6.4 Maxima-Minima, Indeterminate Forms & Taylor Series

Maxima and Minima of a Single Variable

Optimisation is the most directly useful calculus topic for a mechanical engineer: minimum cost of machining, maximum power transmitted by a belt, optimum order quantity in inventory control. All of them follow the same two-step recipe.

Step 1 — locate stationary points. Solve $f'(x) = 0$.

Step 2 — classify them using the second derivative test:

When $f''(x) = 0$ the point may be a maximum, a minimum, or a point of inflection. The reliable fallback is the first-derivative sign test: if $f'$ changes from positive to negative, it is a maximum; negative to positive, a minimum; no change, an inflection.

For a closed interval $[a,b]$, the absolute extremum must be found by comparing the values at all interior stationary points and at both endpoints. Candidates routinely forget the endpoints.

Worked example. A rectangular settling tank of fixed volume $V = 32$ m$^3$ has a square base of side $x$ and height $h$, and is open at the top. Minimise the surface area.

With $h = 32/x^2$, the area is

$$A = x^2 + 4xh = x^2 + \frac{128}{x}$$

Then $dA/dx = 2x - 128/x^2 = 0$ gives $x^3 = 64$, so $x = 4$ m and $h = 2$ m. Check: $d^2A/dx^2 = 2 + 256/x^3 = 6 > 0$, confirming a minimum. The minimum area is $16 + 32 = 48$ m$^2$.

Indeterminate Forms

Seven forms are indeterminate:

$$\frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \times \infty, \quad \infty - \infty, \quad 0^0, \quad \infty^0, \quad 1^{\infty}$$

L'Hopital's rule applies directly only to the first two. If $f(a) = g(a) = 0$ (or both tend to infinity) and $g'(a) \neq 0$, then

$$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$

and the rule may be applied repeatedly while the form remains indeterminate. Applying it to a form that is not indeterminate — for instance $2/0$ — produces a wrong answer, and this is the single most common error on such items.

The other five forms are converted first. A product $0 \times \infty$ is rewritten as a quotient; a difference $\infty - \infty$ is combined over a common denominator; the three exponential forms are handled by taking logarithms.

Worked example. $\lim_{x\to0} \dfrac{e^x - 1 - x}{x^2}$ is of the form $0/0$. Differentiating gives $\dfrac{e^x - 1}{2x}$, still $0/0$. Differentiating again gives $\dfrac{e^x}{2} \to \dfrac{1}{2}$.

Taylor and Maclaurin Series

Any function with sufficiently many derivatives at $x = a$ can be expanded as

$$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2!}f''(a) + \frac{(x-a)^3}{3!}f'''(a) + \cdots$$

Setting $a = 0$ gives the Maclaurin series

$$f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \cdots$$

The expansions you must recall without derivation:

$$\ne^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad (-1 < x \leq 1)$$ $$(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots \quad (|x| < 1)$$

Note the parity: $\sin x$ contains only odd powers, $\cos x$ only even powers. A question offering an even-power term in the sine expansion can be eliminated on inspection.

Why This Matters Downstream in Paper-II

Truncating these series after the first-order term is the formal justification for approximations you will use repeatedly:

• $\sin\theta \approx \theta$ and $\cos\theta \approx 1$ for small $\theta$ — used to linearise the simple pendulum and the governor equations in Theory of Machines.
• $(1+x)^n \approx 1 + nx$ — used for small changes in efficiency, density and clearance calculations.
• The two-term expansion of $e^x$ underlies the linearisation of exponential temperature decay in lumped-capacitance heat transfer.

A Taylor expansion truncated after the linear term is exactly a tangent-line approximation, which connects back to the mean value theorem of the previous section.

Worked Example: Approximating a Root

Estimate $\sqrt{4.1}$ using a two-term expansion. Write $\sqrt{4.1} = 2\sqrt{1 + 0.025}$ and apply $(1+x)^{1/2} \approx 1 + x/2$:

$$\sqrt{4.1} \approx 2\left(1 + 0.0125\right) = 2.025$$

The true value is 2.02485, so the linear term alone gives four-figure accuracy — which is why this technique is worth more than a calculator in a timed paper.