AOS3 Topic 8: Application of Derivatives

Derivatives are essential for measuring rates of change, optimizing functions, and analyzing motion. In physics, they determine velocity and acceleration, while in economics, they help maximize profit and minimize costs. Derivatives also play a key role in curve sketching, identifying points of inflection, and analyzing concavity. Engineers use derivatives in system modeling, while in biology, they model population growth and disease spread. In financial mathematics, derivatives help in pricing and risk management. Additionally, derivatives are used in signal processing to filter and analyze data, and in geometry to find tangent and normal lines to curves.

The use of the derivative to determine instantaneous rates of change is a very important application of calculus. One of the first areas of applied mathematics to be studied in the seventeenth century was motion in a straight line. The problems of kinematics were the motivation for Newton’s work on calculus.

Tangents and Normals

The derivative of a function is a new function that gives the measure of the gradient of the tangent at each point on the curve. With the gradient, we can find the equation of the tangent line at a given point on the curve.

Suppose that \((x_1, y_1)\) is a point on the curve \(y = f(x)\). Then, if \(f\) is differentiable at \(x = x_1\), the equation of the tangent at \((x_1, y_1)\) is given by:

\[ y - y_1 = f'(x_1)(x - x_1) \]

The normal to a curve at a point on the curve is the line that passes through the point and is perpendicular to the tangent at that point.

Two lines with gradients \(m_1\) and \(m_2\) are perpendicular if and only if \(m_1 m_2 = -1\).

Thus, if a tangent has gradient \(m\), the normal has gradient \(-\frac{1}{m}\).

Rates of Change

The process of differentiation can be used to address many problems involving rates of change.

For a function with rule \( f(x) \):

The average rate of change for \( x \in [a, b] \) is given by:

\[ \frac{f(b) - f(a)}{b - a} \]

The instantaneous rate of change of \( f \) with respect to \( x \) when \( x = a \) is given by \( f'(a) \).

The derivative \( \frac{dy}{dx} \) gives the instantaneous rate of change of \( y \) with respect to \( x \).

If \( \frac{dy}{dx} > 0 \), then \( y \) is increasing as \( x \) increases.

If \( \frac{dy}{dx} < 0 \), then \( y \) is decreasing as \( x \) increases.

Stationary Points

In the previous chapter, we have seen that the gradient of the tangent at a point \( (a, f(a)) \) on the curve with rule \( y = f(x) \) is given by \( f'(a) \).

A point \( (a, f(a)) \) on a curve \( y = f(x) \) is said to be a stationary point if \( f'(a) = 0 \).

Equivalently, a point \( (a, f(a)) \) on \( y = f(x) \) is a stationary point if \( \frac{dy}{dx} = 0 \) when \( x = a \).

Types of Stationary Points

The graph of \(y = f(x)\) shown has three stationary points: A, B, and C.

A - Local Maximum Point

Point A is called a local maximum point. Notice that immediately to the left of A, the gradient is positive, and immediately to the right, the gradient is negative.

Gradient: + 0 −

Shape of \(f\): ↗ — ↘

B - Local Minimum Point

Point B is called a local minimum point. Notice that immediately to the left of B, the gradient is negative, and immediately to the right, the gradient is positive.

Gradient: − 0 +

Shape of \(f\): ↘ — ↗

C - Stationary Point of Inflection

Point C is called a stationary point of inflection. The gradient is positive immediately to the left and right of C.

Gradient: + 0 +

Shape of \(f\): ↗ — ↗

Clearly, it is also possible to have stationary points of inflection where the gradient is negative immediately to the left and right.

Gradient: − 0 −

Shape of \(f\): ↘ — ↘

Stationary points of types A and B are referred to as turning points.

Absolute Maximum and Minimum Values

Local maximum and minimum values were discussed in the previous section. These are often not the actual maximum and minimum values of the function.

For a function defined on an interval:

  • The actual maximum value of the function is called the absolute maximum.
  • The actual minimum value of the function is called the absolute minimum.

The corresponding points on the graph of the function are not necessarily stationary points.

More precisely, for a continuous function \( f \) defined on an interval \([a, b]\):

  • If \( M \) is a value of the function such that \( f(x) \leq M \) for all \( x \in [a, b] \), then \( M \) is the absolute maximum value of the function.
  • If \( N \) is a value of the function such that \( f(x) \geq N \) for all \( x \in [a, b] \), then \( N \) is the absolute minimum value of the function.

Maximum Rates of Increase and Decrease

We know that when we take the derivative of a function, we obtain a new function, the derivative, which gives the instantaneous rate of change. We can apply the same technique to the new function to find the maximum rate of increase or decrease.

Remember:

  • If \( \frac{dy}{dx} > 0 \), then \( y \) is increasing as \( x \) increases.
  • If \( \frac{dy}{dx} < 0 \), then \( y \) is decreasing as \( x \) increases.
Example 1

Find the equation of the tangent to the curve \( y = x^3 + \frac{1}{2}x^2 \) at the point \( x = 1 \).

Solution

When \( x = 1 \), \( y = \frac{3}{2} \), and so \( \left( 1, \frac{3}{2} \right) \) is a point on the tangent.

Since \( \frac{dy}{dx} = 3x^2 + x \), the gradient of the tangent at \( x = 1 \) is 4.

Hence, the equation of the tangent is:

\( y - \frac{3}{2} = 4(x - 1) \)

\( y = 4x - \frac{5}{2} \)

Example 2

Find the equation of the tangent to the curve with equation \( y = x^{\frac{3}{2}} - 4x^{\frac{1}{2}} \) at the point on the graph where \( x = 4 \).

Solution

Let \( y = x^{\frac{3}{2}} - 4x^{\frac{1}{2}} \). Then, \( \frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} \).

When \( x = 4 \),

\( y = 4^{\frac{3}{2}} - 4 \times 4^{\frac{1}{2}} = 0 \)

\( \frac{dy}{dx} = \frac{3}{2} \times 4^{\frac{1}{2}} - 2 \times 4^{-\frac{1}{2}} = 2 \)

Hence, the equation of the tangent is:

\( y - 0 = 2(x - 4) \)

i.e., \( y = 2x - 8 \)

Example 3

Find the equations of the tangent and normal to the graph of \( y = -\cos x \) at the point \( \left(\frac{\pi}{2}, 0\right) \).

Solution

First, find the gradient of the curve at this point:

\(\frac{dy}{dx} = \sin x\), and so, when \( x = \frac{\pi}{2} \),

\(\frac{dy}{dx} = 1\).

The equation of the tangent is:

\(y - 0 = 1\left(x - \frac{\pi}{2}\right)\)

i.e.

\(y = x - \frac{\pi}{2}\)

The gradient of the normal is \(-1\), and therefore the equation of the normal is:

\(y - 0 = -1\left(x - \frac{\pi}{2}\right)\)

i.e.

\(y = -x + \frac{\pi}{2}\)

Example 4

Find the equation of the tangent to:

a) \( f(x) = x^{\frac{1}{3}} \) where \( x = 0 \)

b) \( f(x) = x^{\frac{2}{3}} \) where \( x = 0 \).

Solution:

a) The derivative of \( f \) is not defined at \( x = 0 \). For \( x \in \mathbb{R} \setminus \{0\} \), \( f'(x) = \frac{1}{3}x^{-\frac{2}{3}} \).

It is clear that \( f \) is continuous at \( x = 0 \) and that \( f'(x) \to \infty \) as \( x \to 0 \).

The graph has a vertical tangent at \( x = 0 \).

b) \( f(x) = x^{\frac{2}{3}} \)

The derivative of \( f \) is not defined at \( x = 0 \). For \( x \in \mathbb{R} \setminus \{0\} \), \( f'(x) = \frac{2}{3}x^{-\frac{1}{3}} \).

It is clear that \( f \) is continuous at \( x = 0 \) and that \( f'(x) \to \infty \) as \( x \to 0^+ \) and \( f'(x) \to -\infty \) as \( x \to 0^- \).

There is a cusp at \( x = 0 \), and the graph of \( y = f(x) \) has a vertical tangent at \( x = 0 \).

Example 5

A balloon develops a microscopic leak and gradually decreases in volume. Its volume, \(V\) (in \(\text{cm}^3\)), at time \(t\) seconds is given by:

\[ V = 600 - 10t - \frac{1}{100}t^2, \quad t \geq 0. \]

a) Find the rate of change of volume after:

i) 10 seconds

ii) 20 seconds

b) For how long could the model be valid?

Solution

a)

\[ \frac{dV}{dt} = -10 - \frac{t}{50} \]

i) When \(t = 10\):

\[ \frac{dV}{dt} = -10 - \frac{10}{50} = -\frac{101}{5} \]

i.e. the volume is decreasing at a rate of \(\frac{101}{5}\) \(\text{cm}^3\) per second.

ii) When \(t = 20\):

\[ \frac{dV}{dt} = -10 - \frac{20}{50} = -\frac{102}{5} \]

i.e. the volume is decreasing at a rate of \(\frac{102}{5}\) \(\text{cm}^3\) per second.

b) The model will not be meaningful when \(V < 0\). Consider when \(V = 0\):

\[ 600 - 10t - \frac{1}{100}t^2 = 0 \]

\[ \therefore t = 100(\sqrt{31} - 5) \quad \text{or} \quad t = -100(\sqrt{31} + 5) \]

The model may be suitable for \(0 \leq t \leq 100(\sqrt{31} - 5)\).

Example 6

A pot of liquid is put on the stove. When the temperature of the liquid reaches 80°C, the pot is taken off the stove and placed on the kitchen bench. The temperature in the kitchen is 20°C. The temperature of the liquid, \( T \)°C, at time \( t \) minutes is given by:

\[ T = 20 + 60e^{-0.3t} \]

a) Find the rate of change of temperature with respect to time in terms of \( T \).

b) Find the rate of change of temperature with respect to time when:

i) \( T = 80 \)

ii) \( T = 30 \)

Solution

a) By rearranging \( T = 20 + 60e^{-0.3t} \), we see that:

\[ e^{-0.3t} = \frac{T - 20}{60} \]

Now,

\[ \frac{dT}{dt} = -18e^{-0.3t} \]

Substituting,

\[ \frac{dT}{dt} = -18 \times \frac{T - 20}{60} \]

Hence,

\[ \frac{dT}{dt} = -3 \times \frac{T - 20}{10} \]

\[ \frac{dT}{dt} = 0.3(20 - T) \]

b) When \( T = 80 \),

\[ \frac{dT}{dt} = 0.3(20 - 80) \]

\[ \frac{dT}{dt} = -18 \]

The liquid is cooling at a rate of 18°C per minute.

i) When \( T = 30 \),

\[ \frac{dT}{dt} = 0.3(20 - 30) \]

\[ \frac{dT}{dt} = -3 \]

The liquid is cooling at a rate of 3°C per minute.

Example 7

Find the stationary points of the following functions:

1. \( y = 4 + 3x - x^3 \)

2. \( p = 2t^3 - 5t^2 - 4t + 13 \)

Solution:

\[ \frac{dy}{dx} = 3 - 3x^2 \]

Setting \( \frac{dy}{dx} = 0 \) implies:

\[ 3(1 - x^2) = 0 \]

So, \( x = \pm 1 \)

The stationary points are \( (1, 6) \) and \( (-1, 2) \).

2. \( p = 2t^3 - 5t^2 - 4t + 13 \)

\[ \frac{dp}{dt} = 6t^2 - 10t - 4, \quad t > 0 \]

Setting \( \frac{dp}{dt} = 0 \) implies:

\[ 2(3t^2 - 5t - 2) = 0 \]

Factoring gives:

\[ (3t + 1)(t - 2) = 0 \]

So, \( t = -\frac{1}{3} \) or \( t = 2 \)

But since \( t > 0 \), the only acceptable solution is \( t = 2 \).

The corresponding stationary point is \( (2, 1) \).

Example 8

Find the stationary points of the following functions:

a) \( y = \sin(2x), \, x \in [0, 2\pi] \)

b) \( y = e^{2x} - x \)

c) \( y = x \log_e(2x), \, x \in (0, \infty) \)

Solution:

Solution

a) \( y = \sin(2x) \)

\( \frac{dy}{dx} = 2\cos(2x) \)

So

\( \frac{dy}{dx} = 0 \) implies \( 2\cos(2x) = 0 \)

\( \cos(2x) = 0 \)

\( 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \text{ or } \frac{7\pi}{2} \)

\( \therefore x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \text{ or } \frac{7\pi}{4} \)

The stationary points are \( \left(\frac{\pi}{4}, 1\right) \), \( \left(\frac{3\pi}{4}, -1\right) \), \( \left(\frac{5\pi}{4}, 1\right) \), and \( \left(\frac{7\pi}{4}, -1\right) \).

b) \( y = e^{2x} - x \)

\( \frac{dy}{dx} = 2e^{2x} - 1 \)

So

\( \frac{dy}{dx} = 0 \) implies \( 2e^{2x} - 1 = 0 \)

\( e^{2x} = \frac{1}{2} \)

\( \therefore x = \frac{1}{2} \log_e \left(\frac{1}{2}\right) = -\frac{1}{2} \log_e 2 \)

When \( x = -\frac{1}{2} \log_e 2 \),

\( y = e^{2 \times \left(-\frac{1}{2} \log_e 2\right)} + \frac{1}{2} \log_e 2 = \frac{1}{2} + \frac{1}{2} \log_e 2 \)

The coordinates of the stationary point are \( \left(-\frac{1}{2} \log_e 2, \frac{1}{2} + \frac{1}{2} \log_e 2\right) \).

c) \( y = x \log_e (2x) \)

\( \frac{dy}{dx} = \log_e (2x) + 1 \)

So

\( \frac{dy}{dx} = 0 \) implies \( \log_e (2x) + 1 = 0 \)

\( \log_e (2x) = -1 \)

\( 2x = e^{-1} \)

\( \therefore x = \frac{1}{2e} \)

When \( x = \frac{1}{2e} \),

\( y = \frac{1}{2e} \log_e \left(\frac{2}{2e}\right) = -\frac{1}{2e} \)

The coordinates of the stationary point are \( \left(\frac{1}{2e}, -\frac{1}{2e}\right) \).

Example 9

The curve with equation \( y = x^3 + ax^2 + bx + c \) passes through the point (0, 5) and has a stationary point at (2, 7). Find \( a \), \( b \), and \( c \).

Solution:

When \( x = 0 \), \( y = 5 \). Thus \( 5 = c \).

\(\frac{dy}{dx} = 3x^2 + 2ax + b \) and at \( x = 2 \), \( \frac{dy}{dx} = 0 \). Therefore

\( 12 + 4a + b = 0 \)   (1)

The point (2, 7) is on the curve and so

\( 8 + 4a + 2b + 5 = 7 \)

\( \therefore 6 + 4a + 2b = 0 \)   (2)

Subtracting (1) from (2) gives \( -6 + b = 0 \). Thus \( b = 6 \). Substitute in (1):

\( 12 + 4a + 6 = 0 \)

\( 4a = -18 \)

Hence \( a = -\frac{9}{2} \), \( b = 6 \), and \( c = 5 \).

Exercise &&1&& (&&1&& Question)

Find the equation of the normal to the curve with equation \( y = x^3 - 2x^2 \) at the point \((1, -1)\).

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Exercise &&2&& (&&1&& Question)

Find the equation of the tangent to the graph of \( y = \sin x \) at the point where \( x = \frac{\pi}{3} \).

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Exercise &&3&& (&&1&& Question)

For the function with rule \(f(x) = x^2 + 2x\), find the average rate of change for \(x \in [2, 3]\).

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Exercise &&4&& (&&1&& Question)

For the function with rule \(f(x) = x^2 + 2x\), find the average rate of change for the interval \([2, 2 + h]\).

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Exercise &&5&& (&&1&& Question)

Find the stationary points of this function: \(y = 9 + 12x - 2x^2\)

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