6x and find the. y=2x+5 and y=4 -x 2. gradient of the curve at this point. tangent to the curve y = 4 - x 2? Figure 5.11

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1 Figure 5.11 EERCISE 5C 1 For eah part of this question, (a) find: (b) fmd the gradient of the urve at the given point. (il y = x- 2 ; (.25, 16) (iil y= x- 1 + x-4; (-1, ) (iiil y= 4x x- 5 ; (1, 6) (iv) y= 3x 4-4- sx- 3 ; (2, 43) (vi y= {;,; + 3x; (4, 14) (vi) (9, q) 2 m Sketh the urve y = x 2-4. (iil Write down the o-ordinates of the points where the urve rosses the x axis. (iiil Differentiate y= x 2-4. (ivl Find the gradient of the urve at the points where it rosses the x axis. 3 m Sketh the urve y = x 2-6x. (iil Differentiate y = x 2-6x. (iiil Show that the point (3, -9) lies on the urve y = x 2 - gradient of the urve at this point. (ivl Relate your answer to the shape of the urve. 6x and find the 4 m Sketh, on the same axes, the graphs with equations y=2x+5 and y=4 -x 2 (iil Show that the point ( -1, 3) lies on both graphs. (iiil Differentiate y= 4- x 2 and so find its gradient at (-1, 3). (ivl Do you have suffiient evidene to deide whether the line y = 2x + 5 is a tangent to the urve y = 4 - x 2? (vi Is the line joining o) to (, 5) a tangent to the urve y= 4- x 2?

2 5 The urve y = x 3-6x 2 + llx- 6 uts the x axis at x = 1, x = 2 and x = 3. m Sketh the urve. (iil Differentiate y = x 3-6x x- 6. liiil Show that the tangents to the urve at two of the points at whih it uts the x axis are parallel. 6 m Sketh the urve y = x 2 + 3x- 1.!ii) Differentiate y = x 2 + 3x- 1.!iiil Find the o-ordinates of the point on the urve y = x 2 + 3x- 1 at whih it is parallel to the line y = 5x- 1. (ivl Is the line y = 5x- 1 a tangent to the urve y = x 2 + 3x- 1? Give reasons for your answer. m n ;; Cll (') 7 m Sketh, on the same axes, the urves with equations y = x 2-9 and y = 9 - x 2 for -4 x 4. (ii) Differentiate y= x 2-9. (iiil Find the gradient of y = x 2-9 at the points (2, -5) and ( -2, -5). (iv) Find the gradient of the urve y= 9- x 2 at the points (2, 5) and ( -2, 5). (v) The tangents toy= x 2-9 at (2, -5) and (-2, -5), and those toy= 9- x 2 at (2, 5) and (-2, 5) are drawn to form a quadrilateral. Desribe this quadrilateral and give reasons for your answer. 8 m Sketh, on the same axes, the urves with equations y= x 2-1 and y= x for -3 x 3. (iil Find the gradient of the urve y= x 2-1 at the point (2, 3).!iiil Give two explanations, one involving geometry and the other involving alulus, as to why the gradient at the point (2, 7) on the urve y = x should have the same value as your answer to part (ii). (iv) Give the equation of another urve with the same gradient funtion as y=x2-l. 9 The funtion f(x) = ax 3 + bx+ 4, where a and bare onstants, goes through the point (2, 14) with gradient 21.!il Using the fat that (2, 14) lies on the urve, fmd an equation involving a and b.!iil Differentiate f(x) and, using the fat that the gradient is 21 when x= 2, form another equation involving a and b.!iiil By solving these two equations simultaneously find the values of a and b.

3 1 In his book Mathematiian's Delight, W.W. Sawyer observes that the arh of Vitoria Falls Bridge appears to agree with the urve ;. Gl Gl :t: i x 2 y = 12 taking the origin as the point mid-way between the feet of the arh, and taking the distane between its feet as 4. 7 units. n. ' (i) Find:. 11 (i) (iil Evaluate: when x = and when x = (iii) Find the value of x for whih : = Use your knowledge of the shape of the urve y = - to sketh the urve 1 y= (iil Write down the o-ordinates of the point where the urve rosses the x axis. (iii) Differentiate y = l + 2. (ivl Find the gradient of the urve at the point where it rosses the x axis The sketh shows the graph of y= 2 + x. y

4 a. 4 (i) D1uerent1ate y = 2 + x. (iil Show that the point ( -2, -1) lies on the urve. (iiil Find the gradient of the urve at ( -2, -1). livl Show that the point (2, 3) lies on the urve. (vi Find the gradient of the urve at (2, 3). (vi) Relate your answer to part (vi to the shape of the urve. 13 lil Sketh, on the same axes, the graphs with equations 1 y=2+1 and y=-16x+13 for -3 ::::; x::::;3. (iil Show that the point (.5, 5) lies on both graphs. liiil Differentiate y= + 1 and find its gradient at (.5, 5). (ivl What an you dedue about the two graphs? m > " n iii' " en n 14 m Sketh the urve y= Fx for ::::; x ::::; 1. (iil Differentiate y= Fx. (iiil Find the gradient of the urve at the point ( 9, 3) m Sketh the urve y= 2 for -3 ::::; x ::::; 3. liil Differentiate y = 4 2 (iiil Find the gradient of the urve at the point ( -2, 1). (ivl Write down the gradient of the urve at the point (2, 1). Explain why your answer is -1 x your answer to part (iiil. 16 The sketh shows the urve y =! _l 2 y m Differentiate y =! - l 2 (iil Find the gradient of the urve at the point where it rosses the x axis The gradient of the urve y = kx'- at the point x = 9 is 18. Find the value of k. 18 Find the gradient of the urve y = x Fx 2 at the point where x = 4.

5 Tangents and normals Now that you know how to find the gradient of a urve at any point you an use this to find the equation of the tangent at any speified point on the urve... as Gl ::: EAMPLE 5.8 Find the equation of the tangent to the urve y = x 2 + 3x + 2 at the point (2, 12). SOLUTION Calulating : = 2x + 3. Substituting x = 2 into the expression that point: to find the gradient m of the tangent at m=22+3 =7. The equation of the tangent is given by y- y 1 = m(x- x 1 ). In this ase x 1 = 2, y 1 = 12 so y- 12 = 7(x- 2) => y=7x-2. This is the equation of the tangent. y Figure I The normal to a urve at a partiular point is the straight line whih is at right angles to the tangent at that point (see figure 5.13). Remember that for perpendiular lines, m 1 m 2 = -1.

6 tangent urve Figure 5.13 If the gradient of the tangent is m 1, the gradient, m 2, of the normal is given by ;4 :I :I Ill Ill :I Cl. :I 3 Ill iii 1 mz =-m, This enables you to find the equation of the normal at any speified point on a urve. EAMPLE 5.9 A urve has equation y = 16-4-[;. The normal to the urve at the point ( 4, -4) meets the y axis at the point P. Find the o-ordinates of P. SOLUTION You may find it easier to write y = 16-4-[; as y = 16x Differentiating gives : = -16x- 2 - x 4x -! At the point ( 4, -4), x = 4 and dy dx J4 =-1-1= xz-.j; So at the point ( 4, -4) the gradient of the tangent is -2. Gradientofnormal= d" = gra tent o tangent The equation of the normal is given by y- y 1 = m(x- x 1 ) y- (-4) = 4) y 6 P is the point where the normal meets they axis and so where x =. Substituting x = into y = 6 gives y = -6. SoP is the point (, -6).

7 EERCISE 5 1 The graph of y = 6x- x 2 is shown below. y.;: Ill.;: Gl Q The marked point, P, is (1, 5). (i) Find the gradient funtion Z. (iil Find the gradient of the urve at P. (iiil Find the equation of the tangent at P. 2 lil Sketh the urve y= 4x- x 2 5 lil Differentiate y = 4x- x 2. (iiil Find the gradient of y = 4x- x 2 at the point ( 1, 3). (ivl Find the equation of the tangent to the urve y= 4x- x 2 at the point (1, 3). 3 lil Differentiate y = x 3-4x 2. (iil Find the gradient of y = x 3-4x 2 at the point (2, -8). (iiil Find the equation of the tangent to the urve y = x 3 - (2, -8). 4x 2 at the point (ivl Find the o-ordinates of the other point at whih this tangent meets the urve. 4 m Sketh the urve y = 6 - x 2 (iil Find the gradient of the urve at the points ( -1, 5) and ( 1, 5). (iiil Find the equations of the tangents to the urve at these points. (ivl Find the o-ordinates of the point of intersetion of these two tangents. 5 m Sketh the urve y = x and the straight line y = 4x on the same axes. liil Show that both y = x and y = 4x pass through the point ( 2, 8). (iiil Show that y = x and y = 4x have the same gradient at (2, 8), and state what you onlude from this result and that in part (iil. 6 m Find the equation of the tangent z, to the urve y= 2x 3-15x x at (2, 4). (iil Using your expression for find the o-ordinates of another point on the urve at whih the tangent is parallel to the one at (2, 4). (iiil Find the equation of the normal at this point.

8 7 (i) Given that y= x 3-4x 2 + 5x- 2, find:. The point P is on the urve and its x o-ordinate is 3. (iil Calulate the y o-ordinate of P. (iiil Calulate the gradient at P. (iv) Find the equation of the tangent at P. (vi Find the equation of the normal at P. (vi) Find the values of x for whih the urve has a gradient of 5. [MEI] m > n ; en 8 (i) Sketh the urve whose equation is y = x 2-3x + 2 and state the o-ordinates of the points A and B where it rosses the x axis. (iil Find the gradient of the urve at A and at B. (iiil Find the equations of the tangent and normal to the urve at both A and B. (ivl The tangent at A meets the tangent at Bat the point P. The normal at A meets the normal at B at the point Q. What shape is the figure APBQ? 9 (i) Find the points of intersetion of y = 2x2-9x and y = x- 8. (ii) Find for the urve and hene fmd the equation of the tangent to the urve at eah of the points in part (il. (iiil Find the point of intersetion of the two tangents. (ivl The two tangents from a point to a irle are always equal in length. Are the two tangents to the urve y= 2x 2-9x (a parabola) from the point you found in part (iiil equal in length? 1 The equation of a urve is y = Fx. (i) Find the equation of the tangent to the urve at the point (1, 1). (iil Find the equation of the normal to the urve at the point ( 1, 1). (iiil The tangent uts the x axis at A and the normal uts the x axis at B. Find the length of AB. 11 The equation of a urve is y = l. (i) Find the equation of the tangent to the urve at the point (2, (ii) Find the equation of the normal to the urve at the point (2, (iiil Find the area of the triangle formed by the tangent, the normal and the yaxis.

9 12 The sketh shows the graph of y = Jx - 1. y.;: Ill.;: Gl i5-1 m Differentiate y = Jx - 1. (iil Find the o-ordinates of the point on the urve y = Jx - 1 at whih the tangent is parallel to the line y = 2x - 1. liiil Is the line y = 2x -1 a tangent to the urve y = Jx - 1? Give reasons for your answer. 13 The equation of a urve is y = Jx - m Find the equation of the tangent to the urve at the point where x = i (iil Find the equation of the normal to the urve at the point where x = i. (iiil Find the area of the triangle formed by the tangent, the normal and the x axis. 14 The equation of a urve is y = Jx The tangent to the urve at the point (9, 3) meets the x axis at A and they axis at B. Find the length of AB. 15 The equation of a urve is y = m Find the equation of the normal to the urve at the point (2, 4). liil Find the area of the triangle formed by the normal and the axes. 16 The graph of y = 3x- shown below. The point marked Pis (1, 2). Y

10 (i) Find the gradient funtion ix. (ii) Use your answer from part (i) to find the gradient of the urve at P. (iiil Use your answer from part (iil, and the fat that the gradient of the urve at Pis the same as that of the tangent at P, to find the equation of the tangent at P in the form y = mx The graph of y= x 2 +lis shown below. The point marked Q is (1, 2). m > n iii' 1.11 Q (i) Find the gradient funtion liil Find the gradient of the tangent at Q. liiil Show that the equation of the normal to the urve at Q an be written as x+ y=3. (ivl At what other points does the normal ut the urve? 3 18 The equation of a urve is y = x 2 The tangent and normal to the urve at the point x = 4 interset the x axis at A and B respetively. Calulate the length of AB. 19 (i) The diagram shows the line 2 y = x + 5 and the urve y = x2-4x + 7, whih interset at the points A and B. y 2y=x+ 5

11 .;:: ea '.;:: Q) :t: Find (a) the x o-ordinates of A and B, (b) the equation of the tangent to the urve at B, () the aute angle, in degrees orret to 1 deimal plae, between this tangent and the line 2y = x + 5. (ii) Determine the set of values of k for whih the line 2y = x + k does not interset the urve y = :x?-- 4x + 7. [Cambridge AS & A Level Mathematis 979, Paper l2 QlO November 29] 2 The equation of a urve is y = 5 - _. m Show that the equation of the normal to the urve at the point P(2, 1) is 2y+ x= 4. This normal meets the urve again at the point Q. (ii) Find the o-ordinates of Q. (iiil Find the length ofpq. Maximum and minimum points [Cambridge AS & A Level Mathematis 979, Paper 1 Q8 November 28] ACTIVITY 5.6 Plot the graph of y = x 4 - x 3-2:x?-, taking values of x from -2.5 to +2.5 in steps of.5, and answer these questions. m How many stationary points has the graph? (ii) What is the gradient at a stationary point? (iii) One of the stationary points is a maximum and the others are minima. Whih are of eah type? (iv) Is the maximum the highest point of the graph? (v) Do the two minima our exatly at the points you plotted? (vi) Estimate the lowest value that y takes. Gradient at a maximum or minimum point Figure 5.14 shows the graph of y= -x It has a maximum point at (, 16). y Figure 5.14

12 You will see that at the maximum point the gradient is zero the gradient is positive to the left of the maximum and negative to the right of it. This is true for any maximum point (see figure 5.15). Figure : I» > " 3 I» :I CL 3 :; " 3 "Cl :; Ill In the same way, for any minimum point (see figure 5.16): the gradient is zero at the minimum the gradient goes from negative to zero to positive. Figure 5.16 Maximum and minimum points are also known as stationary points as the gradient funtion is zero and so is neither inreasing nor dereasing. EAMPLE 5.1 Find the stationary points on the urve of y = x 3-3x + 1, and sketh the urve. SOLUTION The gradient funtion for this urve is = 3x 2-3. The x values for whih = are given by 3x 2-3 = 3(x 2-1) = 3(x+ 1)(x-l) = ==> x=-1 or x=l. The signs of the gradient funtion just either side of these values tell you the nature of eah stationary point.

13 Forx=-1: x=-2 ::::} dy = 3( -2)2-3 = +9 dx x=o ::::} = 3() 2-3 = -3. -;; 111. G) G) ;: Figure 5.17 ;/ For x= 1: = ::::} dy = -3 dx x= 2::::} = 3(2) 2-3 = +9. /( Figure 5.18 Thus the stationary point at x = -1 is a maximum and the one at x = 1 is a mmrmum. Substituting the x values of the stationary points into the original equation, y= x 3-3x+ 1, gives when x=-1, whenx= 1, y= (-1) 3-3(-1) + 1 = 3 y= (1) 3-3(1) + 1 = -1. There is a maximum at (-1, 3) and a minimum at (1, -1). The sketh an now be drawn (see figure 5.19). maximum (- 1, 3) y m1rnmum (1, - 1) Figure 5.19

14 EAMPLE 5.11 In this ase you knew the general shape of the ubi urve and the positions of all of the maximum and minimum points, so it was easy to selet values of x for whih to test the sign of Z The urve of a more ompliated funtion may have several maxima and minima lose together, and even some points at whih the gradient is undefined. To deide in suh ases whether a partiular stationary point is a maximum or a minimum, you must look at points whih are just either side of it. Find all the stationary points on the urve of y = 2t 4 - SOLUTION dy = 8t 3-2t dt At a stationary point, % =, so 8t 3-2t= 2t(4t 2-1) = 2t(2t-1)(2t+ 1) = :::::} % = when t= -.5, or.5. t and sketh the urve. You may fmd it helpful to summarise your working in a table like the one below. You an find the various signs, + or-, by taking a test point in eah interval, for example t=.25 in the interval < t <.5. t < --o.5 --o.5 --o.5 < t < o < t <.5 o.5 T t> o.5 s: I» )C " 3 I» :I Q. 3 :r " 3 "CC :r Ill Sign of% Stationary point mm max m in There is a maximum point when t= and there are minimum points when t= -.5 and +.5. When t=o: When t = -.5: When t=.5: y= 2() 4 - () = 1. y = 2( -.5) 4 - (-.5) =.875. y= 2(.5) 4 - (.5) =.875. Therefore (, 1) is a maximum point and (-.5,.875) and (.5,.875) are minima.

15 The graph of this funtion is shown in figure 5.2. y. «<. Gl is - I Figure 5.2 Inreasing and dereasing funtions When the gradient is positive, the funtion is desribed as an inreasing funtion. Similarly, when the gradient is negative, it is a dereasing funtion. These terms are often used for funtions that are inreasing or dereasing for all values of x. EAMPLE 5.12 Show that y = x 3 + x is an inreasing funtion. SOLUTION y y = x 3 + x => = 3x Sine x 2 ;;;. for all real values of x, ;;;. 1 => y = x 3 + x is an inreasing funtion. Figure 5.21 shows its graph Figure 5.21

16 EAMPLE 5.13 Find the range of values of x for whih the funtion y = x 2-6x is a dereasing funtion. y SOLUTION y = x 2-6x ==> : = 2x- 6. y dereasing ==> : < m >C Cl ; Ul m ==> 2x- 6 < ==> < 3. Figure 5.22 shows the graph of y=z- 6x. Figure 5.22 EERCISE SE 1 Given that y= x 2 + 8x+ 13 m find :, and the value of x for whih : = liil showing your working learly, deide whether the point orresponding to this xvalue is a maximum or a minimum by onsidering the gradient either side of it (iiil show that the orresponding y value is -3 (iv) sketh the urve. 2 Given that y= x 2 + 5x+ 2 (i) find :, and the value of x for whih : = (ii) lassify the point that orresponds to this x value as a maximum or a m1mmum (iiil find the orresponding y value (iv) sketh the urve. 3 Given that y= x 3-12x+ 2 m find :, and the values of x for whih : = lii) lassify the points that orrespond to these x values liiil find the orresponding yvalues (iv) sketh the urve. 4 m Find the o-ordinates of the stationary points of the urve y = x 3-6x 2, and determine whether eah one is a maximum or a minimum. (iil Use this information to sketh the graph of y= x 3-6x 2 5 Find : when y = x 3 - x and show that y = x 3 - xis an inreasing funtion 1 1 for x <-.J}andx >.J3.

17 6 Given that y = x 3 + 4x (i) liil show that y = x 3 + 4x is an inreasing funtion for all values of x. ;:: «< ;:: f Cl) ::: is 7 Given that y= x 3 + 3x 2-9x+ 6 lil find and fatorise the quadrati expression you obtain liil write down the values of x for whih = liiil show that one of the points orresponding to these x values is a minimum and the other a maximum livl show that the orresponding yvalues are 1 and 33 respetively (v) sketh the urve. 8 Given that y= 9x+ 3x 2 - x 3 (i) find and fatorise the quadrati expression you obtain liil find the values of x for whih the urve has stationary points, and lassify these stationary points liiil find the orresponding y values (ivl sketh the urve. 9 (i) Find the o-ordinates and nature of eah of the stationary points of y= x 3-2x 2-4x+ 3. liil Sketh the urve. 1 o lil Find the o-ordinates and nature of eah of the stationary points of the urve with equation y = x 4 + 4x 3-36x liil Sketh the urve. 11 lil Differentiate y = x 3 + 3x. liil What does this tell you about the number of stationary points of the urve with equation y = x 3 + 3x? liiil Find the values of y orresponding to x = -3, -2, -1,, 1, 2 and 3. livl Hene sketh the urve and explain your answer to part (iil. 12 You are given that y= 2x 3 + 3x 2-72x F" ddy hl m dx. P is the point on the urve where x = 4. (ii) Calulate the y o-ordinate of P. (iii) Calulate the gradient at P and hene find the equation of the tangent to the urve at P. (ivl There are two stationary points on the urve. Find their o-ordinates. [MEI]

18 13 (il Find the o-ordinates of the stationary points of the urve f(x) = 4x + l. (iil Find the set of values of x for whih f(x) is an inreasing funtion. 14 The equation of a urve is y = (2x- 3 ) 3-4x. (il Find dy_ dx (iil Find the equation of the tangent to the urve at the point where the urve intersets the y axis. (iiil Find the set of values of x for whih ( 2x- 3) 3 - funtion of x. 4x is an inreasing [Cambridge AS & A Level Mathematis 979, Paper 12 Q 1 June 21] :r "1:1 lot. ;. (;' n ; :I 15 The equation of a urve is y = x?- - 3x + 4. m Show that the whole of the urve lies above the x axis. (iil Find the set of values of x for whih x?- - 3x + 4 is a dereasing funtion of x. The equation of a line is y + 2x = k, where k is a onstant. liiil In the ase where k = 6, find the o-ordinates of the points of intersetion of the line and the urve. (ivl Find the value of k for whih the line is a tangent to the urve. [Cambridge AS & A Level Mathematis 979, Paper 1 Q 1 June 25] 16 The equation of a urve C is y = 2x?- - 8x + 9 and the equation of a line Lis x+ y= 3. m Find the x o-ordinates of the points of intersetion of L and C. (iil Show that one of these points is also the stationary point of C. [Cambridge AS & A Level Mathematis 979, Paper I Q4 June 28] e Points of infletion It is possible for the value of dy to be zero at a point on a urve without it being a dx maximum or minimum. This is the ase with the urve y= x\ at the point (, ) (see figure 5.23). dy = 3x2 dx and when x =, dy =. dx Figure 5.23

19 This is an example of a point of infletion. In general, a point of infletion ours where the tangent to a urve rosses the urve. This an happen also when *, as shown in figure Ill. Gl Gl :t same sign. Figure 5.24 If you are a driver you may find it helpful to think of a point of infletion as the point at whih you hange from left lok to right lok, or vie versa. Another way of thinking about a point of infletion is to view the urve from one side and see it as the point where the urve hanges from being onave to onvex. The seond derivative Figure 5.25 shows a sketh of a funtion y = f(x), and beneath it a sketh of the orresponding gradient f(x). dy d Figure 5.25

20 ACTIVITY 5.7 Sketh the graph of the gradient of against x for the funtion illustrated in figure Do this by traing the two graphs shown in figure 5.25, and extending the dashed lines downwards on to a third set of axes. You an see that P is a maximum point and Q is a minimum point. What an you say about the gradient of at these points: is it positive, negative or zero? The gradient of any point on the urve is given This is written or f"(x), and is alled the seond derivative. It is found by differentiating the funtion a seond time. -1 ::r Ul n a. = a. <" A The seond derivative, is not the same as ( J. EAMPLE 5.14 Given that y = x 5 + 2x, SOLUTION dy = 5x dx d2y- 3 dx 2-2x. d2 dx Using the seond derivative You an use the seond derivative to identify the nature of a stationary point, instead oflooking at the sign of either side of it. Stationary points Notie that at P, =. This tells you that the is zero and dereasing. It must be going from positive to negative, so P is a maximum point (see figure 5.26). At Q, = and >. This tells you that the gradient, is zero and inreasing. It must be going from negative to positive, so Q is a minimum point (see figure 5.27).

21 d2y - > d_x2 atq ;::; Ill ;::; Ill Ill. i3 d2y - < d_x2 atp Figure 5.26 Figure 5.27 The next example illustrates the use of the seond derivative to identify the nature of stationary points. EAMPLE 5.15 Given that z, y = 2x 3 + 3x 2-12x z (i) find and find the values of x for whih (ii) find the value of = at eah stationary point and hene determine its nature (iii) find they values of eah of the stationary points (iv) sketh the urve given by y= 2x 3 + 3x 2-12x. SOLUTION lil dy = 6x 2 + 6x- 12 dx = 6(x 2 + x- 2) z = 6(x+ 2)(x- 1) = when x=-2 or x= 1. d2y liil dx 2 = 12x + 6. d2y When x= -2, dx 2 = 12 x (-2) + 6 = -18. d2y - < maximum. 2 dx When x= 1, liiil When x = -2, d 2 y = 6( ) = 18. dx 2 d 2 y dx 2 > y = 2(-2) 3 + 3(-2) 2-12(-2) =2 so (-2, 2) is a maximum point. When x= 1, y = =-7 so (1, -7) is a minimum point.

22 (iv) y Figure '=" Ill Cll n :I Cl. Cl. ;;: Cll A Remember that you are looking for the value of at the stationary point. Note On oasions when it is diffiult or laborious to find that you an always determine the nature of a stationary point by looking at the sign points just either side of it. for d2 Take are when J =. Look at these three graphs to see why. y =xj dy dy dx = 3x 2 : at (, ) dx = d2y d2y dx 2 = 6x: at (, ) dx 2 = Figure 5.29 dy dy dx = 4x 3 : at (, ) dx = Figure 5.3

23 y ;: «<. ID Ill ::: Figure 5.31 EERCISE 5F d d You an see that for all three of these funtions both 2 are zero at the on gm. z z Consequently, if both are zero at a point, you still need to hek the values of either side of the point in order to determine its nature. z 1 For eah of the following funtions, find and (i) y= x3 (ii) y=x 5 (iii) y= 4x 2 (iv) y= x- 2 3 (v) y=x 2 (vi) y= x 4 _1._ x3 2 Find any stationary points on the urves of the following funtions and identify their nature. m y = x 2 + 2x + 4 (iii) y= x 3-3x (v) y = x 4 + x 3-2x 2-3x + 1 (vii) y = 16x + 3 (ix) y = 6x- x 2 3 You are given z. that y= x 4-8x 2 m Find. d2y (ii) (ii) y = 6x- x 2 (iv) y = 4x 5 - Sx 4 (vi) y =x +l ( """ ) VIII y= - (iii) Find any stationary points and identify their nature. (iv) Hene sketh the urve. 1581

24 4 Given that y= (x-1) 2 (x- 3) (i) multiply out the right-hand side and (ii) find the position and nature of any stationary points (iiil sketh the urve. 5 Given that y= x 2 (x- 2) 2 m multiply out the right-hand side and find :t (iil find the position and nature of any stationary points (iiil sketh the urve. m )( ;; ;., Cl1 6 The funtion y= px 3 + qx 2, where p and q are onstants, has a stationary point at (1, -1). m Using the fat that ( 1, -1) lies on the urve, form an equation involving pand q.!iil Differentiate y and, using the fat that ( 1, -1) is a stationary point, form another equation involving p and q.!iiil Solve these two equations simultaneously to find the values of p and q. 7 You are given f(x) = 4.x2 + l. (i) Find f'(x) and f"(x).!iil Find the position and nature of any stationary points. 8 For the funtion y = x- 4-fx, dy d 2 y!il find - and - dx dx 2!iil find the o-ordinates of the stationary point and determine its nature. 9 The equation of a urve is y = 6-fx - x Fx. Find the x o-ordinate of the stationary point and show that the turning point is a maximum o For the urve x 2-1 Ox 2, (i) find the values of x for whih y =!iil show that there is a minimum turning point of the urve when x = 6 and alulate the y value of this minimum, giving the answer orret to 1 deimal plae.

25 ;; CO ;; Gl i5 Appliations EAMPLE 5.16 There are many situations in whih you need to find the maximum or minimum value of an expression. The examples whih follow, and those in Exerise 5G, illustrate a few of these. Kelly's father has agreed to let her have part of his garden as a vegetable plot. He says that she an have a retangular plot with one side against an old wall. He hands her a piee of rope 5 m long, and invites her to mark out the part she wants. Kelly wants to enlose the largest area possible. What dimensions would you advise her to use? SOLUTION Let the dimensions of the bed be x m x y m as shown in figure ym 5m xm Figure 5.32 ym xm The area, A m 2, to be enlosed is given by A = xy. Sine the rope is 5m long, 2x+ y= 5 or y= 5-2x. Writing A in terms of x only A= x(5-2x) = 5x- 2x 2 To maximise A, whih is now written as a funtion of x, you differentiate A with respet to x da dx = 5-4x. At a stationary point, =, so 5-4x = 5 = 4 = d 2 A h.... dx 2 = -4 => t e turnmg pomt IS a maximum. The orresponding value of y is 5-2( 1.25) = 2.5. Kelly should mark out a retangle 1.25 m wide and 2.5 m long.

26 EAMPLE 5.17 A stone is projeted vertially upwards with a speed of 3 m s- 1. Its height, hm, above the ground after t seonds (t < 6) is given by: h = 3t- St 2 "I) p m d dh -d an d -d2h t dt 2. ( (ii) Find the maximum height reahed. (iiil Sketh the graph of h against t. SOLUTION m = 3- lot. d 2 h dt2 = -1. ( > F or as t a t wnary. pom. t 'dt dh = 3- lot= 1(3- t) = t=3. d2h h.... dt 2 < t e statwnary pomt 1s a maximum. The maximum height is h = 3(3)- 5(3) 2 = 45m. (iii) h (metres) t (seonds) Figure 5.33 Note For a position-time graph, suh as this one, the gradient, is the aeleration. is the veloity and

27 EERCISE 5G 1 A farmer wants to onstrut a temporary retangular enlosure of length xm and width y m for his prize bull while he works in the field. He has 12 m of fening and wants to give the bull as muh room to graze as possible. ':: Ill ':: Gl = (i) Write down an expression for y in terms of x. (ii) Write down an expression in terms of x for the area, A, to be enlosed. (iiil and and so find the dimensions of the enlosure that give the bull the maximum area in whih to graze. State this maximum area. 2 A square sheet of ard of side 12 m has four equal squares of side xm ut from the orners. The sides are then turned up to make an open retangular box to hold drawing pins as shown in the diagram. xm (i) Form an expression for the volume, V, of the box in terms of x. (iil Find: is 2m. and show that the volume is a maximum when the depth 3 The sum of two numbers, x and y, is 8. m Write down an expression for y in terms of x. (iil Write down an expression for S, the sum of the squares of these two numbers, in terms of x. (iiil By onsidering find the least value of the sum of their squares. 4 A new hildren's slide is to be built with a ross-setion as shown in the diagram. A long strip of metal 8 m wide is available for the shute and will be bent to form the base and two sides. The designer thinks that for maximum safety the area of the ross-setion should be as large as possible.

28 ross-setion..! x rn y rn m >C <» n iii' m Cl1 C) m Write down an equation linking x and y. (iil Using your answer to part (i), form an expression for the ross-setional area, A, in terms of x. (iiil By onsidering safe as possible. find the dimensions whih make the slide as 5 A arpenter wants to make a box to hold toys. The box is to be made so that its volume is as large as possible. A retangular sheet of thin plywood measuring 1.5 m by 1 m is available to ut into piees as shown. x m x m!m x m xm 1.5m (il Write down the dimensions of one of the four retangular faes in terms of x. (iil Form an expression for the volume, V, of the made-up box, in terms of x. (. 1 F. ddv dd2 V m m dx an dx 2. (ivl Hene find the dimensions of a box with maximum volume, and the orresponding volume. 6 A piee of wire 16m long is ut into two piees. One piee is 8xm long and is bent to form a retangle measuring 3xm by xm. The other piee is bent to form a square. (i) Find in terms of x: (a) the length of a side of the square (b) the area of the square. (iil Show that the ombined area of the retangle and the square is A m 2 where A= 7x 2-16x+ 16. (iiil Find the value of x for whih A has its minimum value. (ivl Find the minimum value of A. [MEI]

29 7 A piee of wire 3 m long is going to be made into two frames for blowing bubbles. The wire is to be ut into two parts. One part is bent into a irle of radius r m and the other part is bent into a square of side xm = Q lil Write down an expression for the perimeter of the irle in terms of r, and hene write down an expression for r in terms of x. (iil Show that the ombined area, A, of the two shapes an be written as A = ( 4 + 1t) x 2-6x t liiil Find the lengths that must be ut if the area is to be a minimum. 8 A ylindrial an with a lid is to be made from a thin sheet of metal. Its height is to be hem and its radius rem. The surfae area is to be 25n m 2 (i) Find h in terms of r. (iil Write down an expression for the volume, V, of the an in terms of r.. dv d 2 V (iiil Fmd -d and -- r dr 2 (ivl Use your answers to part (iiil to show that the an's maximum possible volume is 169 m 3 (to 3 signifiant figures), and find the orresponding dimensions of the an. 9 Charlie wants to add an extension with a floor area of 18m 2 to the bak of his house. He wants to use the minimum possible number of briks, so he wants to know the smallest perimeter he an use. The dimensions, in metres, are x and y as shown. y (il Write down an expression for the area in terms of x and y. (iil Write down an expression, in terms of x and y, for the total length, T, of the outside walls. (iiil Show that T = 2x+ 18. (ivl Find dt and d 2 T. dx dx 2 (vi Find the dimensions of the extension that give a minimum value of T, and onfirm that it is a minimum.

30 1 A fish tank with a square base and no top is to be made from a thin sheet of toughened glass. The dimensions are as shown. m )C n iii' Ul Cl (il Write down an expression for the volume V in terms of x and y. (iil Write down an expression for the total surfae area A in terms of x and y. The tank needs a apaity of.5 m 3 and the manufaturer wishes to use the minimum possible amount of glass. (iiil Dedue an expression for A in terms of x only... da d 2 A hvl Fmd -d and - x dx 2. (v) Find the values of x and y that use the smallest amount of glass and onfirm that these give the minimum value. 11 A losed retangular box is made of thin ard, and its length is three times its width. The height is hem and the width is x m. (i) The volume of the box is 972 m 3. Use this to write down an expression for h in terms of x. (ii) Show that the surfae area, A, an be written as A= 6x (iiil Find Find d 2 dx value of A. and use it to fmd a stationary point. use it to verify that the stationary point gives the minimum (ivl Hene find the minimum surfae area and the orresponding dimensions of the box.

31 12 A garden is planned with a lawn area of 24m 2 and a path around the edge. The dimensions of the lawn and path are as shown in the diagram. lm!m H H ym lawn L r l.5m x m (i) Write down an expression for y in terms of x. liil Find an expression for the overall area of the garden, A, in terms of x. liiil Find the smallest possible overall area for the garden. 13 The diagram shows the ross-setion of a hollow one and a irular ylinder. The one has radius 6 m and height 12 m, and the ylinder has radius rem and height hem. The ylinder just fits inside the one with all of its upper edge touhing the surfae of the one. 12m 6m m Express h in terms of r and hene show that the volume, V m 3, of the ylinder is given by V= 12rr 2-2rr 3 liil Given that r varies, find the stationary value of V [Cambridge AS & A Level Mathematis 979, Paper 1 QS November 25]

32 The hain rule -t ":1" Ill n ":1" Dl :; m- Wh.... b dv d dh, atmtormatwms given y dh an dt Wh.. b d V dh, atmtormatwms given y dh x dt: How would you differentiate an expression like y=.jx 2 +1? I Your firstthought may be to write it as y = ( x 2 + 1) 2 and then get rid of the brakets, but that is not possible in this ase beause the power is not a positive integer. Instead you need to think of the expression as a omposite funtion, a 'funtion of a funtion'. You have already met omposite funtions in Chapter 4, using the notation g[f(x)] or gf(x). In this hapter we all the first funtion to be applied u(x), or just u, rather than f(x). In this ase, and u=x y=._(u = This is now in a form whih you an differentiate using the hain rule. Differentiating a omposite funtion To find Z for a funtion of a funtion, you onsider the effet of a small hange in x on the two variables, y and u, as follows. A small hange 8x in x leads to a small hange 8u in u and a orresponding small hange 8y in y, and by simple algebra, 8y 8y 8u -=-bx 8u 8x"

33 ;. ::: EAMPLE 5.18 In the limit, as, oy dy oy dy d ou du dx'ou duan dx and so the relationship above beomes dy _ dy du dx- dux dx' This is known as the hain rule. Differentiate y= (x 2 + 1) 2 I SOLUTION As you saw earlier, you an break down this expression as follows. Differentiating these gives and dy 1 _l u ==== du J x du = 2x. dx By the hain rule dy dy du dx =dux dx 1 x 2x 2-J x Jx2 +1 A Notie that the answer must be given in terms of the same variables as the question, in this ase x and y. The variable u was your invention and so should not appear in the answer. You an see that effetively you have made a substitution, in this ase u = x This transformed the problem into one that ould easily be solved. Note Notie that the substitution gave you two funtions that you ould differentiate. Some substitutions would not have worked. For example, the substitution u = x 2, would give you 1 y= (u+ 1) 2 and u= x 2. You would still not be able to differentiate y, so you would have gained nothing.

34 EAMPLE 5.19 Use the hain rule to when y= (x 2-2) 4 SOLUTION Let u = x 2 - and du=2x dx 2, then y= u 4 dy =4u3 du = 4(x 2-2) 3 -t ::r " n ::r Ill s m dy dy du dx =dux dx =4(x 2-2) 3 x2x = 8x (x 2-2) 3. (i) A student does this question by first multiplying out (x 2-2) 4 to get a polynomial of order 8. Prove that this heavy-handed method gives the same result. A With pratie you may find that you an do some stages of questions like this in your head, and just write down the answer. If you have any doubt, however, you should write down the full method. Differentiation with respet to different variables The hain rule makes it possible to differentiate with respet to a variable whih does not feature in the original expression. For example, the volume V of a sphere of radius r is given by V = the rate of hange of volume with radius, Differentiating this with respet to r gives = 4rr 2. However you might be more interested in finding the rate of hange of volume with time, t. To find this, you would use the hain rule: dv = dv x dr dt dr dt dv = 4 dt rr2dr dt You have now differentiated V with respet to t. Notie that the expression for dv. dr dt rnludes dt' the rate of inrease of radius with time. The use of the hain rule in this way widens the sope of differentiation and this means that you have to be areful how you desribe the proess.

35 . Ill ': f G) :t 'Differentiate y = x 2 ' ould mean differentiation with respet to x, or t, or any other variable. In this book, and others in this series, we have adopted the onvention that, unless otherwise stated, differentiation is with respet to the variable on the right-hand side of the expression. So when we write 'differentiate y= x 2 ' or simply 'differentiate x 2 ', it is to be understood that the differentiation is with respet to x. The expression 'inreasing at a rate of is generally understood to imply differentation with respet to time, t. EAMPLE 5.2 The radius r m of a irular ripple made by dropping a stone into a pond is inreasing at a rate of 8 m s- 1. At what rate is the area Am 2 enlosed by the ripple inreasing when the radius is 25 m? SOLUTION A =nr da =2nr dr The question is asking for the rate of hange of area with respet to time. Now da = da x dr dt dr dt dr = 2nr dt' Whenr = 25and = 8 = 2rrx25x8 = 126m 2 s- 1. no I

36 EERCISE 5H 1 Use the hain rule to differentiate the following funtions. m y=(x+2) 3 (iv) y= (x 3 + 4) 5 d (vii) y= (x 2-1) 2 2 Giventhaty=(3x-5) 3 liil y=(2x+3) 4 (v) y= (3x+ 2)- 1 (viii) y= + xf (iiil y= (x 2-5) 3 (vi) y = (x2 3? (ix) y = ( -Fx - 1 r lil (iil find the equation of the tangent to the urve at (2, 1) liiil show that the equation of the normal to the urve at ( 1, -8) an be written in the form 36y+ x+ 287 =. 3 Given thaty= (2x-1) 4 lil find dy dx (iil find the o-ordinates of any stationary points and determine their nature (iiil sketh the urve. 4 Given that y= (x 2 - x- 2) 4 lil (iil find the o-ordinates of any stationary points and determine their nature liiil sketh the urve. 5 The length of a side of a square is inreasing at a rate of.2 m s- 1 At what rate is the area inreasing when the length of the side is 1 m? 6 The fore F newtons between two magneti poles is given by the formula F = r m is their distane apart. SOOr Find the rate of hange of the fore when the poles are.2 m apart and the distane between them is inreasing at a rate of.3ms The radius of a irular fungus is inreasing at a uniform rate of 5 m per day. At what rate is the area inreasing when the radius is 1 m?

37 . Ill. KEY POINTS 1 y= kxn ::::} dy = knxn-1 dx y= ::::} dy = dx 2 y= f(x) + g(x) ::::} z 3 Tangent and normal at (x 1, y 1 ) = f'(x) + g'(x). Gradient of tangent, m 1 =value of z Gradient of normal, m 2 =- Equation of tangent is } Where k, n and are onstants. when x = x 1 Equation of normal is 4 At a stationary point, z =. The nature of a stationary point an be determined by looking at the sign of the gradient just either side of it. Maximum Minimum 5 The nature of a stationary point an also be determined by onsidering the. d2y sign of dx 2. If d2y h... dx2 <, t e pomt IS a maximum. d2y If dx2 >, the point is a minimum. z 6 =, hek the values of on either side of the point to determine its nature... dy _ dy du 7 Cham rule. dx - du x dx.

38 Integration Many small make a great. f) In what way an you say that these four urves are all parallel to eah other? y Chauer :a Ill ;: :r CO Q. = = ; ; = Reversing differentiation In some situations you know the gradient funtion, z, and want to find the z funtion itself, y. For example, you might know that = 2x and want to find y. z You know from the previous hapter that if y= x 2 then = 2x, but y = x 2 + 1, y = x 2-2 and many other funtions also give z = 2x. Suppose that f(x) is a funtion with f'(x) = 2x. Let g(x) = f(x)- x 2 Then g'(x) = f'(x)- 2x= 2x- 2x= for all x. So the graph of y= g(x) has zero gradient everywhere, i.e. the graph is a horizontal straight line. Thus g(x) = (a onstant). Therefore f(x) = x 2 +. z All that you an say at this point is that if = 2x then y = x 2 + where is desribed as an arbitrary onstant. An arbitrary onstant may take any value.