Applications of Derivatives | ISC Class 12 Maths Notes

This note covers rates of change, related rates, tangents and normals, angles between curves, increasing and decreasing functions, critical and stationary points, local and absolute extrema, derivative tests, and optimisation problems.

How does a derivative measure a rate of change?

A function assigns an output to each allowed input. Write y = f(x), where x is the input, y is the output and f is the rule connecting them. Its derivative, written dy/dx or f′(x), measures the instantaneous rate of change of y with respect to x.

The prime in f′ denotes differentiation. At a specified input x₀, the notation f′(x₀) means the derivative evaluated at that input. The domain is the set of allowed inputs. Differentiability at a point means that the derivative exists there.

If t denotes time, dy/dt measures change per unit time. Its units depend on the quantity being measured. For area measured in square centimetres and time in seconds, the rate has units cm²/s, where cm means centimetre and s means second.

How are related rates connected?

When two quantities depend on time, the chain rule connects their rates: dy/dt = (dy/dx)(dx/dt). Thus dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. A rate with respect to a length differs from a rate with respect to time.

1. Define the changing quantities and write an equation connecting them.
2. Differentiate with respect to the variable named in the question, retaining factors such as dx/dt.
3. Insert the given dimensions and rates after differentiation.
4. State the result with its units and whether the quantity is increasing or decreasing.

Worked example 1. A cube's volume increases at 9 cm³/s. Find the rate of increase of its surface area when its edge is 10 cm.

Let x be the edge length, V the volume and S the surface area. Then V = x³ and S = 6x².

Answer: dV/dt = 3x²(dx/dt) = 9, so dx/dt = 3/x². Therefore dS/dt = 12x(dx/dt) = 36/x. At x = 10, the surface area increases at 3.6 cm²/s.

Worked example 2. Circular waves spread across a quiet lake with radius increasing at 4 cm/s. Find the rate of increase of enclosed area when the radius is 10 cm.

Let r denote radius and A enclosed area. The constant π is the ratio of a circle's circumference to its diameter. Since A = πr², differentiation gives dA/dt = 2πr(dr/dt).

Answer: dA/dt = 2π × 10 × 4 = 80π cm²/s.

Why must the signs of rates be retained?

Worked example 3. A rectangle's length x decreases at 3 cm/min while its width y increases at 2 cm/min. Find the rates of change of perimeter and area when x = 10 cm and y = 6 cm. Here min means minute.

Let P be perimeter and A area. Then P = 2(x + y), A = xy, dx/dt = −3 and dy/dt = 2.

Answer: dP/dt = 2(−3 + 2) = −2 cm/min. By the product rule, dA/dt = y(dx/dt) + x(dy/dt) = 6(−3) + 10(2) = 2 cm²/min.

The perimeter decreases while the area increases at that instant. The dimensions alone do not determine these changes: the rates and their signs are essential. Substituting fixed dimensions before differentiating would remove the dependence on time.

How are equations of tangents and normals obtained?

A tangent gives the local direction of a curve at a point. A normal passes through that point perpendicular to the tangent. The slope of a non-vertical line measures vertical change divided by horizontal change.

Let (x₁, y₁) be the point of contact on y = f(x), so y₁ = f(x₁). Let m denote the tangent's finite slope, given by m = f′(x₁). Coordinates x and y in the following line equations represent a general point on the relevant line.

Result: Point-slope equations

The point-slope form of a line through (x₁, y₁) with slope m is y − y₁ = m(x − x₁). Substituting the derivative gives the tangent equation. For a finite non-zero tangent slope, the perpendicular normal has slope −1/m.

When a tangent is horizontal, its equation is y = y₁. Its normal is vertical, so dividing by the zero tangent slope is invalid. A vertical tangent, when present, has equation x = x₁ and its normal is horizontal.

How should the calculation be organised?

First confirm that the given point satisfies the curve equation. Next differentiate and evaluate the derivative at that point. Finally insert the point and the appropriate slope into the line equation. The tangent and normal share a point, but their directions differ.

Worked example 4. Find the normal to y = sin x at (0, 0). Here sin and cos denote sine and cosine; trigonometric arguments are in radians.

Answer: dy/dx = cos x gives tangent slope 1 at x = 0. The normal slope is −1, so y = −x, or x + y = 0.

A radian is the angle subtended at a circle's centre by an arc equal in length to its radius. For an acute angle in a right-angled triangle, sine is opposite side divided by hypotenuse, and cosine is adjacent side divided by hypotenuse. The hypotenuse is the side opposite the right angle.

How is the angle between two curves calculated?

The angle of intersection of two curves is the angle between their tangents at the intersection. Solve the curve equations together before evaluating their derivatives. Tangent slopes at unrelated points do not give an angle of intersection.

Result: Angle between tangents

Let m₁ and m₂ denote the two finite tangent slopes, and θ the smaller angle between the tangents. For 1 + m₁m₂ ≠ 0, the straight-line angle formula gives tan θ = |(m₂ − m₁)/(1 + m₁m₂)|. Here tan θ = sin θ/cos θ denotes the trigonometric tangent.

The vertical bars mean absolute value, the non-negative magnitude of a real number. They select the smaller non-obtuse angle. For perpendicular lines, the angle is π/2 radians, equivalent to 90 degrees, and the displayed fraction has a zero denominator.

Two curves intersect orthogonally when their tangents are perpendicular. If both slopes are finite, this requires m₁m₂ = −1. When a tangent is vertical, use its direction directly; a vertical line has no finite slope.

Worked example 5. Find the angles of intersection of y² = x and x² = y.

Answer: Substitution gives x⁴ = x, hence the real intersection points are (0, 0) and (1, 1). The derivatives are 1/(2y) and 2x respectively.

At (0, 0), the tangents are vertical and horizontal, giving π/2. At (1, 1), their slopes are 1/2 and 2, giving tan θ = 3/4 and θ = tan⁻¹(3/4). Here tan⁻¹(3/4) denotes the angle whose tangent is 3/4.

For several intersections, evaluate the slopes separately at each point. Keep the exceptional vertical case separate from the finite-slope calculation. An equation obtained by dividing by a coordinate cannot be used at a point where that coordinate is zero.

What do increasing and decreasing functions mean?

Let I denote an interval contained in the domain of a real-valued function f. Choose any two inputs x₁ and x₂ in I with x₁ < x₂. The function is strictly increasing if f(x₁) < f(x₂), and strictly decreasing if f(x₁) > f(x₂).

A constant function has the same output at every input in the interval. A monotonic function is increasing throughout the interval or decreasing throughout it. These descriptions compare outputs as the input moves from left to right.

Theorem: Derivative sign and monotonicity

Let a and b be real numbers with a < b. The notation [a, b] includes both endpoints, while (a, b) excludes them. Suppose f is continuous on [a, b], meaning its value agrees with its limiting value throughout, and differentiable on (a, b).

The sign must hold throughout the relevant interval. Checking the derivative at one isolated input does not establish behaviour across an entire interval. Differentiation supplies a method for comparing function values without calculating every possible pair of outputs.

Worked example 6. Show that f(x) = x³ − 3x² + 4x is increasing for every real x.

Answer: f′(x) = 3x² − 6x + 4 = 3(x − 1)² + 1. Since a square is non-negative, f′(x) > 0 for every real x. Therefore f is increasing on the real line.

The graph of f(x) = x² is an upward-opening parabola with its lowest point at the origin. Its left branch falls towards the origin as x increases; its right branch rises away from the origin.

See Fig. 6.1 in your NCERT textbook

The origin is the coordinate point (0, 0), and a parabola is the quadratic curve shown here. This function decreases for negative inputs and increases for positive inputs. A single function can therefore behave differently on different intervals.

How are intervals of increase and decrease found?

A sign chart records whether a derivative is positive or negative on successive intervals. Begin with the function's domain. Values where the derivative vanishes, or where it is undefined, can separate intervals that require individual examination.

How does factorisation help?

1. Differentiate the given function and factorise the derivative where possible.
2. Find the relevant zeros of the derivative and locate any undefined points.
3. Arrange these dividing points in order within the domain.
4. Determine the sign of the derivative on each resulting interval and state the function's behaviour there.

Worked example 7. Find where f(x) = 4x³ − 6x² − 72x + 30 is increasing and decreasing, for real x.

Answer: f′(x) = 12x² − 12x − 72 = 12(x − 3)(x + 2). Its zeros are −2 and 3. The derivative is positive before −2, negative between −2 and 3, and positive after 3.

Thus f is increasing on (−∞, −2) and (3, ∞), and decreasing on (−2, 3). The symbol ∞ indicates unbounded extent, not a real endpoint.

The positive factor 12 does not change the sign. Each linear factor keeps its sign until its zero is crossed. This explains why the interval chart establishes the derivative's sign throughout each interval, rather than merely reporting isolated sample calculations.

Do not join the two increasing intervals across the intervening decreasing interval. The conclusion must identify where each behaviour occurs. The original function is neither increasing throughout the real line nor decreasing throughout it.

For a restricted domain, retain only dividing points inside that domain. If a finite endpoint belongs to the domain and continuity holds there, the increasing or decreasing behaviour can sometimes be extended to include it using the theorem's endpoint conditions.

How do critical, stationary and extreme points differ?

A critical point is an input c in the domain of f where f′(c) = 0 or f is not differentiable. A stationary point is a point with zero derivative. The letter c here denotes the input being investigated.

A local maximum occurs when f(c) is at least as large as nearby values. A local minimum occurs when f(c) is at most as large as nearby values. Together, maximum and minimum values are called extrema. An extreme point is a point where such a value occurs.

What does “local” specify?

For an interior point c, take a positive number h small enough that (c − h, c + h) lies in the domain. This interval is a neighbourhood of c. A local comparison uses the inputs in such a neighbourhood, rather than every input in the domain.

The input c, the graph point (c, f(c)), and the function value f(c) are different pieces of information. A question asking for the minimum value requires the output. A question asking where it occurs requires the input, or the graph coordinates if requested.

Theorem: A necessary condition for a local extremum

If f has a local maximum or minimum at an interior point c, then either f′(c) = 0 or f is not differentiable there. This is a necessary condition: it identifies candidates but does not establish that each candidate is an extremum.

Worked example 8. Find the local minimum of f(x) = 3 + |x| for real x.

Answer: For x < 0, f(x) = 3 − x and f′(x) = −1. For x > 0, f(x) = 3 + x and f′(x) = 1. The function decreases towards 0 and increases afterwards, so its local minimum value is 3 at x = 0.

This example is not differentiable at zero, yet it has a minimum there. A turning point is where the graph changes from increasing to decreasing, or conversely. Stationary points need not be turning points: the function f(x) = x³ has f′(0) = 0 but continues increasing through zero.

How does the first derivative test classify a critical point?

The first derivative test examines the sign of f′ on each side of a critical input. Let f be defined on an open interval containing c and continuous at c. Determine the derivative's sign at points sufficiently close to c on both sides.

Theorem: Sign changes identify local extrema

Geometrically, positive followed by negative means the graph rises towards a point and then falls away. Negative followed by positive means it falls towards the point and then rises. If the sign stays positive or stays negative, the direction does not reverse.

Worked example 9. Find the local extrema of f(x) = x³ − 3x + 3 for real x.

Answer: f′(x) = 3x² − 3 = 3(x − 1)(x + 1). The critical inputs are −1 and 1.

At −1, f′ changes from positive to negative, so the local maximum value is f(−1) = 5. At 1, f′ changes from negative to positive, so the local minimum value is f(1) = 1.

The sign test establishes the kind of extremum; substitution into the original function gives its value. Substituting into the derivative instead would merely reproduce zero at these stationary points. Keep the classification and the value calculation as separate steps.

The curve marks a smooth local maximum at c₁ and a smooth local minimum at c₂, both with zero derivative. It also marks a pointed local maximum at c₃ and a pointed local minimum at c₄ where differentiability fails.

See Fig. 6.14 in your NCERT textbook

Here c₁, c₂, c₃ and c₄ label successive input positions in the diagram. The picture illustrates why the candidate search includes non-differentiable points. Continuity at the candidate remains part of the test, even when a derivative at the candidate itself does not exist.

When does the second derivative test work?

The second derivative, written f″(x), is the derivative of f′(x). At a stationary input it can classify a local extremum without a full sign chart. This test is often easier to apply than the first derivative test.

Theorem: Second derivative test

Suppose f is twice differentiable at an interior point c and f′(c) = 0. If f″(c) < 0, f has a local maximum at c. If f″(c) > 0, f has a local minimum there. The corresponding value is f(c).

If f″(c) = 0 as well, the test is inconclusive. Return to the first derivative test. Also, a positive or negative second derivative at a point is insufficient on its own: the stationary condition f′(c) = 0 must first be checked.

Why does the test give these conclusions?

By the definition of the second derivative, f″(c) is the limit of [f′(c + h) − f′(c)]/h as the non-zero increment h tends to zero. Since f′(c) = 0, this quotient reduces to f′(c + h)/h.

If the limit is positive, the quotient is positive for sufficiently small non-zero h. Therefore f′(c + h) is negative to the left and positive to the right. The first derivative test gives a minimum. A negative limit reverses these signs and gives a maximum.

Worked example 10. Classify the stationary point of f(x) = 2x³ − 6x² + 6x + 5.

Answer: f′(x) = 6(x − 1)², so the stationary input is 1. Also f″(x) = 12(x − 1), hence f″(1) = 0. The second derivative test is inconclusive.

However, f′ is positive just to both sides of 1. Thus x = 1 is neither a local maximum nor a local minimum.

This calculation shows why an inconclusive test must be followed by further analysis. It does not prove the absence of an extremum. The absence in this particular function follows from the positive derivative on both sides of the stationary point.

The same example also separates zero slope from a change of direction. A graph can flatten momentarily and continue increasing. Use the derivative's behaviour around the candidate, rather than interpreting the single equation f′(c) = 0 as a complete classification.

How are absolute maxima and minima found on a closed interval?

An absolute maximum is a greatest value across the entire specified domain. An absolute minimum is a least value across that domain. These are also called global extrema. Local extrema compare nearby values and need not be the global extrema.

Theorem: Existence on a closed interval

A function continuous on a closed interval [a, b] attains an absolute maximum and an absolute minimum, each at least once. Both continuity and the closed interval matter. Endpoints belong to the interval and must be considered.

1. Find critical inputs inside the interval, including places where the derivative is undefined but the function exists.
2. Add both endpoints to the candidate list.
3. Evaluate the original function at every candidate.
4. Compare all resulting values, stating the greatest and least values and where each occurs.

Worked example 11. Find the absolute extrema of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].

Answer: f′(x) = 6x² − 30x + 36 = 6(x − 3)(x − 2), giving critical inputs 2 and 3. Include the endpoints 1 and 5.

The values are f(1) = 24, f(2) = 29, f(3) = 28 and f(5) = 56. Hence the absolute maximum is 56 at x = 5, and the absolute minimum is 24 at x = 1.

Here both absolute extrema occur at endpoints, even though the interval contains stationary points. A derivative test at the interior candidates cannot replace the endpoint comparison. The final comparison uses function values rather than derivative values.

Why must the domain be stated?

For f(x) = x on the open interval (0, 1), neither endpoint is allowed. Every permitted input has both a smaller and a larger permitted input. Consequently, neither a minimum nor a maximum is attained.

Extending that same function to [0, 1] changes the answer: its minimum is 0 at x = 0 and its maximum is 1 at x = 1. The formula has stayed the same, but the domain has changed the available values.

How are practical optimisation problems translated into functions?

Optimisation means finding a greatest or least value under stated conditions. The objective function represents the quantity to optimise. A constraint is a relation restricting the variables, while the feasible domain consists of values allowed by the problem.

How is one variable eliminated?

Define the quantities before writing equations. Use the constraint to express the objective in one variable. Record restrictions, such as positive lengths, before solving the derivative equation. A root outside the feasible domain cannot provide a physical solution.

Worked example 12. Find two positive numbers with sum 15 whose sum of squares is minimum.

Let one number be x; the other is 15 − x, with 0 < x < 15. Let S(x) denote their sum of squares. Then S(x) = x² + (15 − x)² = 2x² − 30x + 225.

Answer: S′(x) = 4x − 30 vanishes at x = 15/2. Since S″(x) = 4 > 0, this gives a minimum. The derivative is negative before this input and positive afterwards throughout the feasible domain. The numbers are 15/2 and 15/2.

The sum condition supplies the second number. The quantity being minimised is the sum of squares, not the sum itself, which is fixed. Stating the objective explicitly prevents differentiation of an expression that does not answer the question.

How does a geometric constraint restrict the answer?

Worked example 13. Equal squares are removed from the corners of a rectangular aluminium sheet measuring 3 m by 8 m. The sides are folded up to make an open-topped box. Find its greatest volume. Here m means metre.

Let x metres be the cut-out square's side. The box has height x, length 8 − 2x and breadth 3 − 2x. All must be positive, so 0 < x < 3/2. Let V(x) be its volume.

V(x) = x(3 − 2x)(8 − 2x) = 4x³ − 22x² + 24x. Therefore V′(x) = 4(x − 3)(3x − 2), giving candidates 3 and 2/3.

Answer: Reject x = 3 because it makes a dimension negative. At x = 2/3, V″(x) = 24x − 44 gives V″(2/3) = −28 < 0. The greatest volume is 200/27 m³.

For the global conclusion, V′ is positive before 2/3 and negative afterwards within the feasible domain. Volume therefore rises to this candidate and falls away from it. This checks the whole permitted range, rather than stopping at a local classification.

Panel (a) shows corner squares marked with side x and a remaining central rectangle labelled 8 − 2x and 3 − 2x. Panel (b) shows the folded box with those base dimensions and height x.

See Fig. 6.23 in your NCERT textbook

Finish by returning to the quantity requested. A question about the cut-out size needs x, while a question about greatest volume needs V(x). Preserve the units: lengths are in metres and the box's volume is in cubic metres.

Glossary

• Derivative — The instantaneous rate of change of a function's output with respect to its input.
• Related rates — Rates of changing quantities connected by an equation and differentiated with respect to a common variable.
• Tangent — A line giving the local direction of a curve at a specified point.
• Normal — The line through a point on a curve perpendicular to its tangent there.
• Orthogonal intersection — An intersection where the tangents to the two curves are perpendicular to one another.
• Monotonic function — A function that is increasing throughout an interval or decreasing throughout that interval.
• Critical point — An input in the domain where the derivative is zero or the function is not differentiable.
• Stationary point — A point on a differentiable curve where the first derivative has value zero.
• Local maximum — A function value at least as large as the values at all sufficiently nearby inputs.
• Local minimum — A function value at most as large as the values at all sufficiently nearby inputs.
• Absolute extremum — A greatest or least function value over the entire domain under consideration.
• Second derivative — The derivative of the first derivative, used in one test for local extrema.
• Constraint — A stated condition connecting or restricting the variables in an optimisation problem.
• Feasible domain — The set of variable values permitted by every condition of the given problem.

Common errors and misconceptions

• Misconception: A derivative with respect to radius is automatically a rate per second. Correct: A time rate requires differentiation with respect to time, using the radius's time rate where necessary.
• Misconception: A decreasing length has a positive derivative because length is positive. Correct: A positive quantity can have a negative rate of change; retain the sign given by its motion.
• Misconception: The normal has slope 1/m when the tangent slope is m. Correct: Its slope is −1/m for finite non-zero m; a horizontal tangent has a vertical normal.
• Misconception: Every stationary point is a maximum or minimum. Correct: The derivative of x³ vanishes at zero, but the function keeps increasing through that input.
• Misconception: Extrema occur only where the derivative equals zero. Correct: Non-differentiable points can be extrema, and endpoints must be checked for absolute extrema on a closed interval.
• Misconception: A zero second derivative proves that there is no extremum. Correct: The second derivative test is inconclusive when both derivatives vanish; examine the first derivative's signs.
• Misconception: The greatest local maximum must be the absolute maximum on a closed interval. Correct: Compare all candidate values, including endpoint values, before making a global conclusion.
• Misconception: Every root of an optimisation equation is an acceptable answer. Correct: Reject roots outside the feasible domain and verify the surviving candidate's required maximum or minimum behaviour.

Exam-style questions with model answers

Q1. Define a critical point of a function f. Explain, using f(x) = x³ for real x, why a critical point need not be a local extremum. [2 marks]

1. A critical point is an input c in the domain where f′(c) = 0 or f is not differentiable.
2. For f(x) = x³, f′(x) = 3x² vanishes at zero but is positive on both sides. Thus zero is critical, yet is neither a local maximum nor a local minimum.

Q2. A circle's radius r increases at 4 cm/s. Using A = πr² for its area A and t for time, find the rate of increase of area when r = 10 cm. [3 marks]

1. Differentiate the area relation with respect to time, retaining the radius's dependence on time: dA/dt = 2πr(dr/dt).
2. At the stated instant, r = 10 cm and dr/dt = 4 cm/s. Therefore dA/dt = 2π × 10 × 4 = 80π cm²/s.
3. The rate is positive, so the enclosed area is increasing at 80π square centimetres per second at that instant.

Q3. For real x, find the intervals on which f(x) = 4x³ − 6x² − 72x + 30 is increasing or decreasing. [4 marks]

1. Differentiate and factorise: f′(x) = 12x² − 12x − 72 = 12(x − 3)(x + 2).
2. The derivative vanishes at −2 and 3, dividing the real line into three intervals for the sign analysis.
3. Both factors have the same sign before −2 and after 3. Therefore f is increasing on (−∞, −2) and (3, ∞).
4. Between −2 and 3, the factors have opposite signs. Therefore f is decreasing on (−2, 3).

Q4. Find and classify all stationary points of f(x) = 2x³ − 6x² + 6x + 5 for real x. Explain what happens to the second derivative test. [4 marks]

1. The first derivative is f′(x) = 6x² − 12x + 6 = 6(x − 1)², giving the sole stationary input x = 1.
2. The second derivative is f″(x) = 12(x − 1), so f″(1) = 0 and the second derivative test is inconclusive.
3. For inputs sufficiently close to 1 on either side, 6(x − 1)² is positive. Hence the derivative does not change sign there.
4. Therefore x = 1 is neither a local maximum nor a local minimum. Its graph point is (1, 7), since f(1) = 7.

Q5. Find the absolute maximum and absolute minimum of f(x) = 2x³ − 15x² + 36x + 1 on the closed interval [1, 5], stating where they occur. [5 marks]

1. The polynomial is continuous on [1, 5], so it attains absolute maximum and minimum values on this closed interval.
2. Differentiate: f′(x) = 6x² − 30x + 36 = 6(x − 3)(x − 2). The interior critical inputs are 2 and 3.
3. Evaluate these candidates in the original function: f(2) = 29 and f(3) = 28. These values alone do not complete the comparison.
4. Include both endpoints: f(1) = 24 and f(5) = 56. The complete set of candidate values is therefore 24, 29, 28 and 56.
5. The absolute maximum is 56 at x = 5, and the absolute minimum is 24 at x = 1.

Q6. Equal squares are cut from the four corners of a 3 m by 8 m aluminium sheet. The sides are folded to form an open-topped rectangular box. Find the cut-out side and the greatest possible box volume. [6 marks]

1. Let x metres be the cut-out side. The height is x and the base dimensions are 3 − 2x and 8 − 2x metres, requiring 0 < x < 3/2.
2. Let V(x) denote volume. Then V(x) = x(3 − 2x)(8 − 2x) = 4x³ − 22x² + 24x.
3. Differentiating gives V′(x) = 12x² − 44x + 24 = 4(x − 3)(3x − 2), which vanishes at 3 and 2/3.
4. The input 3 is outside the feasible domain and must be rejected. Thus the relevant candidate is x = 2/3 metre.
5. V′ is positive before 2/3 and negative afterwards within the feasible domain. The volume therefore attains its greatest value at this candidate.
6. Substitution gives V(2/3) = (2/3)(5/3)(20/3) = 200/27 m³. Cut out squares of side 2/3 metre.

Key takeaways

• A derivative measures change with respect to a specified variable; use the chain rule to connect related rates.
• Tangent slope comes from the derivative; a finite non-zero tangent slope gives normal slope equal to its negative reciprocal.
• The angle between curves is calculated from their tangents at the same intersection point.
• The derivative's sign across an interval establishes increase or decrease, while its sign change can classify a local extremum.
• A zero derivative identifies a stationary candidate; it does not, by itself, establish a maximum or minimum.
• The second derivative test requires a stationary point and becomes inconclusive when the second derivative also vanishes.
• For absolute extrema on a closed interval, compare original function values at interior critical points and both endpoints.
• In optimisation, define the objective, apply the constraint, restrict the domain, verify the extremum and report the requested quantity.

Test yourself

What additional information connects dA/dr to dA/dt for a changing circular area A with radius r and time t?

The radius's time rate dr/dt is needed: dA/dt = (dA/dr)(dr/dt).

A tangent at (x₁, y₁) is horizontal. What is the normal's equation?

The normal is vertical, so its equation is x = x₁.

For two finite tangent slopes m₁ and m₂ at a common point, what condition gives an orthogonal intersection?

The tangents must be perpendicular, requiring the slope product m₁m₂ = −1.

If a continuous function's derivative changes from positive to negative through a critical input c, what occurs there?

A local maximum occurs at c, with local maximum value f(c).

Why does f(x) = x³ have no local extremum at zero despite f′(0) = 0?

Its derivative is positive on both sides of zero, so the function continues increasing.

What does f′(c) = f″(c) = 0 tell you about the second derivative test?

The test is inconclusive; examine the first derivative's signs on either side.

Which candidates must be compared for absolute extrema of a continuous function on [a, b]?

Compare its values at interior critical points and at both endpoints a and b.

Why is x = 3 inadmissible for a box cut from a 3 m by 8 m sheet using corner squares of side x metres?

It makes the dimension 3 − 2x negative, violating the requirement of positive box dimensions.