Rates of change

Rates of change

The word rate refers to the change of a substance or quantity per unit time. Understanding the nature of change and the rate at which it takes place enables experts to make important predictions and decisions about atmospheric pressure, humidity, wind patterns, temperature, and other phenomena.

Differentiation helps us determine the rate of change of quantities such as volume, population, area, length, and motion.

The chain rule can be used when dealing with problems involving rates of change because in some cases, the relationship between variables is not direct. The following examples illustrate the application of differentiation in real-life situations.

Example

The area of a circle is increasing at the rate of $6 \ \text{cm}^2/\text{s}$. Find the rate of change of the radius of the circle when the radius is 4 cm.

Solution:

Let the area be $A$ and the radius be $r$. Then:

$$A = \pi r^2$$

Differentiating with respect to time:

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

Given $\frac{dA}{dt} = 6$, we substitute:

$$6 = 2\pi r \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{6}{2\pi r} = \frac{3}{\pi r}$$

When $r = 4$, then:

$$\frac{dr}{dt} = \frac{3}{4\pi} \ \text{cm/s}$$

Therefore, the radius is increasing at the rate of $\frac{3}{4\pi} \ \text{cm/s}$ when the radius is 4 cm.

Example

A spherical balloon is being blown up so that its volume increases at a constant rate of $1 \ \text{cm}^3/\text{s}$. Find the rate of increase of the radius when the volume is $6 \ \text{cm}^3$.

Solution:

Let $V$ be the volume and $r$ the radius. Then:

$$V = \frac{4}{3} \pi r^3$$

Differentiating with respect to time:

$$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$$

Given $\frac{dV}{dt} = 1$ and $V = 6$, we solve:

$$\frac{4}{3} \pi r^3 = 6 \Rightarrow r^3 = \frac{18}{4\pi} \Rightarrow r \approx 1.13 \ \text{cm}$$

Then:

$$1 = 4\pi (1.13)^2 \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi (1.13)^2} \approx \frac{1}{5.11\pi} \approx 0.062 \ \text{cm/s}$$

Therefore, the radius is increasing at approximately $0.062 \ \text{cm/s}$ when the volume is $6 \ \text{cm}^3$.

Example

Water runs into a conic tank at a rate of $9 \ \text{m}^3/\text{min}$. The tank slants point-down and has a height of 10 m and base radius of 5 m. How fast is the water level rising when the water is 6 m deep?

Solution:

Let $V$ be the volume of water and $h$ the depth. For a cone:

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

From similar triangles, since $\frac{r}{h} = \frac{5}{10} = \frac{1}{2}$, we have:

$$r = \frac{1}{2} h \Rightarrow V = \frac{1}{3} \pi \left(\frac{1}{2}h\right)^2 h = \frac{1}{12} \pi h^3$$

Differentiating with respect to time:

$$\frac{dV}{dt} = \frac{1}{4} \pi h^2 \cdot \frac{dh}{dt}$$

Given $\frac{dV}{dt} = 9$, and $h = 6$, we substitute:

$$9 = \frac{1}{4} \pi (6)^2 \cdot \frac{dh}{dt} \Rightarrow 9 = 9\pi \cdot \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{\pi} \ \text{m/min}$$

Therefore, the water level is rising at the rate of $\frac{1}{\pi} \ \text{m/min}$.

Critical (stationary) points of a curve

Figure 4.4 illustrates a curve in which the gradient changes from point to point. This change in gradient demonstrates different types of critical points.

Figure: a curve showing critical points and change of signs

In the figure:

• Point P represents a maximum turning point, where the gradient (first derivative) is zero and changes sign from positive to negative.
• Point Q represents a minimum turning point, where the gradient is zero and changes from negative to positive.
• Point R is an inflection point, where the gradient is zero, but there is no change of sign in the gradient.

Differentiation can be used to determine critical points and their nature. The first derivative test and the second derivative test are commonly used:

Tests for critical points

1. If $f'(x) = 0$ and $f''(x) > 0$, the critical point is a minimum.
2. If $f'(x) = 0$ and $f''(x) < 0$, the critical point is a maximum.
3. If $f'(x) = 0$ and $f''(x) = 0$, the critical point may be an inflection point.

Example

Determine the critical points and their nature for the curve: $y = x^3 + 3x^2 - 9x$

Solution

1. First derivative: $\frac{dy}{dx} = 3x^2 + 6x - 9$
2. Set the derivative equal to zero: $3x^2 + 6x - 9 = 0 \quad \Rightarrow \quad x^2 + 2x - 3 = 0$ $(x + 3)(x - 1) = 0 \quad \Rightarrow \quad x = -3, \, x = 1$
3. Second derivative: $\frac{d^2y}{dx^2} = 6x + 6$
   • At $x = -3$: $\frac{d^2y}{dx^2} = 6(-3) + 6 = -12 \Rightarrow$ Maximum
   • At $x = 1$: $\frac{d^2y}{dx^2} = 6(1) + 6 = 12 \Rightarrow$ Minimum
4. Coordinates:
   • $y(-3) = (-3)^3 + 3(-3)^2 - 9(-3) = -27 + 27 + 27 = 27 \Rightarrow (-3, 27)$
   • $y(1) = 1 + 3 - 9 = -5 \Rightarrow (1, -5)$

Conclusion: Maximum point at $(-3, 27)$, minimum point at $(1, -5)$

Example

Find and classify the nature of all the critical points of: $y = 2x^3 + 3x^2 - 12x + 7$

Solution

1. First derivative: $\frac{dy}{dx} = 6x^2 + 6x - 12$ $6x^2 + 6x - 12 = 0 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x - 1)(x + 2) = 0$ $x = -2, \, x = 1$
2. Second derivative: $\frac{d^2y}{dx^2} = 12x + 6$
   • At $x = -2$: $\frac{d^2y}{dx^2} = 12(-2) + 6 = -18 \Rightarrow$ Maximum
   • At $x = 1$: $\frac{d^2y}{dx^2} = 12(1) + 6 = 18 \Rightarrow$ Minimum
3. Coordinates:
   • $y(-2) = 2(-8) + 3(4) - 12(-2) + 7 = -16 + 12 + 24 + 7 = 27 \Rightarrow (-2, 27)$
   • $y(1) = 2(1) + 3(1) - 12(1) + 7 = 2 + 3 - 12 + 7 = 0 \Rightarrow (1, 0)$

Conclusion: Maximum at $(-2, 27)$, minimum at $(1, 0)$

Example

Sketch the graph of the function $f(x) = -x^2 - 6x - 9$ using differentiation.

Solution

1. First derivative: $f'(x) = -2x - 6$ Critical point: $f'(x) = 0 \Rightarrow -2x - 6 = 0 \Rightarrow x = -3$
2. Second derivative: $f''(x) = -2 \quad \text{(Always negative)} \Rightarrow \text{Maximum at } x = -3$
3. Find the corresponding y-value: $f(-3) = -(-3)^2 - 6(-3) - 9 = -9 + 18 - 9 = 0 \Rightarrow \text{Point } (-3, 0)$
4. x- and y-intercepts:
   • When $x = 0$, $f(0) = -9 \Rightarrow (0, -9)$
   • When $f(x) = 0$, solve $-x^2 - 6x - 9 = 0 \Rightarrow x = -3$

Conclusion: The vertex (maximum point) of the parabola is at $(-3, 0)$. The graph opens downward.

Marginal cost and marginal revenue

In the manufacturing of a commodity, three key functions are involved:

1. Cost Function $C(x)$: The total cost of producing $x$ units of a product during a specific period.
2. Revenue Function $R(x)$: The total revenue obtained by selling $x$ units of the product during that period.
3. Profit Function $P(x)$: The total profit obtained from selling $x$ units, calculated as:

$$P(x) = R(x) - C(x)$$

Marginal cost is the rate at which cost increases with an additional unit produced. It is the derivative of the cost function:

$$\text{Marginal Cost} = C'(x) = \frac{d}{dx} C(x)$$

Marginal revenue is the extra revenue generated by selling one more unit of a product. It is the derivative of the revenue function:

$$\text{Marginal Revenue} = R'(x) = \frac{d}{dx} R(x)$$

Example

The cost function for manufacturing $x$ units of a product is given by:

$$C(x) = x^3 - 20x^2 + 300x + 1000$$

If each item is sold for Tsh 300, find the marginal cost and marginal revenue when 1000 units are produced.

Solution

Marginal Cost:

$$C'(x) = \frac{d}{dx}(x^3 - 20x^2 + 300x + 1000) = 3x^2 - 40x + 300$$

At $x = 1000$:

$$C'(1000) = 3(1000)^2 - 40(1000) + 300 = 3,000,000 - 40,000 + 300 = 2,960,300$$

Marginal Cost is Tsh 2,960,300.

Marginal Revenue:

Since each item sells for Tsh 300, the revenue function is:

$$R(x) = 300x \quad \Rightarrow \quad R'(x) = \frac{d}{dx}(300x) = 300$$

Marginal Revenue is Tsh 300.

Example

If the total cost is given by:

$$C(x) = 300 + 40x + 18x^2$$

Find the marginal cost function.

Solution

$$C'(x) = \frac{d}{dx}(300 + 40x + 18x^2) = 40 + 36x$$

Marginal Cost Function is: $C'(x) = 40 + 36x$

Example

The cost function is:

$$C(x) = 300x - 10x^2 + \frac{1}{3}x^3$$

If each unit sells for Tsh 200, and 100 units are produced, find:

1. Marginal cost
2. Marginal revenue, given that:

$$R(x) = 8000x + 40x^2 + x^3$$

Solution

a. Marginal Cost:

$$C'(x) = \frac{d}{dx}(300x - 10x^2 + \frac{1}{3}x^3) = 300 - 20x + x^2$$ $$C'(100) = 300 - 20(100) + 100^2 = 300 - 2000 + 10000 = 8,300$$

Marginal Cost is Tsh 8,300.

b. Marginal Revenue:

$$R'(x) = \frac{d}{dx}(8000x + 40x^2 + x^3) = 8000 + 80x + 3x^2$$ $$R'(100) = 8000 + 80(100) + 3(100)^2 = 8000 + 8000 + 30000 = 46,000$$

Marginal Revenue is Tsh 46,000.