The rate of change describes how one quantity changes with respect to another. In physics, the derivative of a position function with respect to time gives velocity, and the derivative of velocity gives acceleration.

Distance, velocity, and acceleration

If $s(t)$ is the position of a particle at time $t$, then:

• Velocity: $v(t) = s'(t) = \frac{ds}{dt}$
• Acceleration: $a(t) = v'(t) = s''(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Also, $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}$

Example 1

The position of a particle is given by $s(t) = 2t^3 - 6t^2 + 5$. Find the velocity and acceleration at $t = 2$ seconds.

Velocity: $v(t) = s'(t) = 6t^2 - 12t$

At $t = 2$, $v(2) = 6(2)^2 - 12(2) = 24 - 24 = 0$ m/s.

Acceleration: $a(t) = v'(t) = 12t - 12$

At $t = 2$, $a(2) = 12(2) - 12 = 12$ m/s².

Example 2

A particle moves along a line OB such that its distance from O is $s$ meters in $t$ seconds, where $s = (2t - 3)(t - 4) = 2t^2 - 11t + 12$. Describe the motion at $t = 2$ seconds.

Velocity: $v(t) = s'(t) = 4t - 11$

Acceleration: $a(t) = v'(t) = 4$

At $t = 2$:

1. Position: $s(2) = 2(2)^2 - 11(2) + 12 = 8 - 22 + 12 = -2$ m (2m from O in the opposite direction from B)

2. Direction: Since $v(2) = 4(2) - 11 = -3$ m/s, the particle is moving towards O.

3. Speed: $|v(2)| = |-3| = 3$ m/s.

4. Speed is increasing because the acceleration is positive and velocity is negative. The particle is moving towards O and speeding up.

5. Rate of change of speed (acceleration): $a(t) = 4$ m/s².

Example 3

A liquid is running out of a conical funnel at a rate of 5 cm³/s. The radius of the funnel is 10 cm and its height is 20 cm. How fast is the liquid level dropping when the liquid is 10 cm deep?

Let $h$ be the depth, $r$ the radius, and $V$ the volume. $\frac{dV}{dt} = -5$ cm³/s.

By similar triangles, $\frac{r}{h} = \frac{10}{20} \Rightarrow r = \frac{h}{2}$.

Volume of a cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{h}{2})^2 h = \frac{\pi h^3}{12}$

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

Using the chain rule: $\frac{dV}{dt} = \frac{dV}{dh} \frac{dh}{dt}$

$-5 = \frac{\pi h^2}{4} \frac{dh}{dt}$

When $h = 10$:

$-5 = \frac{\pi (10)^2}{4} \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{-20}{100\pi} = \frac{-1}{5\pi}$ cm/s.

Example 4

The volume of a spherical balloon increases at a constant rate of 1.5 cm³/s. When the volume is 62 cm³, find the rate of increase of its radius.

$\frac{dV}{dt} = 1.5$ cm³/s.

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

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

When $V = 62$, $62 = \frac{4}{3}\pi r^3 \Rightarrow r^3 = \frac{3 \cdot 62}{4\pi} \Rightarrow r \approx \sqrt[3]{14.77} \approx 2.455$ cm.

Using the chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$

$1.5 = 4\pi (2.455)^2 \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{1.5}{4\pi (2.455)^2} \approx 0.0197$ cm/s $\approx 0.02$ cm/s.

Turning points are points on a curve where the graph changes direction. A point of inflexion is a point where the curve changes its concavity (from concave up to concave down, or vice versa).

Turning points

A turning point occurs where the gradient of the curve is zero (i.e., $f'(x) = 0$).

Steps to find turning points

1. Differentiate the function to find $f'(x)$.
2. Set $f'(x) = 0$.
3. Solve for $x$. These are the x-coordinates of the turning points.
4. Substitute the x-values back into the original function $f(x)$ to find the corresponding y-coordinates.

Example 1

Find the coordinates of the turning point of $y = x^2 - 4x + 5$.

$\frac{dy}{dx} = 2x - 4$

Set $\frac{dy}{dx} = 0$:

$2x - 4 = 0 \Rightarrow x = 2$

Substitute $x = 2$ into the original equation:

$y = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1$

The turning point is $(2, 1)$.

Example 2

Determine the turning points of the curve $y = 4x^3 - 3x^2 - 60x + 12$.

$\frac{dy}{dx} = 12x^2 - 6x - 60$

Set $\frac{dy}{dx} = 0$:

$12x^2 - 6x - 60 = 0 \Rightarrow 2x^2 - x - 10 = 0$

Solving the quadratic equation (using factoring or the quadratic formula) gives $x = \frac{5}{2} = 2.5$ and $x = -2$.

When $x = 2.5$:

$y = 4(2.5)^3 - 3(2.5)^2 - 60(2.5) + 12 = 62.5 - 18.75 - 150 + 12 = -94.25$

When $x = -2$:

$y = 4(-2)^3 - 3(-2)^2 - 60(-2) + 12 = -32 - 12 + 120 + 12 = 88$

The turning points are $(-2, 88)$ and $(2.5, -94.25)$.

Example 3

Determine the turning points of the curve $f(x) = 2x^3 - 9x^2 + 24x + 7$.

$f'(x) = 6x^2 - 18x + 24$

Set $f'(x) = 0$:

$6x^2 - 18x + 24 = 0 \Rightarrow x^2 - 3x + 4 = 0$

The discriminant of this quadratic is $b^2 - 4ac = (-3)^2 - 4(1)(4) = 9 - 16 = -7$. Since the discriminant is negative, there are no real roots. Therefore, this curve has no turning points.

Points of inflexion

A point of inflexion occurs where the second derivative is zero and changes sign (i.e., $f''(x) = 0$).

To find points of inflexion, you would typically:

1. Find the second derivative, $f''(x)$.
2. Set $f''(x) = 0$ and solve for $x$.
3. Check if the sign of $f''(x)$ changes around those x-values. If it does, those are x-coordinates of points of inflexion.
4. Substitute the x-values back into the original function, $f(x)$, to find the corresponding y-coordinates.

Turning points (maximum, minimum, and points of inflexion) have various applications, including curve sketching and economic analysis (e.g., equilibrium prices, break-even points, marginal costs, and revenues).

Classifying stationary points

Here are the steps to find and classify stationary points:

1. Find the first derivative, $\frac{dy}{dx} = f'(x)$.
2. Set $\frac{dy}{dx} = 0$ and solve for $x$.
3. Substitute the $x$ values back into the original equation $y = f(x)$ to find the corresponding $y$ values. This gives the coordinates of the stationary points.

There are two methods to determine the nature of stationary points:

Method 1: Second derivative test

1. Find the second derivative, $\frac{d^2y}{dx^2} = f''(x)$.
2. Substitute the $x$ values of the stationary points into $f''(x)$:
   • If $f''(x) > 0$, the point is a minimum.
   • If $f''(x) < 0$, the point is a maximum.
   • If $f''(x) = 0$, the test is inconclusive. Use the first derivative test (Method 2).

Method 2: First derivative test (sign change of gradient)

1. Consider the sign of the first derivative $f'(x)$ just before and just after the x-coordinate of the stationary point.
2. If the sign changes from positive to negative, the point is a maximum.
3. If the sign changes from negative to positive, the point is a minimum.
4. If the sign does not change, the point is a point of inflexion.

Example 1

Identify the nature of the stationary points of $f(x) = 3x^4 - 8x^3 + 6x^2 - 5$.

$f'(x) = 12x^3 - 24x^2 + 12x = 12x(x^2 - 2x + 1) = 12x(x-1)^2$

Set $f'(x) = 0$: $12x(x-1)^2 = 0 \Rightarrow x = 0$ or $x = 1$.

When $x = 0$, $f(0) = -5$. When $x = 1$, $f(1) = 3 - 8 + 6 - 5 = -4$.

Stationary points: $(0, -5)$ and $(1, -4)$.

$f''(x) = 36x^2 - 48x + 12$

When $x = 0$, $f''(0) = 12 > 0$. Thus, $(0, -5)$ is a minimum point.

When $x = 1$, $f''(1) = 36 - 48 + 12 = 0$. The second derivative test is inconclusive. Use the first derivative test:

For $x < 1$ (e.g., $x = 0.9$), $f'(x) > 0$.

For $x > 1$ (e.g., $x = 1.1$), $f'(x) > 0$.

Since the gradient does not change sign, $(1, -4)$ is a point of inflexion.

Example 2

A gardener has 400 m of fencing. Find the dimensions of a rectangular field to maximize the area.

Let $x$ be the length and $y$ be the width. The perimeter is $2x + 2y = 400$, so $y = 200 - x$.

Area: $A = xy = x(200 - x) = 200x - x^2$

$\frac{dA}{dx} = 200 - 2x$

Set $\frac{dA}{dx} = 0$: $200 - 2x = 0 \Rightarrow x = 100$.

Then $y = 200 - 100 = 100$.

$\frac{d^2A}{dx^2} = -2 < 0$, so this is a maximum.

Dimensions: 100 m by 100 m. Maximum area: $100 \cdot 100 = 10000$ m².

Example 3

Find and classify the turning points of $y = x^3 - 9x + 6$.

$\frac{dy}{dx} = 3x^2 - 9$

Set $\frac{dy}{dx} = 0$: $3x^2 - 9 = 0 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}$

When $x = \sqrt{3}$, $y = (\sqrt{3})^3 - 9\sqrt{3} + 6 = 3\sqrt{3} - 9\sqrt{3} + 6 = -6\sqrt{3} + 6 \approx -4.39$

When $x = -\sqrt{3}$, $y = (-\sqrt{3})^3 - 9(-\sqrt{3}) + 6 = -3\sqrt{3} + 9\sqrt{3} + 6 = 6\sqrt{3} + 6 \approx 16.39$

Turning points: $(\sqrt{3}, -6\sqrt{3} + 6)$ and $(-\sqrt{3}, 6\sqrt{3} + 6)$

$\frac{d^2y}{dx^2} = 6x$

When $x = \sqrt{3}$, $\frac{d^2y}{dx^2} = 6\sqrt{3} > 0$. Minimum point.

When $x = -\sqrt{3}$, $\frac{d^2y}{dx^2} = -6\sqrt{3} < 0$. Maximum point.

Example 4

Find and classify the nature of the turning points for the function $y = x^4 - 4x^3 + 16x - 16$.

1. Find the first derivative

The first derivative, which represents the gradient of the curve, is:

$\frac{dy}{dx} = 4x^3 - 12x^2 + 16$

2. Find the stationary points

Stationary points occur where the gradient is zero, so we set $\frac{dy}{dx} = 0$:

$4x^3 - 12x^2 + 16 = 0$

Divide by 4 to simplify:

$x^3 - 3x^2 + 4 = 0$

By inspection (trying small integer values), we find that $x = -1$ is a root:

$(-1)^3 - 3(-1)^2 + 4 = -1 - 3 + 4 = 0$

Therefore, $(x + 1)$ is a factor. We can perform polynomial division or factor by grouping to find the other factors:

$x^3 - 3x^2 + 4 = (x + 1)(x^2 - 4x + 4)$

The quadratic factor can be further factored:

$(x + 1)(x - 2)^2 = 0$

This gives us two solutions for $x$: $x = -1$ and $x = 2$.

Now, we find the corresponding $y$ values:

1. When $x = -1$: $y = (-1)^4 - 4(-1)^3 + 16(-1) - 16 = 1 + 4 - 16 - 16 = -27$. So, one stationary point is $(-1, -27)$.
2. When $x = 2$: $y = (2)^4 - 4(2)^3 + 16(2) - 16 = 16 - 32 + 32 - 16 = 0$. So, the other stationary point is $(2, 0)$.

3. Classify the stationary points

We'll use the second derivative test first, and if it's inconclusive, we'll use the first derivative test.

Second derivative

The second derivative is:

$\frac{d^2y}{dx^2} = 12x^2 - 24x$

Applying the second derivative test

• At $x = -1$: $\frac{d^2y}{dx^2} = 12(-1)^2 - 24(-1) = 12 + 24 = 36 > 0$. Since the second derivative is positive, $(-1, -27)$ is a minimum point.
• At $x = 2$: $\frac{d^2y}{dx^2} = 12(2)^2 - 24(2) = 48 - 48 = 0$. Since the second derivative is zero, the second derivative test is inconclusive.

Applying the first derivative test (for x = 2)

Since the second derivative test was inconclusive at $x=2$, we examine the sign of the first derivative around $x = 2$:

• For $x$ slightly less than 2 (e.g., $x = 1.9$): $\frac{dy}{dx} = 4(1.9)^3 - 12(1.9)^2 + 16 \approx 0.724 > 0$ (Positive gradient)
• For $x$ slightly greater than 2 (e.g., $x = 2.1$): $\frac{dy}{dx} = 4(2.1)^3 - 12(2.1)^2 + 16 \approx 0.724 > 0$ (Positive gradient)

Since the gradient does not change sign around $x = 2$, the point $(2, 0)$ is a point of inflexion.