## Local Linear Approximation; Differentials Solutions To Selected ...

Oct 2, 2011 ... Local Linear Approximation; Differentials. Solutions To Selected Problems. Calculus 9 th. Edition Anton, Bivens, Davis. Matthew Staley. October ...

Local Linear Approximation; Differentials Solutions To Selected Problems Calculus 9th Edition Anton, Bivens, Davis Matthew Staley October 2, 2011

1. Confirm that the stated formula is the local linear approximation at x0 = 0. (a) (1 + x)15 ≈ 1 + 15x d Let f (x) = (1 + x)15 , then f 0 (x) = 15(1 + x)14 dx (x) = 15(1 + x)14 . Now use the local linear approximation formula for f (x):

f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ) = (1 + x0 )15 + 15(1 + x0 )14 (x − x0 ) = (1 + 0)15 + 15(1 + 0)14 (x − 0) = (1)15 + 15(1)14 (x) = 1 + 15x (b) tan(x) ≈ x Let f (x) = tan(x), then f 0 (x) = sec2 (x). f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ) = tan(x0 ) + sec2 (x0 )(x − x0 ) = tan(0) + sec2 (0)(x − 0) = 0 + 12 (x) = x (c)

1 1+x

≈1−x

Let f (x) =

1 1+x

1 = (1 + x)−1 , then f 0 (x) = −1(1 + x)−2 (1) = − (1+x) 2

f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ) 1 1 = − (x − x0 ) 1 + x0 (1 + x)2 1 1 = (x − 0) − 1 + 0 (1 + 0)2 = 1−x

1

2. Confirm that the stated formula is the local linear approximation of f at x0 = 1, where 4x = x − 1 (a) f (x) = x4 ;

(1 + 4x)4 ≈ 1 + 4 4 x

Use the local linear approximation on f (x) = x4 , f 0 (x) = 4x3 , x0 = 1 f (x) ≈ f (1) + f 0 (1)(x − 1) x4 ≈ (1)4 + 4(1)3 (x − 1) = 1 + 4(x − 1) Set 4x = x − 1; then x = 1 + 4x. → (1 + 4x)4 ≈ 4 4 x

(b) f (x) =

1 2+x

;

1 3+4x

1 3

− 91 4 x.

1 = (2 + x)−1 2+x f 0 (x) = −1(2 + x)−2 (1) 1 =− (2 + x)2 f (x) =

Use the local linear approximation on f (x), f 0 (x), x0 = 1 f (x) ≈ f (1) + f 0 (1)(x − 1) 1 1 1 ≈ − (x − 1) 2+x 2 + 1 (2 + 1)2 1 1 = − (x − 1) 3 9 Since 4x = x − 1, we can add 3 to both sides to obtain 3 + 4x = x + 2. →

1 1 1 1 = ≈ − 4x 2+x 3 + 4x 3 9

2

3. Use an appropriate local linear approximation to estimate the value of the given quantity. (a) (3.02)4 By inspection we see that can be f (x) = (x)4 . Then x0 = 3 and 4x = 0.02, with f 0 (x) = 4x3 . f (3 + 4x) ≈ f (3) + f 0 (3) 4 x f (3.02) ≈ (3)4 + 4(3)3 (0.02) 2 = 81 + 108 100 1 = 81 + 108 50 54 = 81 + 25 4 = 81 + 2 25 4 = 83 = 83.16 25 (3.02)4 ≈ 83.16 with the actual value to 4 decimals places of 83.181696.

(b)

65

The closest perfect square to this is f 0 (x) = 2√1 x , x0 = 64 and 4x = 1.

64 = 8. So f (x) =

x,

f (64 + 4x) = f (64) + f 0 (64) 4 x √ √ 1 65 ≈ 64 + √ (1) 2 64 1 =8+ 16 1 = 8 = 8.0625 16 √

65 ≈ 8.0625 with the actual value to 5 decimal places of 8.06225. 3

(c) sin(0.1) f (x) = sin(x), f 0 (x) = cos(x), x0 = 0, and 4x = 0.1. f (0 + 4x) ≈ f (0) + f 0 (0) 4 x sin(0.1) ≈ sin(0) + cos(0)(0.1) = 0 + 1(0.1) = 0.1 Actual value to 3 decimal places is 0.099. (d) tan(0.2) f (x) = tan(x), f 0 (x) = sec2 (x), x0 = 0, and 4x = 0.2. f (0 + 4x) ≈ f (0) + f 0 (0) 4 x tan(0.2) ≈ tan(0) + sec2 (0)(0.2) = 0 + 12 (0.2) = 0.2 Actual value to 3 decimal places is 0.202. (e) cos(31◦ ) 31◦ is close to 30◦ = π/6, so choose f (x) = cos(x), f 0 (x) = − sin(x), x0 = π/6 and 4x = 1◦ = π/180. f (30◦ + 4x) ≈ f (π/6) + f 0 (π/6) 4 x cos(31◦ ) ≈ cos(π/6) − sin(π/6)(π/180) √ = ( 3/2) − (1/2)(π/180) √ 3 π = − ≈ 0.8573 2 360 Actual value to 5 decimal places is 0.85716.

4

4. Find formulas for dy and 4y. (a)

y = x2 − 2x + 1 dy = 2x − 2 dx dy = (2x − 2)dx 4y = f (x + 4x) − f (x) = [(x + 4x)2 − 2(x + 4x) + 1] − [x2 − 2x + 1] = x2 + 2x 4 x + (4x)2 − 2x − 2 4 x + 1 − x2 + 2x − 1 = 2x 4 x + (4x)2 − 2 4 x

(b)

y = sin(x) dy = cos(x) dx dy = cos(x)dx 4y = f (x + 4x) − f (x) = sin(x + 4x) − sin(x)

5

5. Find the differential dy. (a)

y = 4x3 − 7x2 dy = d(4x3 − 7x2 ) = 12x2 dx − 14xdx = (12x2 − 14x)dx

(b)

y = x cos(x) dy = d(x cos(x)) = xd cos(x) + cos(x)dx = x(− sin(x))dx + cos(x)dx = (−x sin(x) + cos(x))dx

(c)

y=

1 = x−1 x

dy = d(x−1 ) = −1x−2 dx = −

1 dx x2

6

(d)

√ y = x 1 − x = x(1 − x)1/2 dy = d(x(1 − x)1/2 ) = x d(1 − x)1/2 + (1 − x)1/2 dx   1 −1/2 =x (1 − x) (−1)dx + (1 − x)1/2 dx 2 √ x =− √ dx + 1 − x dx 2 1−x   √ x = 1−x− √ dx 2 1−x √   √ 2 1−x x 1−x· √ − √ = dx 2 1−x 2 1−x   2(1 − x) − x √ = dx 2 x   2 − 3x √ = dx 2 1−x

(e)

y = (1 + x)−17 dy = d(1 + x)−17 = −17(1 + x)−18 dx = −

17 dx (1 + x)18

7

6. Use the differential dy to approximate 4y when x changes as indicated. √ (a) y = 3x − 2; from x = 2 to x = 2.03, → dx = 0.03 1 dy = (3x − 2)−1/2 (3) dx 2 3 = √ dx 2 3x − 2 3 dx 4y ≈ dy = √ 2 3x − 2 3 = p (0.03) 2 3(2) − 2 3 = √ (0.03) 2 4   3 3 = 4 100 =

(b) y =

x ; x2 +1

9 = 0.0225 400

from x = 2 to x = 1.96, → dx = −0.04

(x2 + 1)dx − x(2xdx) (x2 + 1)2 (x2 − 2x2 + 1)dx = (x2 + 1)2 1 − x2 = 2 dx (x + 1)2

dy =

1 − 22 (−0.04) (22 + 1)2   1−4 4 = − 52 100   3 1 − =− 25 25

4y ≈ dy =

=

3 3 = = 0.0048 2 25 625 8

7. The side of a square is measured to be 10 ft, with a possible error of ± 0.1 ft. (a) Use differentials to estimate the error in the calculated area. Let x be the side length of the square. Then area = A = x2 ; dA = 2xdx. In this case x = 10 and dx = ± 0.1. Thus, dA = 2xdx = 2(10)(± 0.1) 1 = ± 20 = ± 2f t 10 (b) Estimate the percentage errors in the side and the area. The relative error in the side, or x, is ≈ dx = ±100.1 = ± x ± 0.01. So the percentage error in x is ≈ ± 1%.

1 1 10 10

1 100

=

The relative error in the Area, or A is ≈ dA = 2xdx = 2 dx = 2(± 0.01) = A x2 x ± 0.02. So the percentage error in x is ≈ ± 2%

8. The electrical resistance R of a certain wire is given by R = k/r2 , where k is a constant and r is the radius of the wire. Assuming that the radius r has a possible error of ± 5%, use differentials to estimate the percentage error in R. (Assume k is exact). . So dR = −2kr−3 dr = − 2k dr. Now find We are asked to find dR R r3 dr with the fact that r = ± 0.05. 2k 1 dR = − 3 dr R r R 2k r2 = − 3 dr r k dr = −2 r = −2(± 0.05) = ± 0.10 So the percentage error in R is ≈ 10%. 9

dR R

along