\(\frac{dP}{dt}\)

\(\int\)

\(\frac{dy}{dx}\)

\(e^x\)

🎪 Differential Equations Carnival 🎪

Where math problems become magical adventures!

🎯 Part 1

🎡 Part 2

🎩 Exercise 10.2 - Magical Transformations

Turn these real-world statements into differential equation spells!

(i) Radium decays at a rate proportional to the amount \( Q \) present.

Step 1: Identify the magical ingredient changing

The amount of radium \( Q \) is our potion that's changing over time.

Step 2: Capture the rate of transformation

The decay rate is the magical derivative \( \frac{dQ}{dt} \).

Step 3: Cast the proportionality spell

The rate is proportional to \( Q \), so we write: \( \frac{dQ}{dt} \propto Q \)

Step 4: Add the magic constant

We use \( k \) (negative for decay): \( \frac{dQ}{dt} = -kQ \)

Final Magic Spell:

\frac{dQ}{dt} = -kQ

Mathematical Verification:

This is the standard exponential decay equation.

Solution: \( Q(t) = Q_0e^{-kt} \)

Verification: Differentiating the solution gives \( \frac{dQ}{dt} = -kQ_0e^{-kt} = -kQ \), which matches our equation.

(ii) The population \( P \) of a city increases at a rate proportional to the product of population and to the difference between 5,00,000 and the population.

Step 1: Find the growing magical creature

Population \( P \) is our magical creature that's growing.

Step 2: Measure its growth rate

The growth rate is \( \frac{dP}{dt} \).

Step 3: Break the magical incantation

The rate depends on two factors:

The current population \( P \)
The "room to grow" \( (500,\!000 - P) \)

Step 4: Combine the magical factors

So \( \frac{dP}{dt} \propto P \times (500,\!000 - P) \)

Final Magic Spell:

\frac{dP}{dt} = kP(500,\!000 - P)

Mathematical Verification:

This is a logistic growth equation with carrying capacity 500,000.

Solution: \( P(t) = \frac{500,\!000}{1 + Ae^{-500,\!000kt}} \) where A is a constant.

Verification: The standard logistic form is \( \frac{dP}{dt} = rP(1-\frac{P}{K}) \), which matches our equation with \( K=500,\!000 \) and \( r=500,\!000k \).

(iii) For a certain substance, the rate of change of vapor pressure \( P \) with respect to temperature \( T \) is proportional to the vapor pressure and inversely proportional to the square of the temperature.

Step 1: Identify the magical variables

Vapor pressure \( P \) changes with temperature \( T \).

Step 2: Find the rate of magical change

The rate is \( \frac{dP}{dT} \) (how pressure changes with temperature).

Step 3: Decipher the magical conditions

The rate depends on:

Directly on \( P \) (more pressure → faster change)
Inversely on \( T^2 \) (hotter → slower change)

Step 4: Combine the magical relationships

So \( \frac{dP}{dT} \propto \frac{P}{T^2} \)

Final Magic Spell:

\frac{dP}{dT} = k\frac{P}{T^2}

Mathematical Verification:

This is a separable differential equation.

Solution: \( P(T) = Ce^{-k/T} \) where C is a constant.

Verification: Differentiating the solution gives \( \frac{dP}{dT} = C\frac{k}{T^2}e^{-k/T} = k\frac{P}{T^2} \), which matches our equation.

🎠 Exercise 10.2 - Enchanted Problems

(iv) A saving amount pays 8% interest per year, compounded continuously. In addition, the income from another investment is credited to the amount continuously at the rate of ₹ 400 per year.

Step 1: Find the growing treasure

The savings amount \( A \) is our treasure that grows.

Step 2: Discover the growth magic

The rate \( \frac{dA}{dt} \) has two magical sources:

Continuous interest: 8% of current treasure \( 0.08A \)
Additional gold: ₹400 per year

Step 3: Combine the treasure growth

The total growth rate is the sum of both magical sources.

Final Magic Spell:

\frac{dA}{dt} = 0.08A + 400

Mathematical Verification:

This is a linear first-order ODE.

Solution: \( A(t) = Ce^{0.08t} - 5000 \) where C is a constant.

Verification: Differentiating gives \( \frac{dA}{dt} = 0.08Ce^{0.08t} = 0.08(A + 5000) = 0.08A + 400 \), which matches our equation.

2. Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

Step 1: Identify the shrinking magical orb

The volume \( V \) and radius \( r \) change as the orb evaporates.

Step 2: Measure the evaporation magic

Evaporation rate is \( \frac{dV}{dt} \) (volume change over time).

Step 3: Relate to surface area

Given: \( \frac{dV}{dt} \propto \) surface area \( = k(4\pi r^2) \)

Step 4: Volume-radius relationship

For a sphere: \( V = \frac{4}{3}\pi r^3 \), so \( \frac{dV}{dr} = 4\pi r^2 \)

Step 5: Chain rule magic

\( \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt} \)

Step 6: Equate the magical expressions

\( 4\pi r^2\frac{dr}{dt} = -k(4\pi r^2) \) (negative for shrinking)

Final Magic Spell:

\frac{dr}{dt} = -k

The radius decreases at a constant magical rate!

Mathematical Verification:

This shows the radius decreases at a constant rate.

Solution: \( r(t) = r_0 - kt \) where \( r_0 \) is initial radius.

Verification: Differentiating gives \( \frac{dr}{dt} = -k \), which matches our equation.

Physical Interpretation: Since \( \frac{dV}{dt} = -k(4\pi r^2) \) and \( V = \frac{4}{3}\pi r^3 \), the chain rule gives \( \frac{dr}{dt} = \frac{dV}{dt}/\frac{dV}{dr} = \frac{-k(4\pi r^2)}{4\pi r^2} = -k \).