Complete Solutions: Exercise 8.1

All problems solved with colorful explanations!

Problem 1: Cube Root Approximation

Let f(x) = ³√x. Find the linear approximation at x = 27. Use the linear approximation to approximate ³√27.2.

Step 1: Find f(27)

f(27) = ³√27 = 3

Step 2: Find the derivative f'(x)

f(x) = x^(1/3)

f'(x) = (1/3)x^(-2/3) = 1/(3x^(2/3))

Step 3: Find f'(27)

f'(27) = 1/(3*(27)^(2/3)) = 1/(3*9) = 1/27 ≈ 0.037037

Step 4: Write the linear approximation formula

L(x) = f(a) + f'(a)(x - a)

L(x) = 3 + (1/27)(x - 27)

Step 5: Approximate ³√27.2

L(27.2) = 3 + (1/27)(27.2 - 27) = 3 + (0.2)/27 ≈ 3.007407

Actual value: ³√27.2 ≈ 3.007408 (our approximation is excellent!)

Try it yourself!

Approximate ³√:

Approximation: ?

Problem 2: Various Approximations

Use linear approximation to find approximate values of:

  1. (123)²
  2. ⁴√15
  3. ³√26

(i) Approximating (123)²

Let f(x) = x², approximate near a = 120 (nearest round number)

f(120) = 120² = 14,400

f'(x) = 2x ⇒ f'(120) = 240

L(x) = f(a) + f'(a)(x - a) = 14,400 + 240(x - 120)

L(123) = 14,400 + 240(3) = 14,400 + 720 = 15,120

Actual value: 123² = 15,129

Error: |15,129 - 15,120| = 9 (0.06% error)

(ii) Approximating ⁴√15

Let f(x) = x^(1/4), approximate near a = 16 (nearest 4th power)

f(16) = 16^(1/4) = 2

f'(x) = (1/4)x^(-3/4) ⇒ f'(16) = (1/4)(16)^(-3/4) = (1/4)(1/8) = 1/32

L(x) = 2 + (1/32)(x - 16)

L(15) = 2 + (1/32)(-1) = 2 - 1/32 ≈ 1.96875

Actual value: ⁴√15 ≈ 1.96799

Error: ≈ 0.00076 (0.04% error)

(iii) Approximating ³√26

Let f(x) = x^(1/3), approximate near a = 27 (nearest perfect cube)

f(27) = 3

f'(x) = (1/3)x^(-2/3) ⇒ f'(27) = (1/3)(27)^(-2/3) = (1/3)(1/9) = 1/27

L(x) = 3 + (1/27)(x - 27)

L(26) = 3 + (1/27)(-1) = 3 - 1/27 ≈ 2.96296

Actual value: ³√26 ≈ 2.9625

Error: ≈ 0.00046 (0.016% error)

Practice Any Approximation

Approximate at x =

Approximation: ?

Problem 3: Linear Approximations at Points

Find linear approximations for the following functions at the indicated points:

  1. f(x) = x³ - 5x + 12, x₀ = 2
  2. g(x) = √(x² + 9), x₀ = -4
  3. h(x) = x/(x+1), x₀ = 1

(i) f(x) = x³ - 5x + 12 at x₀ = 2

1. Find f(2) = (2)³ - 5(2) + 12 = 8 - 10 + 12 = 10

2. Find f'(x) = 3x² - 5 ⇒ f'(2) = 3(4) - 5 = 7

3. Linear approximation: L(x) = f(2) + f'(2)(x - 2)

L(x) = 10 + 7(x - 2) = 7x - 4

Final approximation: L(x) = 7x - 4

(ii) g(x) = √(x² + 9) at x₀ = -4

1. Find g(-4) = √((-4)² + 9) = √(16 + 9) = 5

2. Find g'(x) = (1/2)(x² + 9)^(-1/2) * 2x = x/√(x² + 9)

⇒ g'(-4) = -4/5 = -0.8

3. Linear approximation: L(x) = g(-4) + g'(-4)(x - (-4))

L(x) = 5 - 0.8(x + 4) = -0.8x + 1.8

Final approximation: L(x) = -0.8x + 1.8

(iii) h(x) = x/(x+1) at x₀ = 1

1. Find h(1) = 1/(1+1) = 0.5

2. Find h'(x) using quotient rule: [(1)(x+1) - x(1)]/(x+1)² = 1/(x+1)²

⇒ h'(1) = 1/4 = 0.25

3. Linear approximation: L(x) = h(1) + h'(1)(x - 1)

L(x) = 0.5 + 0.25(x - 1) = 0.25x + 0.25

Final approximation: L(x) = 0.25x + 0.25

Problem 4: Circular Plate Errors

The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. Find:

  1. Absolute error
  2. Relative error
  3. Percentage error

Given:

Actual radius (r) = 12.5 cm

Measured radius (r') = 12.65 cm

(i) Absolute error

Absolute error = |Measured - Actual| = |12.65 - 12.5| = 0.15 cm

(ii) Relative error

Relative error = Absolute error / Actual value = 0.15 / 12.5 = 0.012

(iii) Percentage error

Percentage error = Relative error × 100% = 0.012 × 100 = 1.2%

Change in Area (using linear approximation)

Area A = πr² ⇒ dA = 2πr dr

Change in radius (dr) = 0.15 cm

Approximate change in area ≈ 2π(12.5)(0.15) ≈ 11.78 cm²

Note: This is an approximation. Exact change would be π(12.65² - 12.5²) ≈ 11.89 cm²

Problem 5: Melting Ice Sphere

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for:

  1. Change in the volume
  2. Change in the surface area

Given:

Initial radius (r) = 10 cm

Change in radius (Δr) = -0.2 cm

(i) Change in Volume

Volume V = (4/3)πr³ ⇒ dV = 4πr² dr

Approximate change: ΔV ≈ 4π(10)²(-0.2) = -80π ≈ -251.33 cm³

Note: Exact change would be (4/3)π(9.8³ - 10³) ≈ -246.68 cm³

(ii) Change in Surface Area

Surface Area S = 4πr² ⇒ dS = 8πr dr

Approximate change: ΔS ≈ 8π(10)(-0.2) = -16π ≈ -50.27 cm²

Note: Exact change would be 4π(9.8² - 10²) ≈ -49.05 cm²

Problem 6: Pendulum Error

The time T for a complete oscillation of a pendulum is given by T = 2π√(l/g), where g is constant. Find the approximate percentage error in T corresponding to an error of 2% in l.

Step 1: Rewrite the equation

T = 2π√(l/g) = (2π/√g) * l^(1/2)

Step 2: Find the derivative

dT/dl = (2π/√g) * (1/2)l^(-1/2) = (π/√g) * (1/√l)

Step 3: Express the differential

ΔT ≈ dT = (dT/dl)Δl = (π/√(gl)) * Δl

Step 4: Find relative error

ΔT/T ≈ [(π/√(gl)) * Δl] / [2π√(l/g)] = (1/2) * (Δl/l)

Step 5: Convert to percentage

(ΔT/T) × 100% ≈ (1/2) × (Δl/l) × 100% = (1/2) × 2% = 1%

A 2% error in length l leads to approximately 1% error in period T

Problem 7: Percentage Error in Roots

Show that the percentage error in the nᵗʰ root of a number is approximately (1/n) times the percentage error in the number.

Step 1: Define the function

Let y = x^(1/n)

Step 2: Find the derivative

dy/dx = (1/n)x^((1/n)-1) = (1/n)(x^(1/n))/x = (1/n)(y/x)

Step 3: Express the differential

Δy ≈ dy = (dy/dx)Δx = (1/n)(y/x)Δx

Step 4: Find relative error

Δy/y ≈ (1/n)(Δx/x)

Step 5: Convert to percentage

(Δy/y) × 100% ≈ (1/n) × (Δx/x) × 100%

Percentage error in y ≈ (1/n) × Percentage error in x

Example Verification:

For cube root (n=3), if x has 6% error:

Percentage error in ³√x ≈ (1/3) × 6% = 2%

Let x = 64 (³√x = 4), if x becomes 64 + 6% = 67.84

Actual ³√67.84 ≈ 4.08 (2.04% increase - matches our prediction)