DIFFERENTIATION 15

Related Rates

Related rates problems connect two or more changing quantities through an equation and then differentiate that relationship with respect to time.

The main idea

Suppose two quantities \(x\) and \(y\) both change with time \(t\). If they are related by an equation, their rates of change are also related.

\[ \boxed{ \text{relationship between quantities} \quad\longrightarrow\quad \text{relationship between their rates} } \]

1. Quantities depend on time

Even if time is not written explicitly, we may have

\[ x=x(t), \qquad y=y(t). \]

So both \(x\) and \(y\) can change as time passes.

2. A simple relationship

Suppose

\[ y=x^2. \]

If \(x\) changes with time, then \(y\) also changes with time.

Differentiate both sides with respect to \(t\):

\[ \frac{dy}{dt} = 2x\frac{dx}{dt}. \]

\[ \boxed{ \frac{dy}{dt} = 2x\frac{dx}{dt} } \]

This equation connects the rate of change of \(y\) to the rate of change of \(x\).

3. Why the chain rule appears

Since \(y\) depends on \(x\), and \(x\) depends on \(t\),

\[ y=y(x(t)). \]

\[ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}. \]

This is exactly the chain rule.

4. Example

Suppose

\[ y=x^2 \]

and

\[ \frac{dx}{dt}=3. \]

Find \(dy/dt\) when

\[ x=4. \]

From

\[ \frac{dy}{dt} = 2x\frac{dx}{dt}, \]

we obtain

\[ \frac{dy}{dt} = 2(4)(3) = 24. \]

Therefore,

\[ \boxed{ \frac{dy}{dt}=24 } \]

5. Circle example

The area of a circle is

\[ A=\pi r^2. \]

If the radius changes with time, then the area changes too.

Differentiate with respect to \(t\):

\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]

\[ \boxed{ \frac{dA}{dt} = 2\pi r\frac{dr}{dt} } \]

6. Numerical example

Suppose

\[ \frac{dr}{dt}=2 \]

and

\[ r=5. \]

Then

\[ \frac{dA}{dt} = 2\pi(5)(2) = 20\pi. \]

So the area is increasing at a rate of

\[ 20\pi \]

square units per unit time.

7. Volume of a sphere

For a sphere,

\[ V=\frac43\pi r^3. \]

Differentiate with respect to time:

\[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]

So the rate at which volume changes depends on both the current radius and the rate at which the radius changes.

8. Geometry example

Suppose a right triangle has horizontal side \(x\) and vertical side \(y\), with fixed hypotenuse \(L\).

Then

\[ x^2+y^2=L^2. \]

Differentiate with respect to \(t\):

\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \]

Therefore,

\[ x\frac{dx}{dt} + y\frac{dy}{dt} = 0. \]

Solving for \(dy/dt\),

\[ \boxed{ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} } \]

9. Interpreting the negative sign

In the previous result,

\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}, \]

if \(dx/dt>0\), then \(dy/dt<0\).

So as one side increases, the other decreases.

The sign tells us the direction of change.

10. Diagram

11. The practical method

\[ \boxed{ 1.\; \text{Identify the changing quantities} } \] \[ \boxed{ 2.\; \text{Write an equation relating them} } \] \[ \boxed{ 3.\; \text{Differentiate with respect to time} } \] \[ \boxed{ 4.\; \text{Substitute the known values} } \] \[ \boxed{ 5.\; \text{Solve for the required rate} } \]

12. Do not substitute too early

A common mistake is to substitute numerical values before differentiating.

Keep the variables in the relationship first. Differentiate the general equation, then substitute the values relevant to the instant being considered.

13. Units matter

Rates carry units.

For example,

\[ \frac{dr}{dt} \]

might be measured in centimetres per second.

But

\[ \frac{dA}{dt} \]

would then be measured in square centimetres per second.

Units help check whether the final result makes sense.

14. The deeper idea

Related rates are an application of the chain rule.

\[ \boxed{ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} } \]

A change in one quantity passes through a mathematical relationship and produces a change in another quantity.