Porisma

Applications of differentiation

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By the end, you can

  • Connect derivative signs to increasing and decreasing behaviour
  • Classify stationary points and find tangents
  • Model optimisation and connected rates

about 40 minutes for the lesson · practise at your own pace.

The central idea

The first derivative gives slope and the second describes how that slope changes. Stationary points satisfy f′(x)=0, but their nature needs a sign change or another test. In optimisation, reduce the quantity to one variable, preserve the physical domain, and compare candidates with boundaries. Connected rates differentiate a relation with respect to time.

Increasing versus strictly increasing

On an interval I, for every a<ba<b, increasing means f(a)≤f(b)f(a)\le f(b) and strictly increasing means f(a)<f(b)f(a)<f(b). Decreasing and strictly decreasing reverse those inequalities to ≥\ge and >>.

If f is differentiable and f′(x)>0f^{\prime}(x)>0 throughout I, f is strictly increasing there; a negative derivative throughout gives strict decrease. The source curve increases on (a,b)(a,b) and (c,d)(c,d) and decreases on (b,c)(b,c). These are sufficient tests: strict increase can still allow an isolated zero derivative, as with x3x^3 at zero.

Why applications need a model

The source introduces uses in physics (motion, electricity, heat, light, harmonics, acoustics, astronomy and dynamics), chemistry (reaction rates and radioactive decay), biology (birth/death rates), economics (marginal cost/revenue and profit), engineering, statistics and medicine.

Its supermarket example asks how to choose the height-to-diameter ratio of a cylindrical can to minimise manufacturing surface area at fixed volume.

The original can photograph is retained in the source archive, with its printed DataGenetics attribution; the linked external page has not been reviewed.

The full chapter outline also includes interpreting first/second derivative signs, relating a function’s graph to its derivative’s graph, tangents/normals including implicit curves, calculator extrema, analytical classification, optimisation and connected rates. The complete lecture now connects these ideas through sixteen examples, three self-practices, source diagrams and the linked Discussion/Review questions.

Increasing and decreasing intervals

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Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
Illustrative source-labelled curve through A,B,C,D: rising from a to b, falling from b to c, rising from c to d. B is a local peak and C a local trough.
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Method map: Classify a stationary point

Start at an interior input a where f′(a) = 0. Either test can establish a local maximum or minimum; an inconclusive test needs more information.

  • First-derivative test

    Use signs that hold on small intervals on either side of a. Two isolated sample values are not a proof.

    f′(a)=0f^{\prime}(a)=0

    Positive → zero → negativeLocal maximum

    The curve rises into the point, then falls. Its value is greater than nearby values.

    Negative → zero → positiveLocal minimum

    The curve falls into the point, then rises. Its value is less than nearby values.

    Positive on both sides, or negative on both sidesNo local extremum

    There is no turning point. Confirm a change of concavity before calling this a stationary inflection; the same-sign pattern alone does not prove one.

    These conclusions are local, not a comparison with the whole graph. If the one-sided signs do not settle, this chart is inconclusive; use another local argument.

  • Second-derivative test

    At the same stationary input, apply the test under the usual local differentiability assumptions. This is an alternative test; you do not have to perform both tests.

    f′(a)=0f^{\prime}(a)=0

    Negative second derivativeLocal maximum

    f′′(a)<0f^{\prime\prime}(a)<0 establishes a strict local maximum.

    Positive second derivativeLocal minimum

    f′′(a)>0f^{\prime\prime}(a)>0 establishes a strict local minimum.

    Zero or unavailable second derivativeInconclusive

    Return to the first-derivative test or another valid local argument. A zero second derivative does not prove an inflection.

    The worked comparisons below show three different outcomes with f′′(0)=0f^{\prime\prime}(0)=0. Predict each outcome before opening its reasoning.

    Find the point’s height by substituting a into f, not into a derivative. For an optimisation problem, compare valid candidates with relevant domain boundaries too.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Compare the cubic at zero

For the illustrative function f(x)=x3f(x)=x^3, classify the stationary point at x=0x=0. All three comparisons have f′′(0)=0f^{\prime\prime}(0)=0: what extra evidence decides this one?

Show reasoning
  1. 1. Recognise the inconclusive test

    The input is stationary, but a zero second derivative does not classify the point.

    f′(x)=3x2,f′′(x)=6x,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=3x^2,\\f^{\prime\prime}(x)&=6x,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Check the signs on intervals

    For every x<0x<0 and every x>0x>0, 3x2>03x^2>0. The curve is increasing on both sides, so the point is neither a local maximum nor a local minimum. These are interval statements, not just two sampled slopes.

  3. 3. Establish the change of concavity

    The second derivative 6x6x is negative to the left and positive to the right of zero. Concavity changes, and f(0)=0f(0)=0, so (0,0)(0,0) is a stationary point of inflection. Same-sign first derivatives alone would not establish the concavity change.

Compare the quartic at zero

For the illustrative function f(x)=x4f(x)=x^4, classify the stationary point at x=0x=0. Does f′′(0)=0f^{\prime\prime}(0)=0 force the same outcome as the cubic? Predict the signs of f′f^{\prime} on either side.

Show reasoning
  1. 1. Recognise the same inconclusive result

    Both derivatives vanish at zero. The second-derivative test supplies no classification.

    f′(x)=4x3,f′′(x)=12x2,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=4x^3,\\f^{\prime\prime}(x)&=12x^2,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Use the first-derivative sign change

    For every x<0x<0, 4x3<04x^3<0; for every x>0x>0, 4x3>04x^3>0. The curve falls into the point and rises away, establishing a strict local minimum at (0,0)(0,0). The height comes from f(0)f(0), not from a derivative.

  3. 3. Check with the function values

    Every nonzero nearby input has x4>0=f(0)x^4>0=f(0). This independently confirms the strict minimum. Also, 12x2>012x^2>0 on both sides: concavity does not change at zero, so this is not an inflection.

Reflect the quartic in the x-axis

For the illustrative function f(x)=−x4f(x)=-x^4, classify the stationary point at x=0x=0. The second derivative is still zero there. Predict what reflecting the quartic does to its first-derivative signs.

Show reasoning
  1. 1. Keep the test's limitation

    The negative sign changes the nearby slopes, but both derivatives at zero are still zero.

    f′(x)=−4x3,f′′(x)=−12x2,f′(0)=f′′(0)=0.\begin{aligned}f^{\prime}(x)&=-4x^3,\\f^{\prime\prime}(x)&=-12x^2,\\f^{\prime}(0)&=f^{\prime\prime}(0)=0.\end{aligned}
  2. 2. Reverse the sign change

    For every x<0x<0, −4x3>0-4x^3>0; for every x>0x>0, −4x3<0-4x^3<0. The curve rises into the point and falls away, establishing a strict local maximum at (0,0)(0,0).

  3. 3. Check the value and concavity

    Every nonzero nearby input has −x4<0=f(0)-x^4<0=f(0). The second derivative is negative on both sides, so concavity does not change. Across the three comparisons, the same zero second derivative permits an inflection, a minimum or a maximum.

Compare with the curve sketches

Compare the four local configurations below. All four functions are defined for every real x; each plot shows only the stated window. These functions illustrate the source’s four cases, not equations inferred from its combined A/B/C sketch.

Local maximum: rising, then falling

f(x)=x3−3xf(x)=x^3-3x. Inspect −2≤x≤0-2\le x\le0 around the stationary point (−1,2)(-1,2).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −2<x<−1-2<x<-1, f′(x)>0f^{\prime}(x)>0: rising.

At the point

f′(−1)=0f^{\prime}(-1)=0. Horizontal tangent y=2y=2.

Right of the point

On −1<x<0-1<x<0, f′(x)<0f^{\prime}(x)<0: falling.

Since f′(x)=3(x−1)(x+1)f^{\prime}(x)=3(x-1)(x+1), these signs hold throughout the stated intervals. They establish a strict local maximum at (−1,2)(-1,2). Also f′′(−1)=−6<0f^{\prime\prime}(-1)=-6<0. The full cubic is unbounded above, so this is not a whole-graph maximum.

Local minimum: falling, then rising

f(x)=x3−3xf(x)=x^3-3x. Inspect 0≤x≤20\le x\le2 around the stationary point (1,−2)(1,-2).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On 0<x<10<x<1, f′(x)<0f^{\prime}(x)<0: falling.

At the point

f′(1)=0f^{\prime}(1)=0. Horizontal tangent y=−2y=-2.

Right of the point

On 1<x<21<x<2, f′(x)>0f^{\prime}(x)>0: rising.

The factorisation f′(x)=3(x−1)(x+1)f^{\prime}(x)=3(x-1)(x+1) establishes these signs on whole intervals, giving a strict local minimum at (1,−2)(1,-2). Also f′′(1)=6>0f^{\prime\prime}(1)=6>0. The full cubic is unbounded below, so this is not a whole-graph minimum.

Stationary inflection: rising on both sides

f(x)=x3f(x)=x^3. Inspect −1≤x≤1-1\le x\le1 around (0,0)(0,0).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −1<x<0-1<x<0, f′(x)=3x2>0f^{\prime}(x)=3x^2>0.

At the point

f′(0)=0f^{\prime}(0)=0. Horizontal tangent y=0y=0.

Right of the point

On 0<x<10<x<1, f′(x)=3x2>0f^{\prime}(x)=3x^2>0.

There is no turning point. To establish an inflection as well, check f′′(x)=6xf^{\prime\prime}(x)=6x: it is negative to the left and positive to the right, so concavity changes. The zero value f′′(0)=0f^{\prime\prime}(0)=0 alone is inconclusive.

Stationary inflection: falling on both sides

f(x)=−x3f(x)=-x^3. Inspect −1≤x≤1-1\le x\le1 around (0,0)(0,0).

Dashed: horizontal tangent. Dot: stationary point. Coordinates are stated above.
Left of the point

On −1<x<0-1<x<0, f′(x)=−3x2<0f^{\prime}(x)=-3x^2<0.

At the point

f′(0)=0f^{\prime}(0)=0. Horizontal tangent y=0y=0.

Right of the point

On 0<x<10<x<1, f′(x)=−3x2<0f^{\prime}(x)=-3x^2<0.

The curve keeps falling, so there is no turning point. Here f′′(x)=−6xf^{\prime\prime}(x)=-6x changes from positive to negative, establishing the concavity change. Same-sign first derivatives alone do not prove every function has an inflection.

A local maximum or minimum compares nearby function values. Check relevant domain boundaries separately for optimisation; the displayed window does not establish a whole-graph extremum.

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Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
Combined illustrative curve has a local maximum A at a, a local minimum B at b and a stationary inflection C at c. Four separate panels show first-derivative signs: maximum (+, 0, −); minimum (−, 0, +); rising inflection (+, 0, +); falling inflection (−, 0, −), with horizontal tangent segments and concavity change at both inflections.
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Slope and concavity are different

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Four source configurations with tangent segments: case 1 increasing concave down, tangents above on (a,b); case 2 increasing concave up, tangents below on (c,d); case 3 decreasing concave down, above on (c,d); case 4 decreasing concave up, below on (a,b). Each panel shows both derivative signs.
Four source configurations with tangent segments: case 1 increasing concave down, tangents above on (a,b); case 2 increasing concave up, tangents below on (c,d); case 3 decreasing concave down, above on (c,d); case 4 decreasing concave up, below on (a,b). Each panel shows both derivative signs.
Four source configurations with tangent segments: case 1 increasing concave down, tangents above on (a,b); case 2 increasing concave up, tangents below on (c,d); case 3 decreasing concave down, above on (c,d); case 4 decreasing concave up, below on (a,b). Each panel shows both derivative signs.
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Stationary and non-stationary inflections

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Graphs of x³, x³+x and x⁴. Both cubics change concavity at 0: the first has zero slope there, the second slope 1. The quartic has second derivative zero at 0 but no change of concavity.
Graphs of x³, x³+x and x⁴. Both cubics change concavity at 0: the first has zero slope there, the second slope 1. The quartic has second derivative zero at 0 but no change of concavity.
Graphs of x³, x³+x and x⁴. Both cubics change concavity at 0: the first has zero slope there, the second slope 1. The quartic has second derivative zero at 0 but no change of concavity.
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A cubic and its derivative

One input, two different heights

Compare f(x) = 0.1x(x − 6)(x + 6) with f′(x) = 0.3x² − 3.6. The upper height is the function value; the lower height is its slope. Both graphs use the same horizontal scale.

Compare the same inputs

Input xOriginal f(x)Slope f′(x)Interpretation
−2√34.8√3 ≈ 8.310Local maximum; derivative crosses down through zero.
00−3.6The original crosses the axis while decreasing; it is not stationary.
2√3−4.8√3 ≈ −8.310Local minimum; derivative crosses up through zero.

The table and original diagram remain available without loading the explorer. Displayed decimals are approximate.

Read slope as derivative height
Teaching and source notes

The source prints approximate turning coordinates (−3.46, 8.31) and (3.46, −8.31). The exact inputs are ±2√3 and the exact original heights are ∓4.8√3. Presets use the exact inputs rather than the printed rounded decimals.

The Teacher answer connects original turning inputs to derivative x-intercepts. At the left point the derivative crosses from positive to negative; at the right it crosses from negative to positive. A derivative zero alone is not a general turning-point test.

At x = 0 the derivative is −3.6, so the original point is not stationary. A point on the axis need not have a horizontal tangent.

Aligned graphs of 0.1x(x−6)(x+6) and 0.3x²−3.6. The original turning points align with derivative zeros at inputs ±2√3. Printed rounded labels ±3.46 and heights ±8.31 are retained.
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Method map: Read slope as derivative height

Keep the same input. Transfer the original curve's slope to the derivative's height, then inspect both sides of each stationary input. Crossings and touches carry different information.

  1. Keep the input; change the height

    Work where the original function is differentiable. Its height and its slope are different quantities; the derivative graph records the slope.

    x=a,y=f′(a)x=a,\quad y=f^{\prime}(a)

    Positive original slopeDerivative above the axis

    Where f′(x)>0f^{\prime}(x)>0, the derivative graph has positive height. This does not require the original graph itself to be above the axis.

    Zero original slopeDerivative on the axis

    A stationary original point (a,b)(a,b) gives the derivative point (a,0)(a,0). The original height bb does not carry across.

    Negative original slopeDerivative below the axis

    Where f′(x)<0f^{\prime}(x)<0, the derivative graph has negative height. Read the original curve’s direction, not its position relative to the axis.

    A corner or another input where the derivative does not exist is not automatically a derivative zero. The full sketch also needs the function’s domain and justified boundary/asymptote information; this companion covers the local slope transfers.

  2. A turning point gives a crossing in these cases

    Use signs that hold on intervals on either side of the stationary input, not two sampled slopes. The source's local maximum and minimum configurations have opposite signs across the point.

    f′(a)=0f^{\prime}(a)=0

    Positive → zero → negativeOriginal maximum; derivative crosses down

    The original rises into the point and falls away. The derivative is above the axis on the left and below it on the right, passing through (a,0)(a,0).

    Negative → zero → positiveOriginal minimum; derivative crosses up

    The original falls into the point and rises away. The derivative changes from negative to positive through (a,0)(a,0).

    These are the sign-changing local configurations shown in the source. A derivative zero by itself does not establish a turning point.

  3. Same signs give a touch at zero

    Here the derivative is zero at the stationary input and has one strict sign throughout each neighbouring interval. The original keeps its direction through the point.

    f′(a)=0f^{\prime}(a)=0

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    Positive → zero → positiveDerivative minimum at zero

    The derivative is positive nearby and zero at the point: it touches from above. The source’s rising stationary-inflection configuration has this pattern.

    Negative → zero → negativeDerivative maximum at zero

    The derivative is negative nearby and zero at the point: it touches from below. The source’s falling stationary-inflection configuration has this pattern.

    Same-sign first derivatives show no original turning point. Confirm a change of the original curve’s concavity before calling it an inflection; the sign pattern alone is insufficient. The worked cubics below establish that extra condition.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

A maximum above the axis

Consider this illustrative function:

f(x)=2−x2.f(x)=2-x^2.

At input zero, predict the derivative’s height and crossing direction. Does the original height 2 carry across?

Show reasoning
  1. 1. Transfer the slope, keeping the input

    The original point is (0, 2). Its slope is zero, so the derivative point is (0, 0).

    f′(x)=−2x,f′(0)=0.\begin{aligned}f^{\prime}(x)&=-2x,\\f^{\prime}(0)&=0.\end{aligned}
  2. 2. Read both intervals

    For every x<0x<0, −2x>0-2x>0; for every x>0x>0, −2x<0-2x<0. The derivative crosses down through zero. The original curve has a strict local maximum at (0,2)(0,2).

A minimum below the axis

Consider this illustrative function:

f(x)=x2−2.f(x)=x^2-2.

At input zero, predict where the derivative meets the axis. Does an original point below the axis force a negative derivative?

Show reasoning
  1. 1. Differentiate independently of the height

    The original point is (0, -2), while its horizontal tangent gives derivative point (0, 0).

    f′(x)=2x,f′(0)=0.\begin{aligned}f^{\prime}(x)&=2x,\\f^{\prime}(0)&=0.\end{aligned}
  2. 2. Read the upward crossing

    For every x<0x<0, 2x<02x<0; for every x>0x>0, 2x>02x>0. The derivative crosses from below to above. The original curve has a strict local minimum at (0,−2)(0,-2). At x=1x=1, the original height is −1-1 but the derivative height is 22: the original is below the axis while its derivative is above.

Rising on both sides

Consider this illustrative function:

f(x)=x3+1.f(x)=x^3+1.

At input zero, predict whether the derivative crosses or touches the axis. Then check whether the original has an inflection.

Show reasoning
  1. 1. Locate the derivative's minimum

    The squared term is positive at every nonzero input, while the derivative is zero at input zero. The derivative touches zero from above and has a strict minimum there. The original point is (0,1)(0,1), not (0,0)(0,0).

    f′(x)=3x2,f′(0)=0.\begin{aligned}f^{\prime}(x)&=3x^2,\\f^{\prime}(0)&=0.\end{aligned}
  2. 2. Check the original concavity separately

    The second derivative 6x6x is negative to the left of zero and positive to the right. This change of concavity establishes the original stationary inflection. Positive first derivatives on both sides alone would not prove that change.

    f′′(x)=6x.f^{\prime\prime}(x)=6x.

Falling on both sides

Consider this illustrative function:

f(x)=−x3−1.f(x)=-x^3-1.

Near input zero, predict which side of the axis contains the derivative. Is its zero a maximum or a minimum of the derivative?

Show reasoning
  1. 1. Locate the derivative's maximum

    The negative squared term is negative at every nonzero input, while the derivative is zero at input zero. The derivative touches zero from below and has a strict maximum there. The original point is (0,−1)(0,-1).

    f′(x)=−3x2,f′(0)=0.\begin{aligned}f^{\prime}(x)&=-3x^2,\\f^{\prime}(0)&=0.\end{aligned}
  2. 2. Confirm the original inflection

    The second derivative −6x-6x is positive to the left of zero and negative to the right. Concavity changes, so the original point is a stationary inflection. The curve keeps falling through it; it is not an original turning point.

    f′′(x)=−6x.f^{\prime\prime}(x)=-6x.

Compare with the paired source sketches

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Four paired local sketches: a maximum gives a positive-to-negative derivative crossing; a minimum gives a negative-to-positive crossing. A decreasing stationary inflection gives a derivative maximum touching 0 from below; an increasing one gives a derivative minimum touching 0 from above.
Four paired local sketches: a maximum gives a positive-to-negative derivative crossing; a minimum gives a negative-to-positive crossing. A decreasing stationary inflection gives a derivative maximum touching 0 from below; an increasing one gives a derivative minimum touching 0 from above.
Four paired local sketches: a maximum gives a positive-to-negative derivative crossing; a minimum gives a negative-to-positive crossing. A decreasing stationary inflection gives a derivative maximum touching 0 from below; an increasing one gives a derivative minimum touching 0 from above.
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Tangent and normal through the same point

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Illustrative source configuration: rising curve through P(x₁,y₁), tangent following its local direction, and normal through P perpendicular to the tangent, with a right-angle marker.
Illustrative source configuration: rising curve through P(x₁,y₁), tangent following its local direction, and normal through P perpendicular to the tangent, with a right-angle marker.
Illustrative source configuration: rising curve through P(x₁,y₁), tangent following its local direction, and normal through P perpendicular to the tangent, with a right-angle marker.
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Explore

Plot f(x)=x3−3xf(x)=x^3-3x, g(x)=f(x)+5g(x)=f(x)+5 and their derivatives. Before inspecting the plots, predict the stationary inputs, the increasing/decreasing intervals and which coordinates move under the translation. Compare your predictions with exact derivatives. Explain why identical derivative graphs do not determine a unique function graph without a height condition.

Related interactive example

Explore this related example, then return to the prediction task above.

How fast does the level rise at each depth?

dh/dt = 1920/(9πh²) falls steeply as the water gets deeper. Slide the line to find the depth for a given rate.

Height of the horizontal linek = 5.65
Try an example
Number of distinct depths
1

Each rate corresponds to one depth. At 5.65 cm/s the depth is h≈3.47h\approx3.47 cm; the deeper the water, the slower the rise.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
3.466815.65
  • Dh/dt against the depth h
  • Chosen rate
  • Depth with that rate
  • Intersection selected in the coordinate table
Key idea

Constant inflow in a widening container means a slowing rise.

Ideas to try
At what depth is the level rising at 1 cm/s?
Open the full interactive problem →

Worked example

A rectangular pen stands against a straight wall, so 24 m of fencing is used for its other three sides. Let xx be each side perpendicular to the wall and yy the remaining side.

The constraint is 2x+y=242x+y=24 and the area is A=xy=x(24−2x)A=xy=x(24-2x), with 0<x<120<x<12. Thus A′=24−4xA^{\prime}=24-4x, giving x=6x=6 and y=12y=12.

The derivative is positive before 6 and negative after 6 on the feasible domain.

Together with area tending to zero at both ends, this establishes the greatest area of 72 m272\ \mathrm{m}^2.

A stationary value is a candidate. Check the physical domain and allowed endpoints or limiting boundary behaviour before claiming an overall optimum.

Check your understanding

An open box is made from a 24 cm square sheet by cutting a square of side xx cm from each corner and folding up the four sides. Construct a model and find the greatest possible volume. State the feasible domain and justify that your dimensions give the greatest volume, rather than merely a stationary value.

Mastery learning objectives

At the end of this chapter, I should be able to:

Intervals and concavity

Determine intervals where a given function is strictly increasing.

Determine intervals where a given function is strictly decreasing.

Determine intervals where a given function is concave upwards.

Determine intervals where a given function is concave downwards.

Find stationary points

Find stationary points analytically.

Find stationary points graphically.

Classify stationary points

Determine the nature of stationary points using the first derivative test.

Determine the nature of stationary points using the second derivative test.

Sketch a derivative from a function graph

Determine the derivative graph’s x-intercepts.

Determine the derivative graph’s asymptotes, checking the source curve’s slope behaviour.

Determine the derivative graph’s stationary points.

Tangents and normals

Find the tangent equation at a given point of a curve.

Find the normal equation at a given point of a curve.

Applications

Solve local maxima and minima problems.

Solve connected rates of change problems.

Choose a practice question

The five rows follow the printed page 53 question-type map, with links to the worked lecture examples, Discussion questions and Review problems. Use the lecture and tutorial checkpoints above to track progress.

Common question typeLecture examplesDiscussion questionsReview follow-up
Graph of a function and its derivative3, 5, 6, 71, 2, 3, 4, 6(iii)1, 2, 3, 4, 5(iii), 6
Strict increase/decrease, concavity and stationary points (maximum/minimum/inflection)1, 2, 45, 6(i), 6(ii)5(i), 5(ii)
Tangents and normals8, 9, 10, 117, 8, 97
Connected rate of change12, 1310, 13(a)8, 9(iv)
Maximisation and minimisation14, 15, 1611, 12, 13(b)9(i), 9(ii), 9(iii), 10, 11, 12

Source qualifications: Discussion 4 asks to recover hh from ∣h∣|h|; its placement in the printed graph row is retained. The Discussion key on page 60 supplies selected answers only; the Review key on page 66 also omits some asks, so the linked reviewed solutions retain the complete working.

Review 5’s printed strict-decrease ranges exclude the isolated zero derivative at x=−3x=-3, while its function still strictly decreases through that point.

Review 9’s page 66 key prints +12+12 square units/s, but y′(0)=−1/2y^{\prime}(0)=-1/2 and dx/dt=8dx/dt=8 give dA/dt=−12dA/dt=-12: a decrease of 12. The original key is retained in the native text.

Review 12’s right-endpoint test and missing minimum area 24 are qualified and completed in its linked reviewed solution. Printed source evidence and independent corrections remain distinct.

Practice questions

Practice section 1 · Questions 1–6

Classify the turning points of a rational curve · Classify stationary points on two implicit branches · +4 more

Practice section 2 · Questions 7–12

Sketch derivative branches around two vertical asymptotes · Transfer an oblique asymptote to the derivative sketch · +4 more

Practice section 3 · Questions 13–18

Find the area between an exponential tangent and normal · Find an exact triangle area from an implicit normal · +4 more

Practice section 4 · Questions 19–24

Minimise the metal in a fixed-volume cylindrical can · Match a function to its derivative · +4 more

Practice section 5 · Questions 25–30

Recover a function from its gradient and one value · Find horizontal and vertical tangents to a closed curve · +4 more

Practice section 6 · Questions 31–36

Minimise the cone circumscribed about a sphere · Combine a leaking cone with trapezium optimisation · +4 more

Practice section 7 · Questions 37–42

Reconstruct a decreasing function with an isolated zero gradient · Determine a rational curve and sketch its derivative exactly · +4 more

Practice section 8 · Questions 43–44

Maximise a complete globe on a closed cylindrical base · Optimise corner flower crates around a central pond

More practice

Draw the situation and define the quantities before choosing a derivative. Check units, the feasible domain and the justification for the requested maximum, minimum or rate.

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