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 e ae ++= = =

22 22 11 1 2 1 Equation of tangent at is 2 ( ) 2 (1 2 ) aa aa xa y e ae x a y axe a e ++ ++ = − = − = + − iii) Since tangent passes through origin 2 2 21 12 (1 2 ) 0 1Since 0 1 2 0 2 a a ae e a a + +  − =   − =  = 5) The curve C has its equation defined as 2 21ln 21 xxy x ++= − , for 1 2x . (i) Find the equation of the tangent to the curve at P, where x = 1. [5] (ii) If the tangent to the curve at P meets the x-axis at Q, calculate the exact coordinates of Q. [2] 5) At 1x= , 1 2 dy dx =− . 1ln 2 ( 1)2yx− =− − 1 ( 1 2ln 2)2yx=− − − At y = 0, 1 2ln 2x=+ (1 2ln 2 , 0 )+ 2 ( 1)(2 1) dy x dx x x −= +− At x = k, 0dy dx = 2 2 x k = = 2 2 2 2 21(ln ) 21 (ln 2 1 ln 2 1) 11( ln( 2 1) ln(2 1))22 2 2 (1)(2) 2( 1) 2(2 1) 11 ( 1) (2 1) 2 ( 1)(2 1) d x x dx x d x x xdx d x x xdx x xx xx x xx ++ − = + + − − = + + − − +=− +− =− +− −= +−

6) (i) Find the numerical value of the derivative of xx when 2x= , correct to 1 decimal place. [1] (ii) Hence find the equation of the tangent to the graph of xyx= at the point where 2x= , giving your answer in the form .y mx c=+ [3] 6) 7) Given that 2h( ) ln , , 0x x x x x= +   . (i) Find h'( )x and hence show that h(x) is an increasing function for x > 0. [2] (ii) Find the equation of the tangent to the curve h( )yx= at the point where 1x= . [2] (iii) The same tangent to the curve h( )yx= in 8(ii), is also the tangent of another curve 2y px qx=+ at the point where 1x=− . Determine the constants p and q. [4] 7i) ( ) 2h ln 2h '( ) 1 x x x x x =+ =+ For x > 0, h '( ) 0x   h (x) is an increasing function for x > 0.

ii) When x = 1, y = h(1) = 1, 2h '(1) 1 3 1= + = Eqn of tangent : 1 3( 1) 32 yx yx − = − =− iii) 2 2 d 2d At 1, 3( 1) 2 5, d 23d 2 3 ..........(1) Sub ( 1, 5)into : 5 5 ..........(2) Solve(1) & (2) : 2, 7 y px qx y px qx xy y pqx qp y px qx pq qp pq =+ =+ =− = − − =− =− + =  = + − − = + − = −  = + == 8(i) Find, in terms of k, the coordinates of the stationary points of the curve kxxy +−= 62 3 , where k is a constant. (ii) Determine the nature of each of the stationary points found in (i). (iii) Sketch the curve for the case k = 0, labelling clearly all stationary points and intercepts. State the range of values of x for which y is (a) strictly increasing, (b) strictly decreasing. [5] [4] [3] [2]

4)k (-1, and 4)-k (1, are points stationary of Coordiates 4 4 )1(6)1(2 or )1(6)1(2 1 or 1 066 0 Let 66 62 33 2 2 3 + +−= +−−−+−= −= =− = −= +−= kk kky x x dx dy xdx dy kxxy 8) (ii) Therefore, (1, k - 4) is a minimum point and (-1, k+4) is a maximum point. (iii) (a) 1or 1 − xx (b) 11 − x x -1- -1 -1+ dy/dx +ve 0 -ve / – \ x 1- 1 1+ dy/dx -ve 0 +ve \ – / 3or 3,0 0 :intercepts-x )3)(3(2 62 so ,0 3 −=== = +−= −= = xxx y xxx xxy k 0 1 -1 -4 4 √3 -√3 (-1, 4) (1, -4)

b) Rate of change 1) The height of a rain tree can be expressed by 810 4h t=− + m, where t is the number of years after the tree is planted from an established juvenile tree. (i) How high was the tree when it was planted? [1] (ii) Show that d 0d h t  for all 0t . What is the significance of this result? [3] (iii) State the maximum height of the rain tree. [1] 1a) When 0t= , b) ( ) 2 8 4 dh dt t = + Since ( ) 2 44t+ or 2( 4) 0t+ , thus 0dh dt  The tree is always growing or its height increases with time or 0dh dt → as t increases c) Maximum height = 10 m When t→ , 8 04t →+ . Height = 10 0 10−= m 810 8m04h= − = +

2) Variables x and y are related by the equation 2xye= . Given that the rate of change of y is 0.3 units per second, find the corresponding rate of change of x when y = 1. [4] 2) 2 0.3x dygiven y e and dt== 22 xdy edx = 20.3 2 xdy dy dx dx edt dx dt dt=   =  when y = 1 , x = 0 0.3 0.152 dx dt = = units/s 3) The diagram shows a hemispherical bowl of radius 12 cm. Water is poured into the bowl and at any time t , the height of the water level from the lowest point of the hemisphere is h cm. The rate of change of the height of the water level is 0.4 cm/s. (i) Show that the area of the water surface, A, is given by )24( 2hhA −= . [2] (ii) Find the rate of change of A at instant when 5=h cm. (Leave your an