10 9 8 20 7 18 18 16 6 16 14 14 5 12 12 18 1610 4 20 10 f(x) = x2 14 18 g(x) = 2·(x – 1) + 1 8 10 h(x) = sin(x) 12 16 8 9 q(x) = 1 14 3 10 8 6 f(x) = x2 12 6 g(x) = 2·(x – 1) + 1 7 8 10 18 6 16 6 4 14 20 5 8 2 12 4 18 10 4 2 f(x) = x 16 4 g(x) = 2·(x – 1) + 1 8 Lecture 14 :Linear Approximations and Differentials Consider a point on a smooth curve y = f(6x), say P = (a; f(a)), If we draw a tangent line to the curve at the point P146, we can see from the pictures below that as we zoom in towards the point P , the path 10 h(x) = sin(x) of the curve is very close to that of the tangent line. If we zoom in far enough the curve looks almost 2 linear near the9 point P . (See below the tangent to the curve y = x at the point (1; 1) and the tangent f(x) = x π to the curve y = sin x at the point ( ; 1). ) q(x) = 1 3 22 4 128 7 2 62 10 5 4 4 2 h(x) = sin(x) 2 15 – 10 – 5 5 10 15 q(x) = 1 3 8 f(x) = x – 2 1 6 3! 5! – 2! 3! – ! ! – 4 ! ! 3! 2! 5! 3! 7! 4! 9! – – – 2 2 2 2 2 2 2 2 15 – 10 – 5 – 1 1 2 5 10 15 1 –– 26 4 On the other– hand,3 if we pick a sharp point on a curve, the curve does not look linear near the point as we zoom in. See the graph of y = jxj below at the point (0; 0). – 4 f(x) = x – 8 – 5 2 – 6 15 – 10 – 5 – 10 – 2 5 10 15 3! 5! – 2! 3! – ! ! – 7 ! ! 3! 2! 5! 3! 7! 4! 9! – – – – 8 15 – 10 2 15 – 10 2 – 5 – 5 2 – 12 2 5 10 2 15 5 2 2 10 2 15 The difference between the nature of sharp points and points where the curve is smooth is differentiabil- ity. If a curve– 14– is2 differentiable– 1 on an interval containing the point x = a (smooth at x = a), the points on the curve, with x values close to a, are very close to the points on the tangent line to the curve at P . The equation of the tangent line to the curve y = f(x) at x = a is given by, 15 – 10 – 5 – 16 – 4 L(x) = f 0(a)(x − a) + f(a): 5 10 15 – 4 – 2– 2 – 18 1 – 2 – 6 – 8 – 6 – 4 – 3 – 10 – 2 – 12 3! 5! – 2! 3! – ! ! – 4 – 4– 6 ! ! 3! 2! 5! 3! 7! 4! 9! – 14 – 8 – – – – 16 – 5 2 2 2 – 8 2 2 2 2 2 – 10 – 4 – 6 – –610 – 1 – –127 – 12 – 8 – 14 – 6 – 8 – 14 – 16 – 16 – 2 – 8 ––1810 – 12 – 10 – 3 – 14 – 12 – 4 – 16 – 14 – 18 – 16 – 5 – 6 – 7 – 8 We can conclude that if f is differentiable in an interval containing a, then f(x) ≈ L(x) = f(a) + f 0(a)(x − a): This is called the linear approximation or Tangent Line Approximation to f(x) at x = a. The linear function, whose graph is the tangent line to the curve y = f(x) at x = a is called the Linearization of f at a. p Example (a) Find the linearization of the function f(x) = 3 x at a = 27. f(x) f 0(x) a f(a) f 0(a) L(x) = f(a) + f 0(a)(x − a) = p p (b) Use the linearization above to approximate the numbers 3 27:01 and 3 26:99. p (c) We can get an approximation for 3 x from the linearization of the function, L(x), above, for any x in the interval 26 ≤ x ≤ 28. We can see, from the tablep below, that the closer the value of x gets to 27, the better the approximation to the actual value of 3 x. p 3 pf(x) x From L(x) Actual Value x 3 26:5 26:5 2:9814815 2:9813650 p 3 26:9 26:9 2:996296 2:996292 p 3 26:99 26:99 2:9996296 2:9996296 p 3 27 27 3 3 p 3 27:01 27:01 3:0003704 3:0003703 p 3 27:1 27:1 3:0037037 3:0036991 p 3 27:5 27:5 3:0185185 3:0184054 p Error of approximation In fact by zooming in on the graph of f(x) = 3 x, you will see that p p j 3 x − L(x)j < 0:001 or − 0:001 < 3 x − L(x) < 0:001 2 when x is in the interval 26:5 ≤ x ≤ 27:5. Such bounds on the error are useful when using approximations. We will be able to derive such estimates later when we study Newton's method. p Example (a) Find the linearization of the function f(x) = x + 9 at a = 7. f(x) f 0(x) a f(a) f 0(a) L(x) = p p (b) Use the linearization above to approximate the numbers 16:03 and 15:98. Example (commonly used linearizations) (a) Find the Linearization of the functions sin θ and cos θ at θ = 0. f(θ) f 0(θ) a f(a) f 0(a) cos(θ) ≈ sin(θ) ≈ π π o π (b) Estimate the value of sin( 100 ), cos 95 and sin 2 = sin 90 . 3 New Notation : ∆y. Suppose f is differentiable on an interval containing the point a. Linear approximation says that the 6.5 function f can be approximated by 6 f(x) ≈ f(a) + f 0(a)(x − a) where a is fixed and x is a5.5 point (in the interval) nearby. This can gives us the following approximation for the change in function values, when we have a small change in the value of x: 5 f(x) − f(a) ≈ f 0(a)(x − a): 4.5 As before, we use ∆x to denote4 a small change in x values. In this case ∆x = x−a. and ∆y = f(x) − f(a) to denote the corresponding change in the values of y or f(x). This gives us: 3.5 ∆y ≈ f 0(a)∆x 3 where ∆y denotes the change in the value of f between two points a and a + ∆x. 2.5 g(x) = x 2 L(x) h(x) = 1 + 0.5·(x – 1) f(x) 1.5 f ' (a) x y (a, f(a)) 1 x 0.5 Example Approximate the change in the surface area of a spherical hot air balloon when the radius changes from 4 to 3.9 meters. (The surface area of a sphere of radius r is given by S = 4πr2.) – 3 – 2 – 1 1 2 3 4 5 0 – 0.5 ∆S ≈ S (a)∆r – 1 S(r) = 4πr2;S0(r) = 8πr: – 1.5 ∆S ≈ S0(4)∆r = 32π(−0:1) = −3:2π: – 2 Differentials, dy We also use the notation– 2.5 of differentials to denote changes in L(x), the linear approximation to f(x) near a. – 3 • The change in the function values on the curve y = f(x) as x changes from a to a + ∆x is denoted by ∆y as before. (∆– 3.5y = f(a + ∆x) − f(a)). • The differential dx –is4 defined as the change in x,(dx = ∆x.) • The differential dy is defined as the change in the values of the linear approximation L(x) as x changes from a to– 4.5a + ∆x; dy = ∆L = L(a + ∆x) − L(a) = f(a) + f 0(a)∆x − f(a) = f 0(a)∆x. Therefore – 5 dy = f 0(a)∆x in terms of dx: dy = f 0(a)dx: 4 6.5 6 5.5 5 4.5 4 3.5 The differential dy is a dependent variable, depending on the independent variable dx. Check out the difference between dy and3 ∆y on the graph below. 2.5 g(x) = x 2 L(x) h(x) = 1 + 0.5·(x – 1) f(x) 1.5 f ' (a) x = L = dy y (a, f(a)) 1 x = dx 0.5 We can rephrase our results on linear approximation as – 3 – 2 – 1 1 2 3 4 5 – 0.5 dy ≈ ∆y 4 Example Compare the– 1 values of ∆y and dy if y = 3x + 2x + 1 and x changes from 2 to 2:04. a =2 ∆y = f(a + ∆a) − f(a) = f(2:04) − f(2) = – 1.5 57:0367 − 53 = 4:0367 = ∆y ∆x = :04 – 2 0 – 2.5 dy = f (a)dx = (12(4) + 2)(:04) = 98(:04) f(x) =3 x4 + 2x + 1 f 0(x) = 12x3 + 2 = 3:92 = dy – 3 Example (a) Find the– 3.5 differential for the function y = 3 cos2 x. – 4 dy = f 0(x)dx = (6 cos x)(− sin x)dx = −6 cos x sin xdx (b) Use the differential– 4.5 to approximate the change in the values of the function f(x) = 3 cos2 x when we have a small change in the value of x, dx at x = π . – 5 4 π π π 1 1 f(x + ∆x) − f(x) ≈ dy = f 0( )dx = −6 cos( ) sin( )dx = −6p p dx = −3dx: 4 4 4 2 2 2 o 2 π π (c) Use differentials to estimate 3 cos (44 ) = 3 cos ( 4 − 180 ).