P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 CHAPTER 19 Differentiation of Polynomials Objectives To understand the concept of limit. To understand the definition of differentiation. To understand and use the notation for the derivative of a polynomial function. To be able to find the gradient of a curve of a polynomial function by calculating its derivative. To apply the rules for differentiating polynomials to solving problems. In the previous chapter the rate of change of one quantity with respect to a second quantity has been considered. In this chapter a technique will be developed for calculating the rate of change for polynomial functions. To illustrate this an introductory example will be considered. On planet X,anobject falls a distance of y metres in t seconds, where y = 0.8t2. (Note: On Earth the commonly used model is y = 4.9t2.) Can a general expression for the speed of such an object after t seconds be found? In the previous chapter it was y found that the gradient of the curve y = 0.8t 2 at a given point P can be approximated by finding the gradient of a chord PQ, where Q is a point on the curve as close Q as possible to P. The gradient of chord PQ approximates SAMPLEP the speed of the object at P. The ‘shorter’ this chord is made the better 0 5 (5 + h) t the approximation. 536 Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 Chapter 19 — Differentiation of Polynomials 537 Let P be the point on the curve where t = 5. Let Q be the point on the curve corresponding to h seconds after t = 5, i.e. Q is the point on the curve where t = 5 + h. 0.8 (5 + h)2 − 52 The gradient of chord PQ = (5 + h) − 5 0.8 (5 + h)2 − 52 = h = 0.8(10 + h) The table gives the gradient for different values of h. h Gradient of PQ Use your calculator to check these. If values of h of smaller and smaller magnitude are 0.7 8.56 taken, it is found that the gradient of chord PQ gets 0.6 8.48 = closer and closer to 8. At the point where t 5 the 0.5 8.40 gradient is 8. Thus the speed of the object at the 0.4 8.32 = moment t 5is8metres per second. 0.3 8.24 = The speed of the object at the moment t 5isthe 0.2 8.16 limiting value of the gradients of PQ as Q approaches P. 0.1 8.08 A formula for the speed of the object at any time t is required. Let P be the point with coordinates (t, 0.8t2)onthecurve and Q be the point with coordinates (t + h, 0.8(t + h)2). 0.8 (t + h)2 − t2 The gradient of chord PQ = (t + h) − t = 0.8(2t + h) From this an expression for the speed can be found. Consider the limit as h approaches 0, that is, the value of 0.8(2t + h)ash becomes arbitrarily small. The speed at time t is 1.6t metres per second. (The gradient of the curve at the point corresponding to time t is 1.6t.) This technique can be used to investigate the gradient of similar functions. 19.1 The gradient of a curve at a point, and the gradient function Consider the function f : R →R, f (x) = x2. y The gradient of the chord PQ in the f (x) = x2 adjacent graph (a + h, (a + h)2) Q + 2 − 2 = (a h) a a + h − a a2 + 2ah + h2 − a2 SAMPLE= 2 a + h − a P(a, a ) x = 2a + h 0 and the gradient at P can be seen to be 2a. The limit of (2a + h)ash approaches 0 is 2a. It can be seen that there is nothing special about a.Soifx is a real number a similar formula holds. Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 538 Essential Mathematical Methods1&2CAS The gradient of the function y = x2 at any point x is equal to 2x. It is said that 2x is the derivative of x2 with respect to x or, more briefly, the derivative of x2 is 2x. This itself is clearly the rule for a function of x and we refer to this function as the gradient (or derived) function. The straight line that passes through P and has gradient 2a is called the tangent to the curve at P. From the discussion at the beginning of the chapter it can be seen that the derivative of 0.8t2 is 1.6t. Example 1 By first considering the gradient of the chord QP, find the gradient of y = x2 − 2x at the point Q with coordinates (3, 3). Solution y Consider chord PQ: y = x2 – 2x Gradient of PQ + 2 − + − (3 + h, (3 + h)2 – 2(3 + h)) = (3 h) 2(3 h) 3 P 3 + h − 3 + + 2 − − − = 9 6h h 6 2h 3 3 + h − 3 Q(3, 3) + 2 = 4h h h x = 4 + h 0 2 Consider h as it approaches zero. The gradient at the point (3, 3) is 4. Example 2 Find the gradient of chord PQ and hence the derivative of x2 + x. y y = x2 + x P(x + h, (x + h)2 + (x + h)) Q(x, x2 + x) SAMPLEx –1 0 Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 Chapter 19 — Differentiation of Polynomials 539 Solution (x + h)2 + (x + h) − (x2 + x) The gradient of chord PQ = x + h − x 2 + + 2 + + − 2 − = x 2xh h x h x x h + 2 + = 2xh h h h = 2x + h + 1 From this it is seen that the derivative of x2 + x is 2x + 1. The notation for limit of 2x + h + 1ash approaches 0 is lim 2x + h + 1. h→0 The derivative of a function with rule f (x)may be found by first finding an expression for the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) and then finding the limit of this expression as h approaches 0. Example 3 By first considering the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) for the curve f (x) = x3, find the derivative of x3. Solution f (x) = x3 f (x + h) = (x + h)3 f (x + h) − f (x) The gradient of chord PQ = (x + h) − x + 3 − 3 = (x h) x (x + h) − x (x + h)3 − x3 The derivative of f (x) = lim h→0 (x + h) − x x3 + 3x2h + 3h2x + h3 − x3 = lim h→0 h 3x2h + 3h2x + h3 = lim h→0 h = lim 3x2 + 3hx + h2 h→0 SAMPLE= 3x2 We have found that the derivative of x3 is 3x2. The following example provides practice in determining limits. Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 540 Essential Mathematical Methods1&2CAS Example 4 Find: 3x2h + 2h2 a lim 22x2 + 20xh + h b lim h→0 h→0 h c lim 3x d lim 4 h→0 h→0 Solution 3x2h + 2h2 a lim 22x2 + 20xh + h = 22x2 b lim = lim 3x2 + 2h h→0 h→0 h h→0 = 3x2 c lim 3x = 3x d lim 4 = 4 h→0 h→0 Using the TI-Nspire b 11 Use Define√ ( ) with the function f (x) = x and then complete as shown. Using the Casio ClassPad For Example 4b, enter and highlight the expression (3x2h + h2)/h then from the keyboard select 2D—Calc— and enter h and SAMPLE0inthe spaces provided, as shown. Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard P1: FXS/ABE P2: FXS 9780521740524c19.xml CUAU021-EVANS August 23, 2008 14:3 Chapter 19 — Differentiation of Polynomials 541 Exercise 19A 1 A space vehicle moves so that the distance travelled over its first minute of motion is given by y = 4t4,where y is the distance travelled in metres and t the time in seconds. By finding the gradient of the chord between the points where t = 4 and t = 5, estimate the speed of the space vehicle when t = 5. 2 A population of insects grows so that the population, P,attime t (days) is given by P = 1000 + t2 + t,where t > 0. By finding the gradient of the chord between the points where t = 3 and t = 3 + h, find an estimate for the rate of growth of the insect population at time t = 3. Example 4 3 Find: 2x2h3 + xh2 + h 3x2h − 2xh2 + h a lim b lim c lim 20 − 10h h→0 h h→0 h h→0 30hx2 + 2h2 + h d lim e lim 5 h→0 h h→0 Example 3 4 Find: (x + h)2 + 2(x + h) − (x2 + 2x) a lim i.e.