Differentiation measures the instantaneous rate of change of a function. In AI SL, the focus is on the power rule and its application to real-world contexts. The derivative $f'(x)$ gives the gradient of the tangent at any point.
Power rule: If $f(x) = ax^n$, then $f'(x) = nax^{n-1}$
Constant rule: If $f(x) = c$, then $f'(x) = 0$
Sum rule: $(f+g)' = f'+g'$
Key derivatives: $\dfrac{d}{dx}(e^x) = e^x$, $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$, $\dfrac{d}{dx}(\sin x) = \cos x$, $\dfrac{d}{dx}(\cos x) = -\sin x$
Gradient at a point: substitute $x$-value into $f'(x)$
In context: if $P(t)$ is population at time $t$, then $P'(t)$ is the rate of change of population (people per year). If $C(x)$ is total cost at output $x$, then $C'(x)$ is marginal cost. The sign of the derivative tells us whether the quantity is increasing ($f'>0$) or decreasing ($f'<0$).
To find maximum or minimum values: differentiate, set $f'(x)=0$, solve, verify using the second derivative or sign chart, then interpret in context.
A company sells $x$ units per day at price $p = 60 - 0.5x$ dollars. Revenue $R(x) = px$. Find the quantity that maximises revenue and the maximum revenue.
Integration is the reverse of differentiation. The definite integral gives the area between a curve and the $x$-axis.
Power rule: $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ (for $n \ne -1$)
Key integrals: $\displaystyle\int e^x\,dx = e^x+C$, $\displaystyle\int \frac{1}{x}\,dx = \ln|x|+C$
Definite integral: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ where $F$ is the antiderivative
Area interpretation: If $f(x) \ge 0$ on $[a,b]$, then $\displaystyle\int_a^b f(x)\,dx$ = area between curve and $x$-axis.
When a function is difficult to integrate analytically, the trapezoidal rule approximates the definite integral numerically by dividing the region into trapezoids.
With $n$ equal strips of width $h = \dfrac{b-a}{n}$:
$$\int_a^b f(x)\,dx \approx \frac{h}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)]$$
The outer values $f(x_0)$ and $f(x_n)$ are counted once; all intermediate values are counted twice.
More strips $\Rightarrow$ more accurate approximation.
Approximate $\displaystyle\int_0^4 (12-x^2)\,dx$ using the trapezoidal rule with $n=4$ strips.
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $f(x) = 12-x^2$ | 12 | 11 | 8 | 3 | -4 |
Kinematics applies calculus to motion along a straight line. Displacement $s$, velocity $v$, and acceleration $a$ are related by differentiation and integration.
$v = \dfrac{ds}{dt} = s'(t)$ (velocity = rate of change of displacement)
$a = \dfrac{dv}{dt} = v'(t) = s''(t)$ (acceleration = rate of change of velocity)
Reversing: $s = \displaystyle\int v\,dt$, $v = \displaystyle\int a\,dt$
Particle at rest: $v = 0$. Changes direction: $v$ changes sign.
Distance (not displacement) $= \displaystyle\int_a^b |v|\,dt$
A particle moves with velocity $v(t) = 3t^2 - 12t + 9$ m/s. Find: (a) when the particle is at rest; (b) the displacement from $t=0$ to $t=3$; (c) the total distance travelled from $t=0$ to $t=3$.
Perimeter: $2x+2y=40 \Rightarrow y = 20-x$. Area $A = xy = x(20-x) = 20x-x^2$.
$A'(x) = 20-2x = 0 \Rightarrow x = 10$ m. Maximum area $= 10\times10 = 100$ m² (a square).
$h=0.5$. Values: $f(1)=1$, $f(1.5)=\frac{2}{3}$, $f(2)=0.5$, $f(2.5)=0.4$, $f(3)=\frac{1}{3}$.
Approx $= \dfrac{0.5}{2}\left[1+2(\frac{2}{3})+2(0.5)+2(0.4)+\frac{1}{3}\right]=0.25[1+1.333+1+0.8+0.333]=0.25\times4.467\approx1.117$
Exact: $[\ln x]_1^3 = \ln 3 \approx 1.099$. The trapezoidal approximation overestimates since $1/x$ is concave.
(a) $v(t) = 3t^2-12t+9$. $v(2) = 12-24+9 = -3$ m/s (moving in negative direction).
(b) $a(t) = 6t-12 = 0 \Rightarrow t=2$ s.
(c) First direction change at $t=1$ (where $v$ first equals 0). $s(1) = 1-6+9 = 4$ m.