# Detour of Variable Changes: A Complete Exploration of Related Rates

Related rates involve using derivatives to understand how different quantities change together over time. By connecting these changing quantities through an equation and differentiating with respect to time, one can find the rate of change of one quantity based on the known rate of another. This calculus technique is crucial in physics, engineering, and real-world problem-solving.

**What are related rates like?**

Related rates are used by first identifying the relationship between variables in a situation, like the radius of a balloon and its volume. Then, understanding how one variable changes (like the rate of air entering the balloon) helps predict the change in another variable (like the balloon’s radius). This approach is widely applied in physics to predict movements, in engineering for design dynamics, and in everyday problem-solving where different factors are interconnected and change simultaneously.

**Example 1:**

A circular oil spill is expanding on water. The radius of the oil spill is increasing at a constant rate of \( 2 \) meters per minute. Find the rate at which the area of the spill is increasing when the radius is \( 5 \) meters.

**Solution: **

to define the variable, let \( r \) be the radius of the oil spill and \( A \) be the area of the spill. Given that \( \frac{dr}{dt} = 2 \) meters/minute. The area of a circle is given by \( A = \pi r^2 \). Differentiate both sides of the equation with respect to time \( t \) to find \( \frac{dA}{dt} \):

\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)

Applying the chain rule, \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).

Now, substituting values, when \( r=5 \) meters and \( \frac{dr}{dt} = 2 \) meters per minute:

\( \frac{dA}{dt} = 2\pi \times 5 \times 2 \)

\( \frac{dA}{dt} = 20\pi \) \( \frac{meters^2}{minutes} \)

So, The area of the oil spill is increasing at a rate of \( 20\pi \) square meters per minute when the radius is 5 meters.

**Example 2:**

A gas cloud diffuses so that its radius increases at a rate of \( 3 \frac{cm}{s} \). Determine the rate at which the volume of the cloud is increasing when the radius is \( 10 \) cm.

**Solution:**

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

\( \frac{dV}{dt} = 4\pi \times 10^2 \times 3 = 1200\pi \) cm³/s.

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