11 Differentiation Application TutSol

Page 1 of 25 Differentiation Applications Given that 4 3 23 16 24 9y x x x    . Without the use of a GC, (a) find the stationary point(s) of its graph and determine their nature. (0, 9) min pt, (2,7) stat pt of inflex (b) find the range of x, if any, for which the curve is (i) strictly increasing, 0 2 or 2xx   (ii) concave downwards. 2 3 2x (c) sketch the curve showing all the important features. (a) 4 3 2 3 2 2 23 16 24 9 12 48 48 12 ( 4 4) 12 ( 2)dyy x x x x x x x x x x x dx             At stationary points, 0 0, 2dy xdx    . Also, 2 2 2 36 96 48dy xxdx    When 2 20, 9 and 48 0. dyxy dx    So (0, 9) is a minimum point. When 2, 7xy . Note that 2 , 0 dyx dx  , and, when 2 , 0 dyx dx  . Hence, (2, 7) is a stationary point of inflexion. (b) & (c) (i) For strictly increasing, 0 0 2 or 2dy xxdx      (ii) For concave downwards, 2 2 2 2 36 96 48 12(3 8 4) 12(3 2)( 2) 0 2 2 3 dy xxdx xx xx x              N10/I/Q4 Given that 22 2 4 0x y xy    , find d d y x in terms of x and y . For the curve with equation 22 2 4 0x y xy    , find the coordinates of each point at which the tangent is parallel to x-axis.  2, 2 ,  2, 2

Page 2 of 25 Differentiation Applications 22 2 4 0x y xy    Differentiate with respect to x   dd2 2 2 2 0dd d2 2 2 2 d d 2 2 d 2 2 yyx y x y xx yx y y x x y x y x y x          For tangents parallel to x-axis, d 0d y x  22 0 22 xy xyyx     Substitute into equation of curve     2 2 2 2 4 0 2 4 0 2 y y y y y y          Hence the coordinates of the points whose tangent are parallel to the x-axis are  2, 2 and  2, 2 N2012/I/8 The curve C has equation 2()x y x y   It is given that C has only one turning point. (i) Show that d21 d 2 2 1 y x x y  . (ii) Hence, or otherwise, show that 32 2 dd 1dd yy xx   (iii) Hence state, with a reason, whether the turning point is a maximium or a minimium. (i)       2 differentiate wrt : 1 2 1 2 2 1 1 2 2 1 2 2 2 2 1 2 2 1 1 2 2 21 2 2 1 2 2 1 x y x y x dy dy xydx dx dyx y x y dx dy x y dx x y dy x y x y dx x y x y                          

Page 3 of 25 Differentiation Applications (ii)     2 22 2 2 3 Differentiate wrt : 2 22 2 2 1 14 212 2 2 1 11 1 x d y dy dx dx xy dy dxxy dy dy dx dx dy dx                        (iii) 2 2 When 0, 10 dy dx dy dx    Therefore, the turning point is a maximum point. The parametric equations of a curve are 2, yx ct t c , where c is a constant. (i) Find the equation of the tangent to the curve at the point P 2,cp c p   . 3 23p y x cp (ii) Hence find the coordinates of the points Q and R where the tangent meets the x- and y-axes respectively. Q 3 , 02 cp  , R 2 30, c p   (iii) Find the Cartesian equation of the curve. 23x y c (iv) Find a Cartesian equation of the locus of the mid-point of QR as p varies. 3 2 27 32 cxy  (i) 23 2, and,dx dy ccydt t c ttx ct d     3 3 2 2 c dy t dx c t     At point P, tp , we have 3 2dy dx p .

Page 4 of 25 Differentiation Applications Thus, the equation of the tangent to the curve at the point P 2,cp c p   is:   33 23 2 2 2 2 3cy x cp p y cp x cp p y x cppp          (ii) This tangent meets the x- axes at point Q, to find the coordinates of Q, we let 0y , 3 3( ) 2 3 2 cpp o x cp x    . That is Q 3 , 02 cp  . Also this tangent meets the y- axes at point R, to find the coordinates of R, we let 0x , 3 2 32(0) 3 cp y cp y p    . That is R 2 30, c p   . (iii) The Cartesian equation of the curve can be easily obtained by eliminating the parameter t from the given parametric equations 2 2 3 22, () cy x y cx ctt tcc t     The Cartesian equation of the curve is : 23x y c (iv) To find a Cartesian equation of the locus of the mid-point of QR as p varies, we let (x, y) be the coordinate of the mid-point , then 3 0 32 24 cp cpx   , and , 2 2 3 0 3 22 c cpy p   . By eliminating the variable p, we have 2 3 2 2 3 3 27 4 2 32 cp c cxy p    . Therefore, the Cartesian equation of the locus of the mid-point of QR is: 3 2 27 32 cxy  . The line l is tangent to the curve 11y x , where 0x , at the point where xa . (i) Show that the equation of l may be expressed in the form 22 2a y x a a   . (ii) The tangent line l passes through a fixed point  ,XY . Give a brief argument to explain why there cannot be more than 2 tangents passing through  ,XY . (iii) Find the value of a for whic h the line l passes through the origin, and find the equation of l in this case. 2a , 4 xy

Page 5 of 25 Differentiation Applications (i) 1111yx x     2 d1 d y xx At xa , 2 11 d1 d y a y xa   Equation of line l,     2 2 22 22 111 111 2 (shown) 1 y x a aa xy a a a a y a a x a a y x a a                (ii) Since for any fixed point  ,XY , 22 2a Y X a a   is a quadratic equation in a, there are at most 2 real solutions for a. Therefore, there cannot be more than 2 tangents p assing through  ,XY . (iii) Substitute 0, 0xy into (1),   2 20 20 0 (rej) or 2 aa aa aa    Equation of line l, 220 4 yx xy   Find the equations of the tangents to the curve with equation 22 22 1, , 0xy ab ab    , which makes an angle of 45 with the positive x-axis, measured in an anticlockwise sense. 22y x a b   or 22y x a b   At the points on the curve where the tangents make an angle of with the positive x – axis, the gradient of the tangents = = 1. Given equation of curve: 22 22 1xy ab  Differentiating w.r.t. x, 22 22 0 dyyx dx ab  Substitute into the above, we have 2 2 2 2 2 2 (1) 0x y b x y a b a     To find the coordinates of the points at which the gradient is : 45 tan 45dy dx  1dy dx  1dy dx 

Page 6 of 25 Differentiation Applications Substitute 2 2 bxy a  into equation of the curve, 22 2 2 2 2 1 ( ) 1x b x a b a    2 2 2 24 1x b x aa  2 22 ax ab   When 2 22 ax ab   , 22 2 22 b x by a ab    . When 2 22 ax ab   , 22 2 22 b x by a ab    . At 22 2 2 2 2 ,ab a b a b    , the equation of the tangent is : 22 22 2 2 2 2 1bay x y x a b a b a b                      At 22 2 2 2 2 ,ab a b a b    , the equation of the tangent is : 22 22 2 2 2 2 1bay x y x a b a b a b             NJC Prelim 09/I/7 (Anchor Q – Connected rate of change) In a triangle ABC, 3AB cm and 2AC cm. If angle BAC is increasing at a constant rate of 0.1 radians per second, find the rate of increase of the length BC at the instant where π 3BAC radians. 0.196 cm/s [4] Let cm and radians. By cosine rule, When and , 13 12cos 73x    , we have   BC x BAC  2 2 2 3 2 2(3)(2)cosx    2 13 12cosx   dd2 12sindd xx tt  π 3  d 0.1dt   d32 7 12 (0.1)d2 x t  

Page 7 of 25 Differentiation Applications Hence, . Therefore, the length BC is increasing at the rate of 0.196 cm/s. HCI Prelim 08/I/3 A right circular cone with radius 3 cm and height 9 cm is initially full of water. Water is leaking from the circular base of the cone at a constant rate of 2 cm3 s1, find the exact rate of change of the depth of water when the depth of water is