Rate of change

Introduction

We now return to the concept of the rate of change of a function with respect to time, which we introduced at the beginning of this chapter. The study of derivatives provides a powerful tool for analyzing how one quantity changes as another varies, particularly in real-world applications involving motion. Many natural phenomena involve quantities that change with time—such as the growth of populations, the velocity of moving objects, and the expansion or contraction of physical dimensions.

In particular, problems concerning motion often require us to determine how the rate of change of one quantity is related to the rate of change of another. This idea is fundamental in physics and engineering, where relationships between distance, velocity, and acceleration must be carefully examined. In this section, we apply the derivative to study how lengths, areas, and other measurable quantities change dynamically over time, providing insight into various practical problems.

Let us begin with a classic example that illustrates this concept clearly.

A lamp post of height \( H \) meters stands on the ground. A boy of height \( h \) meters moves away from the lamp post in a straight line with a velocity of \( u \) m/s. We seek to determine the rate of change of the height of his shadow.

Let the length of the shadow be \( l \), and let the distance of the boy from the lamp post be \( x \). Clearly, the segment \( NM = x \), and since the boy is moving away from the lamp post, we have:

\[ \frac{dx}{dt} = u \]

We need to determine \( \frac{dl}{dt} \). To do so, we establish a relation between \( l \) and \( x \).

By the similarity of triangles \( \triangle SNP \) and \( \triangle TMP \), we write:

\[ \frac{TM}{MP} = \frac{SN}{NP} \]

which simplifies to:

\[ \frac{h}{l} = \frac{H}{x + l} \]

Rearranging,

\[ \frac{l}{h} = \frac{x}{H} + \frac{l}{H} \]

or equivalently,

\[ l \left( \frac{1}{h} - \frac{1}{H} \right) = \frac{x}{H} \]

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

\[ \left( \frac{1}{h} - \frac{1}{H} \right) \frac{dl}{dt} = \frac{1}{H} \frac{dx}{dt} \]

Substituting \( \frac{dx}{dt} = u \),

\[ \left( \frac{1}{h} - \frac{1}{H} \right) \frac{dl}{dt} = \frac{u}{H} \]

Solving for \( \frac{dl}{dt} \),

\[ \frac{dl}{dt} = \frac{u}{H \left( \frac{1}{h} - \frac{1}{H} \right)} \]

\[ = \frac{uh}{H - h} \]

Thus, the rate of change of the height of the shadow is \( \frac{uh}{H - h} \) m/s.

The Central Idea

The main idea in rate of change problems is that we are given one or more quantities that are changing with respect to time, and we need to find how another related quantity changes over time. The change in one quantity often affects the change in another, and our goal is to determine this relationship.

For example, in the case of the boy walking away from the lamp post, his position is changing with time at a constant speed. This means the distance between him and the lamp post is increasing. Because of this, the length of his shadow is also changing.

Sometimes, more than one quantity may be changing independently. For instance, if the lamp post itself were moving upwards at a certain speed, that would introduce another rate of change into the problem. In such cases, we need to carefully account for how each moving part influences the overall situation.

To solve these problems, we follow a clear step-by-step approach:

1. Identify the given information – Determine which quantities are changing and what their rates of change (derivatives) are. Assign appropriate variables to them. In the example above, the boy’s distance from the lamp post we denoted by \(x\) that is changing at a rate of \( u \) meters per second.

2. Determine what needs to be found – Identify the quantity whose rate of change we want to calculate and denote it by some proper variable. In this case, we need to find the rate at which the length of the shadow is changing that we denoted by \(l\).

3. Establish a relationship between the variables – Find an equation that connects all the changing quantities. Often, this equation comes from geometry (like similar triangles), physics, or other mathematical properties.

4. Differentiate the equation with respect to time – Since we are dealing with rates, we take the derivative of both sides of the equation using implicit differentiation. This step gives us a relationship between the rates of change of different quantities.

5. Substitute known values and solve for the unknown rate – Once we have the equation involving different rates of change, we plug in the given information and solve for the unknown rate.

By following these steps, we can systematically determine how one quantity changes in response to another.

Example

Flying Kite Problem

A boy is flying a kite that stays at a constant height of 100 meters above the ground. The kite moves horizontally away from him at a steady speed of 2 meters per second while maintaining the same height. The boy continuously releases the string to allow the kite to move further.

We want to determine the rate at which the boy should release the string at the moment when the horizontal distance between the kite and the boy is 200 meters.

Solution:

We are given that a kite is flying at a constant height of \( h = 100 \) meters. The boy stands on the ground and releases the string as the kite moves horizontally away from him. We neglect the height of the boy in this calculation.

At some instant of time, let \( x \) be the horizontal distance between the boy and the kite, and let \( l \) be the length of the string. The height of the kite from the ground is constant equal to \(h=100\). Also the rate at which the kite moves away from the boy is \(\frac{dx}{dt} = 2\). From the geometry of the figure:

\[ l^2 = x^2 + h^2 \]

where \( h = 100 \) meters is constant. Differentiating both sides with respect to time \( t \),

\[ 2l \frac{dl}{dt} = 2x \frac{dx}{dt} \]

which simplifies to

\[ l \frac{dl}{dt} = x \frac{dx}{dt} \]

or,

\[ \frac{dl}{dt} = \frac{x}{l} \cdot \frac{dx}{dt} \]

At the moment when \( x = 200 \), the length of the string is given by

\[ l = \sqrt{x^2 + h^2} = \sqrt{200^2 + 100^2} = \sqrt{40000 + 10000} = \sqrt{50000} = 100\sqrt{5} \]

Since the kite moves horizontally at a speed of \( \frac{dx}{dt} = 2 \) m/s, substituting the values gives

\[ \frac{dl}{dt} = \frac{200}{100\sqrt{5}} \times 2 \]

\[ = \frac{400}{100\sqrt{5}} \]

\[ = \frac{4}{\sqrt{5}} \]

\[ = \frac{4\sqrt{5}}{5} \]

Approximating,

\[ \frac{4\sqrt{5}}{5} \approx \frac{4(2.236)}{5} \approx \frac{8.944}{5} \approx 1.79 \text{ m/s} \]

Thus, when the horizontal distance of the kite from the boy is 200 meters, he should release the string at a rate of \( \frac{4\sqrt{5}}{5} \) m/s or approximately 1.79 m/s.

Example

A ladder of fixed length 3 meters rests against a vertical wall. The lower end of the ladder slides away from the wall along the ground at a constant rate of 0.3 m/s. We want to determine how fast the upper end of the ladder slides down the wall at the moment when the lower end is 2.99 meters away from the wall.

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Solution:

Let \( x \) be the horizontal distance of the lower end of the ladder from the wall, and let \( y \) be the vertical height of the upper end of the ladder above the ground. Since the ladder forms a right triangle with the ground and the wall, we use the Pythagorean theorem:

\[ x^2 + y^2 = 3^2 \]

Differentiating both sides with respect to time \( t \),

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

which simplifies to,

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

Solving for \( \frac{dy}{dt} \),

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

At the moment when \( x = 2.99 \) meters, we calculate \( y \):

\[ y = \sqrt{3^2 - 2.99^2} = \sqrt{9 - 8.9401} = \sqrt{0.0599} \]

Approximating,

\[ y \approx 0.2447 \text{ m} \]

Since \( \frac{dx}{dt} = 0.3 \) m/s, we substitute the values:

\[ \frac{dy}{dt} = - \frac{2.99}{0.2447} \times 0.3 \]

\[ = - \frac{0.897}{0.2447} \]

\[ \approx -3.67 \text{ m/s} \]

Thus, when the lower end of the ladder is 2.99 meters from the wall, the upper end is sliding downward at approximately 3.67 m/s.

Sign of Rate

The sign of the rate of a quantity tells us whether the quantity is increasing or decreasing.

Consider a quantity \( x \) that depends on time \( t \). If \( \frac{dx}{dt} > 0 \), this means that as time increases, \( x \) increases as well. That is, the value of \( x \) is getting larger over time.

If \( \frac{dx}{dt} < 0 \), this means that as time increases, \( x \) decreases. That is, the value of \( x \) is getting smaller over time.

To understand why \( \frac{dx}{dt} \) behaves in this way, recall the definition of a derivative:

\[ \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} \]

This expression represents the instantaneous rate of change of \( x \) with respect to \( t \). It is obtained by taking the limit of the average rate of change:

\[ \frac{\Delta x}{\Delta t} = \frac{x(t + \Delta t) - x(t)}{\Delta t} \]

as \( \Delta t \) approaches zero. This measures how much \( x \) changes per unit time over an infinitely small interval.

If \( \frac{dx}{dt} > 0 \), then for small positive \( \Delta t \), we have \( x(t + \Delta t) > x(t) \), meaning \( x \) is increasing as time progresses.

If \( \frac{dx}{dt} < 0 \), then for small positive \( \Delta t \), we have \( x(t + \Delta t) < x(t) \), meaning \( x \) is decreasing as time progresses.

For example, consider an object moving along a straight line. If \( x(t) \) represents its position, then \( \frac{dx}{dt} \) represents its velocity. If \( \frac{dx}{dt} > 0 \), the object is moving forward. If \( \frac{dx}{dt} < 0 \), the object is moving backward.

Similarly, if \( x(t) \) represents the height of a ball thrown into the air, then while the ball is rising, \( \frac{dx}{dt} > 0 \), and when it is falling, \( \frac{dx}{dt} < 0 \). Thus, the sign of \( \frac{dx}{dt} \) tells us the direction in which the quantity is changing over time.

Example

If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, while the cube retains its shape, what is the rate of change of its volume (in cm³/sec) when the length of a side of the cube is 10 cm?

Solution:

Let \( a \) be the length of a side of the cube, \( A \) be its surface area, and \( V \) be its volume. Since the cube retains its shape while its surface area increases, all its sides grow uniformly.

The surface area of a cube is given by

\[ A = 6a^2 \]

Differentiating both sides with respect to time \( t \),

\[ \frac{dA}{dt} = 12a \frac{da}{dt} \]

Since it is given that \( \frac{dA}{dt} = 3.6 \) cm²/sec, we substitute this value,

\[ 12a \frac{da}{dt} = 3.6 \]

which gives

\[ \frac{da}{dt} = \frac{3.6}{12a} \]

Next, we consider the volume of the cube, which is given by

\[ V = a^3 \]

Differentiating both sides with respect to \( t \),

\[ \frac{dV}{dt} = 3a^2 \frac{da}{dt} \]

Substituting \( \frac{da}{dt} = \frac{3.6}{12a} \),

\[ \frac{dV}{dt} = 3a^2 \times \frac{3.6}{12a} \]

\[ = \frac{3 \times 3.6 \times a^2}{12a} \]

\[ = \frac{10.8 a}{12} \]

\[ = 0.9 a \]

Now, when \( a = 10 \) cm,

\[ \frac{dV}{dt} = 0.9 \times 10 = 9 \text{ cm}^3/\text{sec} \]

Thus, when the side length of the cube is 10 cm, the volume of the cube is increasing at a rate of 9 cm³/sec. The reasoning follows directly from relating the changing surface area to the changing side length and then using it to determine the rate of change of volume. Since both surface area and volume depend on \( a \), their rates of change are connected, and this approach systematically determines how one affects the other.

Example

A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness, which melts at a rate of 50 cm³/min. When the thickness of the ice is 5 cm, at what rate (in cm/min) is the thickness of the ice decreasing?

Solution:

Let the thickness of the ice layer be \( x \) cm. Since the iron ball has a fixed radius of 10 cm, the total radius of the sphere (iron ball + ice layer) is \( 10 + x \).

The volume of the entire sphere is given by the standard formula for the volume of a sphere,

\[ V = \frac{4}{3} \pi (10 + x)^3 \]

Differentiating both sides with respect to time \( t \),

\[ \frac{dV}{dt} = 4\pi (10 + x)^2 \frac{dx}{dt} \]

It is given that the ice melts at a rate of 50 cm³/min, meaning the volume is decreasing, so

\[ \frac{dV}{dt} = -50 \]

Substituting this in,

\[ -50 = 4\pi (10 + x)^2 \frac{dx}{dt} \]

When the ice layer is 5 cm thick, we substitute \( x = 5 \),

\[ -50 = 4\pi (10 + 5)^2 \frac{dx}{dt} \]

\[ -50 = 4\pi (15)^2 \frac{dx}{dt} \]

\[ -50 = 4\pi (225) \frac{dx}{dt} \]

\[ \frac{dx}{dt} = \frac{-50}{4\pi (225)} \]

\[ = \frac{-1}{18\pi} \text{ cm/min} \]

Since the result is negative, it confirms that the thickness of the ice layer is decreasing.

Thus, the thickness of the ice decreases at a rate of \( \frac{1}{18\pi} \) cm/min.

Example

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is \( \tan^{-1} \left( \frac{1}{2} \right) \). Water is poured into the tank at a constant rate of 5 m³/min.

At the instant when the depth of water in the tank is 10 m, at what rate (in m/min) is the water level rising?

Solution:

Let the radius of the surface of the water be \(r\) and the depth of the water be \(h\) at some instance.

The given water tank has the shape of an inverted right circular cone with a semi-vertical angle \( \alpha \) such that

\[ \alpha = \tan^{-1} \left( \frac{1}{2} \right) \]

From the definition of tangent,

\[ \tan \alpha = \frac{r}{h} = \frac{1}{2} \]

which gives the relation

\[ r = \frac{1}{2} h \]

The volume of water in the conical tank at any instant is

\[ V = \frac{1}{3} \pi r^2 h \]

Substituting \( r = \frac{1}{2} h \),

\[ V = \frac{1}{3} \pi \left(\frac{1}{2} h\right)^2 h \]

\[ = \frac{1}{3} \pi \frac{1}{4} h^2 h \]

\[ = \frac{1}{12} \pi h^3 \]

Differentiating both sides with respect to time \( t \),

\[ \frac{dV}{dt} = \frac{1}{12} \pi (3h^2) \frac{dh}{dt} \]

\[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

Solving for \( \frac{dh}{dt} \),

\[ \frac{dh}{dt} = \frac{4}{\pi h^2} \frac{dV}{dt} \]

Since it is given that water is poured at a constant rate of 5 m³/min, we substitute \( \frac{dV}{dt} = 5 \),

\[ \frac{dh}{dt} = \frac{4}{\pi h^2} \times 5 \]

Now, when \( h = 10 \) m,

\[ \frac{dh}{dt} = \frac{4}{\pi (10)^2} \times 5 \]

\[ = \frac{4}{100\pi} \times 5 \]

\[ = \frac{20}{100\pi} \]

\[ = \frac{1}{5\pi} \text{ m/min} \]

Thus, at the instant when the depth of water in the tank is 10 m, the rate at which the water level is rising is \( \frac{1}{5\pi} \) m/min.