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