10J

Rates of change

Average and instantaneous rates of change, the derivative as a rate, and where a function is increasing, decreasing or stationary.

Average and instantaneous rates

Example 1●●

A tank holds $V = 2000 - 40t + 0.2t^2$ litres after $t$ minutes, for $0 \le t \le 100$. Find the rate of change at $t = 50$ and when the flow stops.

$\dfrac{dV}{dt} = -40 + 0.4t$

At $t = 50$: $-20$ L/min, so water is leaving at $20$ litres per minute.

$-40 + 0.4t = 0$ at $t = 100$, when $V = 0$: the tank is empty.

Reading the sign of the derivative

Orange: y = x³ − 3x. Faint green: y = f′(x). Move P to see where the function increases and decreases.

Try this

For which $x$-values is the function decreasing?

Example 2●●

Where is $f(x) = x^3 - 3x$ increasing?

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

$f'(x) > 0$ for $x < -1$ or $x > 1$.

An upright parabola is positive outside its zeroes.

Flashcards

Average rate

Gradient of a secant: $\dfrac{\Delta y}{\Delta x}$

Instantaneous rate

$\dfrac{dy}{dx}$ at that moment

Increasing

$f'(x) > 0$

Stationary

$f'(x) = 0$

Practice

Find the average rate of change of $f(x) = x^2$ from $x = 1$ to $x = 4$.
$V = 2000 - 40t + 0.2t^2$. Find $\dfrac{dV}{dt}$ at $t = 50$.
For the same tank, when is $\dfrac{dV}{dt} = 0$?
$f(x) = x^3 - 3x$ is decreasing for:
Find where $y = x^2 - 8x + 3$ is stationary.
If $f'(2) = -3$, then at $x = 2$ the function is:
A population is $P = 500 + 30t + 2t^2$ after $t$ years. Find the growth rate at $t = 5$, in people per year.
For $f(x) = x^3$, at $x = 0$ the function is:
The area of a circle is $A = \pi r^2$. Find $\dfrac{dA}{dr}$ when $r = 3$, correct to 2 decimal places.