Related rates, implicit differentiation, slope fields, Euler's method, and logistic growth
Math AI HL Topic 5 ExtensionIn related rates problems, two or more quantities are related by an equation, and all change with time. We use the chain rule to relate their rates of change.
Chain rule reminder: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$
An oil spill spreads in a circular shape. The radius is increasing at 0.5 m/min. Find the rate at which the area is increasing when the radius is 20 m.
When a relationship between $x$ and $y$ is given implicitly (e.g., $x^2 + y^2 = 25$), we cannot always write $y$ explicitly as a function of $x$. Implicit differentiation differentiates both sides with respect to $x$, treating $y$ as a function of $x$.
When differentiating a term involving $y$ with respect to $x$, use the chain rule:
$$\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$$Examples:
Find $\dfrac{dy}{dx}$ for the curve $x^2 + 3xy + y^2 = 7$, and find the slope at the point $(1, 1)$.
A slope field (direction field) is a visualisation of a first-order differential equation $\dfrac{dy}{dx} = f(x,y)$. At each point $(x,y)$, a short line segment is drawn with slope $f(x,y)$.
When an exact solution to a DE is difficult or impossible, we use Euler's method to approximate the solution numerically.
Given $\dfrac{dy}{dx} = f(x,y)$ with initial condition $y(x_0) = y_0$ and step size $h$:
$$y_{n+1} = y_n + h \cdot f(x_n, y_n), \quad x_{n+1} = x_n + h$$Use Euler's method with step size $h=0.5$ to approximate $y(1.5)$, given $\dfrac{dy}{dx} = x+y$ and $y(0) = 1$.
| Step | $x_n$ | $y_n$ | $f(x_n,y_n) = x_n+y_n$ | $y_{n+1} = y_n + 0.5f$ |
|---|---|---|---|---|
| 0 | 0.0 | 1.000 | 0+1 = 1.000 | 1.000 + 0.5(1.000) = 1.500 |
| 1 | 0.5 | 1.500 | 0.5+1.5 = 2.000 | 1.500 + 0.5(2.000) = 2.500 |
| 2 | 1.0 | 2.500 | 1.0+2.5 = 3.500 | 2.500 + 0.5(3.500) = 4.250 |
| 3 | 1.5 | 4.250 | — | — |
The logistic model describes constrained population growth — growth slows as the population approaches a carrying capacity $L$.
Differential equation:
$$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$$where $k$ = intrinsic growth rate, $L$ = carrying capacity.
Key features:
Solution by separation of variables:
$$P(t) = \frac{L}{1 + Ae^{-kt}}, \quad A = \frac{L - P_0}{P_0}$$Solve $\dfrac{dP}{dt} = 0.2P\!\left(1 - \dfrac{P}{500}\right)$ with $P(0) = 50$. Find $P(10)$.
When the region under $y = f(x)$ from $x=a$ to $x=b$ is rotated $360°$ about the $x$-axis, the resulting solid has volume:
Each cross-section perpendicular to the $x$-axis is a disk of radius $y = f(x)$ and thickness $dx$, contributing volume $\pi y^2 \, dx$.
About the $y$-axis: $V = \pi \int_c^d [g(y)]^2\, dy$ where $x = g(y)$.
1. Water flows into a conical tank (vertex down) of height 4 m and radius 2 m at 0.5 m³/min. Find the rate at which the water level rises when the depth is 1 m. [Note: Volume of cone = $\frac{1}{3}\pi r^2 h$; at depth $h$, the radius of water surface = $h/2$.]
$r = h/2$, so $V = \frac{1}{3}\pi (h/2)^2 h = \frac{\pi h^3}{12}$.
$\frac{dV}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$
At $h=1$: $0.5 = \frac{\pi(1)}{4} \cdot \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{2}{\pi} \approx 0.637$ m/min.
2. Use Euler's method with $h = 0.2$ to estimate $y(0.6)$ given $\dfrac{dy}{dx} = 2x - y$ and $y(0) = 1$. Give answers to 4 decimal places.
$f(x,y) = 2x - y$:
Step 0: $x=0$, $y=1$, $f=0-1=-1$, $y_1 = 1+0.2(-1) = 0.8000$
Step 1: $x=0.2$, $y=0.8$, $f=0.4-0.8=-0.4$, $y_2 = 0.8+0.2(-0.4) = 0.7200$
Step 2: $x=0.4$, $y=0.72$, $f=0.8-0.72=0.08$, $y_3 = 0.72+0.2(0.08) = 0.7360$
$y(0.6) \approx 0.7360$
3. Find the volume of the solid formed when $y = \sqrt{x}$ for $0 \leq x \leq 4$ is rotated about the $x$-axis.
$V = \pi \int_0^4 (\sqrt{x})^2\, dx = \pi \int_0^4 x\, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi \approx 25.1$ cubic units.