rr  When 4 5 r  , 2 2 d4 8 0 20 0d 55 V r    Alternatively, to show that V is maximum using the 1st derivative test we need to factorise the expression of d d V r and examine the factors that contribute to the sign of d d V r because the question already mentioned V is to be maximized. So     22d (32 10 ) 2 16 5 2 4 5 4 5d V rr r rr     . Since 0r  , we have   45 0 r . Therefore the only factor that is going to determine the sign of d d V r at 4 5 r  is   45 r . Hence the volume is a maximum when the radius is 4 cm 5 . r   444 555    45 r ve 0 ve d d V r ve 0 ve

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ____________________________________ Chapter 6B: Applications of Differentiation Page 7 of 12 Some guidelines to solve problems involving maximization and minimization :  Denote each changing quantity by a variable.  Write down a formula for the quantity to be maximized or minimized.  From the given or implied information (usua lly making reference to what is fixed or remains as a constant) in the question, express the quantity to be maximized or minimized in terms of 1 variable only.  Differe