5.14 Implicit differentiation, related rates and optimisationIB Maths: Analysis and Approaches HL: Revision notes

When $x$ and $y$ are linked by an equation such as $x^{2} + xy + y^{2} = 7$, differentiate every term with respect to $x$, treating $y$ as a function of $x$. By the chain rule, $\frac{d}{dx}\left(y^{2}\right) = 2y\frac{dy}{dx}, \qquad \frac{d}{dx}(xy) = y + x\frac{dy}{dx}\ \text{(product rule)}.$ Then collect the $\dfrac{dy}{dx}$ terms and factorise: $2x + y + (x + 2y)\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{2x + y}{x + 2y}.$ The gradient usually depends on both $x$ and $y$, so you need the coordinates of the point. Tangents and normals then follow as usual.

When several quantities change with time, link their rates with the chain rule: $\frac{dV}{dt} = \frac{dV}{dr}\times\frac{dr}{dt}.$ Method:

1. Write down the rate you know and the rate you want, with units.
2. Find an equation linking the two variables (use similar triangles or Pythagoras to remove extra variables).
3. Differentiate, then substitute the values at the instant in question — only after differentiating.

Example: a balloon gains volume at $50\text{ cm}^{3}\text{ s}^{-1}$. At $r = 5$, $50 = 4\pi(25)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{1}{2\pi}\text{ cm s}^{-1}$.

For a cone of water (vertex down) with height $6$ and top radius $3$, similar triangles give $r = \dfrac{h}{2}$, so $V = \frac{1}{3}\pi r^{2}h = \frac{\pi h^{3}}{12}, \qquad \frac{dV}{dh} = \frac{\pi h^{2}}{4}.$ With water entering at $0.8\text{ m}^{3}$ per minute, at $h = 2$: $0.8 = \pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt} = \dfrac{4}{5\pi}$ m per minute. Because $r = \dfrac{h}{2}$ always, $\dfrac{dr}{dt} = \dfrac{1}{2}\dfrac{dh}{dt}$.

You can also differentiate an equation with respect to $t$ implicitly: from $x^{2} + y^{2} = 25$, $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.

To optimise $f(x)$ on a closed interval $a \le x \le b$:

1. Build the function from the context (e.g. time $=$ distance $\div$ speed).
2. Solve $f'(x) = 0$ and keep only solutions inside the interval.
3. Compare $f$ at these stationary points and at the end points $a$ and $b$.

The optimum can be at an end point. If a boat is fast enough, the time $T(x)$ is decreasing on the whole interval ($T'(x) < 0$), so the quickest route is to go all the way by boat ($x = 4$). Solving $T'(x) = 0$ then gives a value outside the domain, which must be rejected.