Enchanted Math Garden

1

Fairy's Differential Potion

Find differential \( dy \) for each magical function:

(i) \( y = \frac{(1 - 2x)^3}{3 - 4x} \)

(ii) \( y = (3 + \sin(2x))^{2/3} \)

(iii) \( y = e^{x^2 - 5x + 7} \cos(x^2 - 1) \)

✨ Solution (i):

Use the Magic Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \)

Let \( u = (1-2x)^3 \), \( v = 3-4x \)

Find \( u' = 3(1-2x)^2(-2) = -6(1-2x)^2 \)

Find \( v' = -4 \)

Apply magic: \( y' = \frac{(3-4x)(-6(1-2x)^2) - (1-2x)^3(-4)}{(3-4x)^2} \)

Simplify with fairy dust: \( dy = \frac{(1-2x)^2[-6(3-4x) + 4(1-2x)]}{(3-4x)^2} dx \)

Final spell: \( dy = \frac{14(1-2x)^2(x - 1)}{(3-4x)^2} dx \)

🧜‍♀️ Solution (ii):

Use the Mermaid's Chain Rule: \( \frac{d}{dx}u^n = n u^{n-1} u' \)

Let \( u = 3 + \sin(2x) \), \( n = 2/3 \)

Find \( u' = \cos(2x) \cdot 2 = 2\cos(2x) \)

Cast the spell: \( y' = \frac{2}{3}(3+\sin(2x))^{-1/3} \cdot 2\cos(2x) \)

Ocean-approved answer: \( dy = \frac{4\cos(2x)}{3(3+\sin(2x))^{1/3}} dx \)

🔥 Solution (iii):

Use the Phoenix Product Rule: \( \frac{d}{dx}(uv) = u'v + uv' \)

Let \( u = e^{x^2-5x+7} \), \( v = \cos(x^2-1) \)

Find \( u' = e^{x^2-5x+7}(2x-5) \)

Find \( v' = -\sin(x^2-1) \cdot 2x \)

Rise from ashes: \( y' = e^{x^2-5x+7}(2x-5)\cos(x^2-1) + e^{x^2-5x+7}(-2x\sin(x^2-1)) \)

Final flame: \( dy = e^{x^2-5x+7}[(2x-5)\cos(x^2-1) - 2x\sin(x^2-1)] dx \)

2

Dragon's Treasure Evaluation

Find \( df \) for \( f(x) = x^2 + 3x \) and evaluate for dragon hoards:

(i) \( x = 2 \) and \( dx = 0.1 \) (Emerald hoard)

(ii) \( x = 3 \) and \( dx = 0.02 \) (Ruby hoard)

🐉 Dragon's Wisdom:

First find derivative: \( f'(x) = 2x + 3 \)

Differential is \( df = (2x + 3) dx \)

💎 Emerald Hoard (i):

At \( x = 2 \), \( f'(2) = 2(2) + 3 = 7 \)

With \( dx = 0.1 \), \( df = 7 \times 0.1 = 0.7 \)

❤️ Ruby Hoard (ii):

At \( x = 3 \), \( f'(3) = 2(3) + 3 = 9 \)

With \( dx = 0.02 \), \( df = 9 \times 0.02 = 0.18 \)

3

Unicorn's Δf and df Comparison

Find \( \Delta f \) and \( df \) for these enchanted functions:

(i) \( f(x) = x^3 - 2x^2; x = 2, \Delta x = dx = 0.5 \)

(ii) \( f(x) = x^2 + 2x + 3; x = -0.5, \Delta x = dx = 0.1 \)

🦄 Solution (i):

First find \( f(2) = 8 - 8 = 0 \)

Find \( f(2.5) = 15.625 - 12.5 = 3.125 \)

Actual change \( \Delta f = 3.125 - 0 = 3.125 \)

Find derivative \( f'(x) = 3x^2 - 4x \)

At \( x=2 \), \( f'(2) = 12 - 8 = 4 \)

Differential \( df = f'(2)dx = 4 \times 0.5 = 2 \)

Comparison: \( \Delta f = 3.125 \), \( df = 2 \)

🧚 Solution (ii):

First find \( f(-0.5) = 0.25 - 1 + 3 = 2.25 \)

Find \( f(-0.4) = 0.16 - 0.8 + 3 = 2.36 \)

Actual change \( \Delta f = 2.36 - 2.25 = 0.11 \)

Find derivative \( f'(x) = 2x + 2 \)

At \( x=-0.5 \), \( f'(-0.5) = -1 + 2 = 1 \)

Differential \( df = f'(-0.5)dx = 1 \times 0.1 = 0.1 \)

Comparison: \( \Delta f = 0.11 \), \( df = 0.1 \)

4

Wizard's Logarithmic Spell

With \( \log_{10} e \approx 0.4343 \), find an approximate value of \( \log_{10} 1003 \).

🧙‍♂️ Wizard's Solution:

Let \( f(x) = \log_{10} x \), we know \( f(1000) = 3 \)

Find derivative \( f'(x) = \frac{1}{x \ln 10} = \frac{\log_{10} e}{x} \)

At \( x=1000 \), \( f'(1000) = \frac{0.4343}{1000} = 0.0004343 \)

For \( x=1003 \), \( dx = 3 \)

Approximate change \( df = f'(1000)dx = 0.0004343 \times 3 \approx 0.0013029 \)

Final approximation: \( \log_{10} 1003 \approx 3 + 0.0013029 \approx 3.0013029 \)

5

Ancient Tree's Growth

The trunk of an ancient tree has diameter 30 cm. During the year, its circumference grew 6 cm.

(i) Approximately how much did the diameter grow?

(ii) What's the percentage increase in cross-section area?

🌳 Solution (i):

Circumference \( C = \pi d \), so \( d = C/\pi \)

Change in circumference \( dC = 6 \) cm

Differential of diameter \( dd = dC/\pi = 6/\pi \approx 1.91 \) cm

🍃 Solution (ii):

Area \( A = \pi r^2 = \pi (d/2)^2 = \pi d^2/4 \)

Differential of area \( dA = (\pi d/2) dd \)

Original area \( A_0 = \pi (30)^2 / 4 = 225\pi \) cm²

Change in area \( dA = (\pi \times 30/2) \times 1.91 \approx 90 \) cm²

Percentage increase \( = (dA/A_0) \times 100 \approx (90/706.86) \times 100 \approx 12.73\% \)

6

Phoenix Egg Shell Volume

A phoenix egg is nearly spherical. Inside radius is 5 mm, outside radius is 5.3 mm.

Find the approximate volume of the shell.

🔥 Solution:

Volume of sphere \( V = \frac{4}{3}\pi r^3 \)

Differential \( dV = 4\pi r^2 dr \)

At \( r=5 \) mm, \( dr = 0.3 \) mm

Approximate shell volume \( dV = 4\pi (5)^2 (0.3) = 30\pi \approx 94.25 \) mm³

Exact calculation would be \( V_{outer} - V_{inner} = \frac{4}{3}\pi(5.3^3 - 5^3) \approx 98.96 \) mm³

Our approximation is quite close to the exact value!

7

Healing Artery Expansion

A potion dilates arteries from radius 2 mm to 2.1 mm.

How much does the cross-sectional area increase approximately?

⚕️ Solution:

Area \( A = \pi r^2 \)

Differential \( dA = 2\pi r dr \)

At \( r=2 \) mm, \( dr = 0.1 \) mm

Approximate area increase \( dA = 2\pi (2)(0.1) = 0.4\pi \approx 1.2566 \) mm²

Exact increase would be \( \pi(2.1^2 - 2^2) = 0.41\pi \approx 1.2885 \) mm²

8

Enchanted City Voter Growth

Voting population (thousands) grows by: \( V(t) = 30 + 12t^2 - t^3 \), \( 0 \leq t \leq 8 \)

Find approximate change from year 4 to 4⅙ (4.1667).

🏰 Solution:

Find derivative \( V'(t) = 24t - 3t^2 \)

At \( t=4 \), \( V'(4) = 96 - 48 = 48 \) thousand/year

Time change \( dt = \frac{1}{6} \approx 0.1667 \) year

Approximate voter change \( dV = V'(4)dt = 48 \times \frac{1}{6} = 8 \) thousand

Exact change would be \( V(4.1667) - V(4) \approx 7.34 \) thousand

9

Memory Elf's Word Learning

Words learned in \( x \) hours: \( y = 52\sqrt{x} \), \( 0 \leq x \leq 9 \)

Approximate words learned when \( x \) changes from:

(i) 1 to 1.1 hour

(ii) 4 to 4.1 hour

🧝 Solution:

Find derivative: \( y' = 52 \times \frac{1}{2\sqrt{x}} = \frac{26}{\sqrt{x}} \)

Differential \( dy = \frac{26}{\sqrt{x}} dx \)

📚 (i) From 1 to 1.1 hour:

At \( x=1 \), \( dx=0.1 \): \( dy = \frac{26}{1} \times 0.1 = 2.6 \) words

🎓 (ii) From 4 to 4.1 hour:

At \( x=4 \), \( dx=0.1 \): \( dy = \frac{26}{2} \times 0.1 = 1.3 \) words

10

Expanding Magic Plate

A circular plate expands from radius 10.5 cm to 10.75 cm.

Find approximate area change and percentage change.

🔥 Solution:

Area \( A = \pi r^2 \)

Differential \( dA = 2\pi r dr \)

At \( r=10.5 \) cm, \( dr=0.25 \) cm

Approximate area change \( dA = 2\pi(10.5)(0.25) = 5.25\pi \approx 16.49 \) cm²

Original area \( A_0 = \pi(10.5)^2 = 110.25\pi \) cm²

Percentage change \( \approx \frac{5.25\pi}{110.25\pi} \times 100 \approx 4.76\% \)

11

Painting the Wizard's Cube

A 10 cm cube gets 0.2 cm paint on all faces.

Approximate paint volume used, then calculate exact amount.

🧙‍♂️ Solution:

Volume \( V = x^3 \), Surface area \( S = 6x^2 \)

Differential \( dV = S dx = 6x^2 dx \)

At \( x=10 \) cm, \( dx=0.2 \) cm

Approximate paint volume \( dV = 6(10)^2(0.2) = 120 \) cm³

Exact calculation:

New volume \( V_{new} = (10.4)^3 = 1124.864 \) cm³ (0.2 cm on each side → +0.4 cm total)

Original volume \( V_{old} = 1000 \) cm³

Exact paint volume \( = 1124.864 - 1000 = 124.864 \) cm³

Our approximation was close but undercounted by about 4 cm³