SAJC Check your Understanding_Applications of differentiation_teacher

Check your Understanding (Applications of Differentiation) a) Graphing and tangent 1) (i) Use a non-calculator method to find the coordinates of each stationary point on the graph of 326 27y x x=− and determine the nature of each stationary point. [4] (ii) Sketch the graph of 326 27y x x=− . [2] 1i) 326 27y x x=− 218 54dy xxdx =− For stationary points, d 0d y x = 218 54 0xx−= 18 ( 3) 0xx −= 0 or 3xx = = Coordinates of the turning points are (0,0),(3, 81)− when 0,x= 0 o dy dx −  and 0 o dy dx +  max pt when 3,x= 3 0dy dx −  and 3 0dy dx +  min pt Hence (0,0) is a maximum point and (3, 81)− is a minimum point. Alternative method 2 2 d 36 54d y xx =− When x = 0, 2 2 d 54 0 Max ptd y x =−  

When x = 3, 2 2 d 54 > 0 Min ptd y x = Hence (0,0) is a maximum point and (3, 81)− is a minimum point. ii) 326 27y x x=− (0,0) 9 2 (3, 81)− 2) Use a non-calculator method to find the coordinates of each stationary point on the graph of 2(2 9 12)y x x x= − + and determine the nature of each stationary point. [5] 2) 2(2 9 12)y x x x= − + 26 18 12dy xxdx = − + For stationary points, d 0d y x = 26 18 12 0xx− + = 6( 1)( 2) 0xx− − =

1 or 2xx = = Coordinates of the turning points are (1,5),(2,4) when 1,x= 1 0dy dx −  and 1 0dy dx +  max pt when 2,x= 2 0dy dx −  and 2 0dy dx +  min pt Hence (1,5) is a maximum point and (2,4) is a minimum point. Alternative method 2 2 d 12 18d y xx =− When x = 1, 2 2 d 6 0 Max ptd y x =−   When x = 2, 2 2 d 6 > 0 Min ptd y x = Hence (1,5) is a maximum point and (2,4) is a minimum point. 3) The curve C has its equation defined as ln 41 xy x= + . (i) Find the gradient of the tangent to the curve at P where x = 1. [3] (ii) Find the equation of the tangent at P. [3] (iii) If the tangent to the curve at P meets the x-axis at Q, calculate the exact coordinates of Q. [2] 3) 1ln ln ln(4 1)4 1 2 xy x x x= = − ++ 14 2 4 1 dy dx x x=− +

(i) when x = 1 1 4 3 2 5 10 dy dx = − =− (ii) when x = 1, ln 5y=− 33 ln 510 10 3 ln 510 33 ln 510 10 y x C C C yx =− + − =− + =− =− + − (iii) y = 0 3 3 10ln 5 1 ln 510 10 3 xx− =−  = − 101 ln 5, 03Q− 4) The equation of a curve is given by 21 xye += . (i) Find dy dx in terms of x. [2] (ii) Find the equation of the tangent to the curve at the point x = a. [3] If the tangent passes through the origin, find the value of a. [3] 4i) 21 xye += 21dy 2 . dx xxe += ii) 2211 dywhen , , 2 dx aax a y