2A_section3_9

# 2A_section3_9 - Linear Approxima-on of a func-on...

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Unformatted text preview: Linear Approxima-on of a func-on Sec\$on 3.9 at point (0,1): y=1+x At 0, the func\$on and the tangent have the same value. Around 0, the values are very close. f’(a) exists Let f be a diﬀeren\$able func\$on at a. What Iinear func\$on best approximates f near a? L(x)=mx+b (a,f(a)) y=f(x) The tangent line at the point (a,f(a)) is the best approxima-on for f near a The lineariza\$on of f at a is a func\$on L(x) whose graph is the tangent line of f at a. (a,f(a)) y=f(x) y=L(x) The lineariza-on of f at a is the best approxima-on for f near a Find the equa.on: The tangent line to the graph of f at the point (a,f(a)) has equa\$on y ­f(a) =m(x ­a) with m=f’(a). Write y = f(a) + f ’(a)(x ­a) . This is L(x)! Lineariza-on: L(x)=f(a)+f’(a)(x ­a) The lineariza\$on error in the approxima\$on: |f(x) ­L(x)| approxima\$on y=L(x) (tangent line) original func\$on y=f(x) L(x2) error in the approxima\$on: |f(x) ­L(x)| y=L(x) The closer x is to a the beRer approxima\$on L(x) is to f(x). 1 y=f(x) 1 x2 EXAMPLE 1: Approximate the value of sin(61o) Use the lineariza\$on of f at a: L(x)=f(a)+f’(a)(x ­a) Sin(a) cos(a) L(x)=sin(a)+cos(a)(x ­a) How do we choose a? a should be a point where we know both cos(a) and sin(a) (so that L(x) can be computed) EXAMPLE 1: Approximate the value of sin(61o) First guess: use a=0 L(x) = sin(a)+cos(a)(x ­a) = x sin(0)=0 cos(0)=1 0 ⎛ 61π ⎞ ⎛ 61π ⎞ 61π sin(61 ) = sin⎜ ≈ 1.06456 ⎟ ≈ L⎜ ⎟= ⎝ 180 ⎠ ⎝ 180 ⎠ 180 o Is this a good approxima-on? EXAMPLE 1: Approximate the value of sin(61o) Lineariza\$on at a=0 gives sin(61o ) ≈ 1.06456 Not a good approxima-on, given that sin(61o)=0.87462 error in the approxima\$on: Lineariza\$on at a=0: y=x original func\$on: y=sin(x) EXAMPLE 1: Approximate the value of sin(61o) Second guess: use a=60o= π / 3 3 1⎛ π ⎞ L(x) = sin(a)+cos(a)(x ­a) = + ⎜ x − ⎟ 2 2⎝ 3⎠ 3 /2 1/ 2 π /3 ⎛ 61π ⎞ ⎛ 61π ⎞ 3 1 ⎛ 61π π ⎞ sin(61 ) = sin⎜ +⎜ − ⎟ ≡ 0.87475 ⎟ ≈ L⎜ ⎟= ⎝ 180 ⎠ ⎝ 180 ⎠ 2 2 ⎝ 180 3 ⎠ o Is this a good approxima-on? EXAMPLE 1: Approximate the value of sin(61o) Lineariza.on at a= π / 3 gives sin(61o ) ≈ 1.87475 Very good approxima\$on, given that sin(61o)=0.87462 LINEARIZATION AT 0 LINEARIZATION AT π / 3 error in the approxima\$on: EXAMPLE 2: Estimate 10 We can use the lineariza\$on of f at a. The best choice for a is: A) a = 0 B) a = 1 C) a = 9 D) a = 16 f’(x)=(1/2)x ­1/2 so f’(9)=(1/2) 9 ­1/2=(1/2)(1/3) Very close to 10 Diﬀeren-al of a func-on If y=f(x), the diﬀeren\$al of f is dy = f ’(x) dx Example: If f(x)=sin(3x2+2)+2x then df= f ’(x) dx= [cos(3x2+2)6x+2] dx Compute the deriva-ve and mul-ply by dx Diﬀeren-al of a func-on If f(x)=2/x, the diﬀeren\$al df is equal to: a) df=  ­2/x2 b) df= [2/x] dx c) df= [ ­2/x2] dx Diﬀeren-al of a func-on: dy=f’(x)dx y=L(x) y=f(x) dy :change in L(x) Δ y : change in f(x) Δ x : change in x Diﬀeren-al of a func-on: dy=f’(x)dx Suppose Δx is a small change in x. Write Δx=dx. The line L(x) has slope f ’(x). We get: f’(x)= (change in L(x))/(change in x) dy :change in L(x) Δ y : change in f(x) Δ x : change in x dy= f ’(x) dx y=f(x) y=L(x) f ’(x)=dy / dx Lineariza\$on at 1 (1.01, 1.0201) Original func\$on f(x)=x2 Δx= change in x = 1.01 – 1 = .01 Δy= change in f(x) = (1.01)2 – 12 = 1.0201 ­1=.0201 dy= change in L(x) = [2(1.01) ­1] – [2(1) ­1] =1.02 ­ 1= .02 Check: dy=f’(1) dx= f ’(1) Δx = 2(1) (.01) = .02 Because Δx is small, dy is close to Δy! ...
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