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:
- (123)²
- ⁴√15
- 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:
- f(x) = x³ - 5x + 12, x₀ = 2
- g(x) = √(x² + 9), x₀ = -4
- 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:
- Absolute error
- Relative error
- 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:
- Change in the volume
- 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)