Problems connecting several ideas

Read, represent and decide

A good solution starts by distinguishing data, unknowns and conditions. Rewrite the question in one sentence: are we seeking a distance, an instant, a maximum or an accumulated quantity? Name the quantities and include units. A clearly labelled drawing may reveal a relationship hidden in the text, but must not add right angles or symmetries that the statement does not provide.

Then choose a working representation. Coordinates make distances and intersections calculable; a parameter tracks two coordinates; an objective function organises optimisation; a rate graph suggests accumulation. You can change representation partway through, provided you explain the correspondence. The initial plan is a proposed solution: if a calculation becomes complicated, return to the problem's structure before persisting with a sequence that has lost its meaning.

The domain is a solution condition

Constraints may be algebraic, geometric or contextual. A denominator must be nonzero; a real square root requires a nonnegative radicand; a length requires positivity; an instant may be limited by an experiment's duration. The final domain is the intersection of these conditions. Keeping it visible prevents an algebraic root from being confused with a valid answer.

In a rectangle inscribed in a semicircle, for example, the height may be y=√(R²−x²), with 0≤x≤R. The area uses the full width 2x, not just x. The endpoints x=0 and x=R represent degenerate figures with zero area. Including them in the mathematical analysis may help compare candidates, even when the context requires a positive-area rectangle. The conclusion must distinguish the region used in the proof from the physical choices actually permitted.

Use structure before calculation

Symmetry, factors and identities can simplify a problem more than a general rule. To minimise the distance between two points, it may be easier to minimise the squared distance, since the square-root function is increasing on nonnegative values. The minimisers are the same and a root disappears from the expression to differentiate. The equivalence depends on nonnegativity and must be stated.

If a quantity is the product of two lengths with fixed sum, completing the square can prove a maximum directly. If sin²θ+cos²θ appears, the circle identity may remove a variable. These simplifications do not compete with differential calculus: they make clearer what the calculus later confirms. A short solution is valuable when every step is justified, not when it omits the conditions making it valid.

Combine rate and total

A rate answers how a quantity changes at an instant; the integral answers the accumulated change over an interval. If velocity is known, position also requires an initial position. Integration produces a family of antiderivatives, and the initial condition fixes the constant. Differentiating the position obtained and checking the initial point are two distinct checks.

If the question changes to distance travelled, the instants when velocity changes sign become essential. Split the interval at these points and add magnitudes of the contributions. The same reasoning applies to balance and volume: inflow minus outflow can be negative although the existing volume remains positive. A model predicting negative volume at later times must be stopped or changed when the tank empties. The mathematical expression may remain defined after losing physical validity.

Prove existence and compare candidates

Before seeking an optimum, check whether it must exist. A continuous function on a closed bounded interval attains extrema, but a function on an open interval may only approach a value. Then collect interior candidates, points without derivatives and included bounds. A derivative sign in a neighbourhood proves a local extremum; a global comparison or an inequality valid throughout the domain proves an absolute optimum.

In equations, existence of a solution and its numerical approximation are different questions. A sign change in a continuous function guarantees a root; a calculator can approximate it. To prove uniqueness, add monotonicity or another argument. A numerically found root must be substituted into the equation with a coherent tolerance. The number of decimal places in the answer must reflect the permitted error, not just the screen's capacity.

Write a solution that can be audited

Organise the answer into relationships, operations and interpretation. Each equality must be true under the stated conditions; each approximation must be marked as approximate. Avoid rounding intermediate steps when final precision matters. Keep exact forms such as √2 and π until the end, and only then present a decimal with its unit.

Verification may use a second representation, an order-of-magnitude estimate, a physical bound or substitution. If a calculated time is negative, ask whether it corresponds to a mathematical past excluded by the context. If an area exceeds the containing figure's area, there is an error. If an answer satisfies the transformed equation but not the original, it may be an extra solution introduced by the operations. Finish by answering the initial question in everyday language and stating any decisive assumption remaining in the model.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

A lengthy calculation demonstrates a correct solution.

Validity depends on relationships, the domain and verification, not length.

One numerical root proves there are no others.

Uniqueness requires an additional argument, such as monotonicity.

The expression remains valid in context as long as it is algebraically defined.

The phenomenon may impose a boundary, such as emptying or contact with the ground.