This note covers three applications of differentiation that go beyond simple functions of the form \(y=f(x)\). Implicit differentiation handles curves where \(y\) is mixed up with \(x\), related rates link quantities that change together over time, and optimisation finds the best possible value of a quantity. Together they model real situations such as growth, motion and design.
Key ideas
Implicit differentiation. Some curves, such as \(x^2+y^2=25\), cannot easily be written as \(y=f(x)\). Instead, differentiate every term with respect to \(x\). Remember that \(y\) depends on \(x\), so by the chain rule \[\frac{d}{dx}\big(y^n\big)=n\,y^{n-1}\frac{dy}{dx}.\] A product such as \(xy\) needs the product rule: \(\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}\). After differentiating, collect all terms containing \(\dfrac{dy}{dx}\) on one side, factorise, and divide. The gradient will usually involve both \(x\) and \(y\). The tangent at a point has this gradient, and the normal has gradient equal to the negative reciprocal.
Related rates of change. Often several quantities depend on time \(t\), and they are linked by a formula. The method is:
- Draw a diagram and name the variables. Mark which rate is given and which is wanted.
- Write an equation connecting the variables. It must hold at all times, so use only quantities that change. Do not substitute fixed values early.
- Differentiate both sides with respect to \(t\), using the chain rule: \(\dfrac{dV}{dt}=\dfrac{dV}{dr}\times\dfrac{dr}{dt}\).
- Now substitute the values at the moment of interest and solve for the unknown rate.
A positive rate means the quantity is increasing, a negative rate means it is decreasing. State this in words in your final answer.
Optimisation. To find a maximum or minimum of a quantity \(Q\):
- Write \(Q\) in terms of one variable only. If there are two variables, use the given constraint to eliminate one of them.
- Find the domain: which values make physical sense (for example lengths must be positive).
- Differentiate and solve \(\dfrac{dQ}{dx}=0\).
- Show that it is a maximum or minimum, using the second derivative (\(\gt0\) for a minimum, \(\lt0\) for a maximum) or the sign change of the first derivative.
- Give the actual value of \(Q\), with units, and answer the question that was asked.
If the question is about a closed interval, also compare the values of \(Q\) at the end points, because the greatest or least value can occur there.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{d}{dx}\big(f(y)\big)=f'(y)\dfrac{dy}{dx}\) | Chain rule when \(y\) is a function of \(x\) that is not written out. |
| \(\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}\) | Product rule inside implicit differentiation. |
| \(\dfrac{dA}{dt}=\dfrac{dA}{dr}\cdot\dfrac{dr}{dt}\) | Related rates, linking rates through a common formula. |
| \(V=\tfrac43\pi r^3\), \(A=4\pi r^2\) | Sphere volume and surface area, often needed for rates. |
| \(\dfrac{dQ}{dx}=0\) | Stationary point condition for optimisation. |
| \(\dfrac{d^2Q}{dx^2}\gt0\) or \(\lt0\) | Minimum or maximum test. |
Worked examples
Example 1 A curve has equation \(x^2+xy+y^2=7\). Show that \((1,2)\) lies on it, and find the equation of the tangent at that point.
Check the point: \(1+2+4=7\). It lies on the curve.
Differentiate with respect to \(x\): \(2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0\).
Collect the \(\dfrac{dy}{dx}\) terms: \((x+2y)\dfrac{dy}{dx}=-(2x+y)\), so \(\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}\).
At \((1,2)\): \(\dfrac{dy}{dx}=-\dfrac{2+2}{1+4}=-\dfrac45\).
Tangent: \(y-2=-\dfrac45(x-1)\), which gives \(4x+5y=14\).
Answer: tangent \(4x+5y=14\) (gradient \(-0.8\)).
Example 2 A spherical balloon is inflated so that its volume increases at a constant \(40\ \text{cm}^3\,\text{s}^{-1}\). Find the rate at which the radius is increasing when \(r=5\) cm.
We know \(\dfrac{dV}{dt}=40\) and want \(\dfrac{dr}{dt}\) when \(r=5\).
\(V=\tfrac43\pi r^3\), so \(\dfrac{dV}{dr}=4\pi r^2\).
Chain rule: \(\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\).
Substitute: \(40=4\pi(25)\dfrac{dr}{dt}=100\pi\dfrac{dr}{dt}\).
So \(\dfrac{dr}{dt}=\dfrac{40}{100\pi}=0.127\) (3 s.f.).
Answer: the radius is increasing at \(0.127\ \text{cm s}^{-1}\).
Example 3 A closed cylindrical can must hold \(500\ \text{cm}^3\). Find the radius that makes the surface area least, and the least surface area.
Volume: \(\pi r^2h=500\), so \(h=\dfrac{500}{\pi r^2}\).
Surface area: \(A=2\pi r^2+2\pi rh=2\pi r^2+\dfrac{1000}{r}\), for \(r\gt0\).
Differentiate: \(\dfrac{dA}{dr}=4\pi r-\dfrac{1000}{r^2}\). Set to zero: \(4\pi r^3=1000\), so \(r^3=\dfrac{250}{\pi}=79.58\) and \(r=4.30\).
Second derivative: \(\dfrac{d^2A}{dr^2}=4\pi+\dfrac{2000}{r^3}\gt0\) for all \(r\gt0\), so this is a minimum.
Least area: \(A=2\pi(4.3013)^2+\dfrac{1000}{4.3013}=116.25+232.49=349\) cm\(^2\) (3 s.f.).
Answer: \(r=4.30\) cm, minimum surface area \(349\ \text{cm}^2\). (The height is then \(8.60\) cm, equal to \(2r\).)
Common errors
- Common error: forgetting the factor \(\dfrac{dy}{dx}\) when differentiating a term in \(y\), for example writing \(\dfrac{d}{dx}(y^2)=2y\).
- Common error: missing the product rule on \(xy\) or \(x^2y\) terms.
- Common error: substituting a number for a changing variable (such as \(r=5\)) before differentiating with respect to \(t\). The derivative of a constant is zero, which loses the rate.
- Common error: giving a related-rates answer without a sign, or without saying whether the quantity is increasing or decreasing.
- Common error: in optimisation, finding the stationary point but not showing whether it is a maximum or minimum, or stopping before giving the value of the quantity asked for.
Exam tip: In a related-rates question, write the given rate and the wanted rate as derivatives at the start, for example "\(\dfrac{dV}{dt}=40\), find \(\dfrac{dr}{dt}\)". It keeps the chain rule organised.
Check yourself
- For \(x^2+y^2=25\), find \(\dfrac{dy}{dx}\) at \((3,4)\).
Answer
\(2x+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac xy=-\frac34\). - For \(x^3+y^3=9xy\), show that \((2,4)\) is on the curve and find \(\dfrac{dy}{dx}\) there.
Answer
\(8+64=72=9\cdot8\). Differentiating: \(3x^2+3y^2y'=9y+9xy'\), so \(y'=\frac{9y-3x^2}{3y^2-9x}\). At \((2,4)\): \(\frac{36-12}{48-18}=\frac{24}{30}=\frac45\). - A 5 m ladder leans on a wall. Its foot slides away from the wall at \(0.5\ \text{m s}^{-1}\). How fast is the top moving down when the foot is 3 m from the wall?
Answer
\(x^2+y^2=25\), so \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). When \(x=3\), \(y=4\): \(\frac{dy}{dt}=-\frac{3(0.5)}{4}=-0.375\). The top moves down at \(0.375\ \text{m s}^{-1}\). - A farmer has 60 m of fencing for three sides of a rectangular pen, with a wall as the fourth side. Find the greatest area.
Answer
Let the sides at right angles to the wall be \(x\). Then the side parallel to the wall is \(60-2x\), and \(A=x(60-2x)=60x-2x^2\). \(A'=60-4x=0\) gives \(x=15\); \(A''=-4\lt0\), so a maximum. \(A=15\times30=450\ \text{m}^2\).