Higher: understand variable rates and local gradients
Distinguish a finite-interval average from the local rate at a selected input, with signs, graph directions and appropriate units.
Exam scope and assessment notes
Higher only: Foundation has no section3.4 Calculus requirement. Higher also includes all shared Foundation content. These six original lessons teach variable/local versus interval rates, termwise integer-power differentiation, gradients/rates, stationary coordinates and graph-shape classification, one-dimensional kinematics and other simple practical models. H:3.4C requires both the gradient/rate and stationary-point lessons; H:3.4E requires both the motion and other-practical lessons with worked examples and all Guided, Independent and Transfer stages. The six-lesson split is StudyForA’s teaching design, not a syllabus-prescribed count. Small negative-integer powers with x≠0 conservatively retain integer-power breadth; fractional/trigonometric differentiation, integration and product/quotient/chain rules are not inferred requirements. H:3.4D classifies maxima and minima by general original-graph shape only; a second derivative is used for acceleration in motion, not as a substitute classification test. A derivative zero supplies an input, not the original stationary ordinate or an automatic turning point. Keep every original domain, physical restriction, admissible integer count and endpoint. Continuous relaxation supports differentiation for discrete counts; it does not make noninteger production feasible. Accepted boundary maxima need not be stationary. The separate H3.3D drawn-tangent construction is not replaced or newly credited by calculus. Fixed interactive examples support exact algebraic self-checks and approximate graph exploration; no general expression/root solver or written-work marker is supplied. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks, prove a construction or establish mastery. All five Calculus source rows have structural teaching and staged-practice links; exhaustive subskill approval, specialist approval and cumulative formal assessment remain unestablished. The wider Maths A course and other Pearson qualifications remain incomplete. Publisher modular-map tags and uncorrected proof navigation are separate evidence: Book2 Graphs9 map pages386–415 differs from proof start385 and its later calculus subheadings. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Read signed coordinates, graph scales and units.
- Calculate a line gradient as output change divided by input change.
- Recognise tangent and secant directions on nonlinear graphs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A record rises from 5 units at time 1 s to 17 units at time 4 s. What is its average rate over that interval?
What you will learn
- Compare constant and variable rates using original graphs and finite secants.
- Interpret a local rate as the tangent direction at a chosen input rather than an interval average.
- Use signed output-per-input units and respect the stated model interval.
f(x)=x² + 1. Original domain: x in all real inputs. Original values use output units; first rates use output units per input unit.
The panels share the same numbered horizontal scale and have independent vertical scales. Original height, derivative value and derivative zero have different meanings. No input, derivative solution or reference point is preselected.
No input is preselected. Choose a point on your own drawing or a rate input you want to check.
Blue: original; purple: first derivative. Vertical scales differ. Filled boundary markers are included; open markers show excluded finite limits or holes. Traces split at original exclusions and are clipped to each panel; offscreen points are omitted rather than moved to a border. A wider view does not change a domain.
An average belongs to an interval
The average rate of change from input a to input b is [f(b)−f(a)]/(b−a), with a≠b. On the graph it is the gradient of the secant joining those two curve points. Calculate the output difference with its sign, then divide by the input change. It measures the net change over that whole interval, not every direction the curve takes within it.
A straight line has one constant gradient, so all its finite secants have that rate. A nonlinear graph can have different averages on different intervals. Equal input steps need not produce equal output changes. Compare the actual coordinates on the original graph; a bigger output alone does not say how quickly that output is changing.
A local rate belongs to a chosen input
The local rate at a selected input describes the curve’s direction there. Mark the point and draw its tangent direction before checking any numerical reference. A rising tangent has a positive rate, a falling tangent a negative rate, and a horizontal tangent a zero rate. A horizontal direction at one point does not make the whole function constant.
The notation f′(x), or dy/dx when y=f(x), names the local rate as a function of the input. At one chosen input c, f′(c) is one rate value. The next lesson gives algebraic rules for finding it; no formal limit proof is needed here. An average can happen to equal a local rate at some point, but that does not turn an interval result into a general local-rate rule.
Keep rate units, domains and accuracy
Rate units are output units per input unit. A temperature change measured in degrees Celsius over minutes has rate °C/min. A negative value means temperature is falling as time increases; it does not mean that the temperature itself must be negative. Use the axis numbers when reading a tangent or secant, rather than its screen angle.
A model used only on a stated interval makes no claim about earlier or later inputs. Use local directions inside that interval, and treat any boundary rate through the permitted side or an explicitly stated model extension. Finite coordinate calculations can be exact, while a freehand tangent reading is approximate. This concept work does not replace the separate requirement to draw and measure a tangent in the graph lesson.
Sources for this lesson
Pause and explain
A temperature graph is falling at a selected instant. Which local-rate statement is justified?
Worked example
Draw the original curve y=x²+1 and the secants for the intervals [1,3] and [3,5]. Calculate both average rates. At x=1 and x=3, draw the local tangent directions before inspecting numerical references. Compare their local rates with the two averages, and explain why no single interval slope is the general local rate. The function has all real inputs.
Show the worked solution
- The original points at x=1,3,5 are (1,2), (3,10), (5,26). The first secant gives (10−2)/(3−1)=8/2=4; the second gives (26−10)/(5−3)=16/2=8. Draw and label these two different secants on the curve.
- Attempt the two local tangent directions first. After the attempt, reference local gradients are 2 at x=1 and 6 at x=3. Your drawn readings should be about those values; they describe directions at individual points. They are not the same as the interval averages 4 and 8.
- The graph becomes steeper as these inputs increase, so a variable local rate is consistent with the changing shape. The original heights 2 and 10 are not the local gradients 2 and 6, even when one number happens to coincide. Keep the all-real function domain, and distinguish exact finite averages from approximate tangent readings.
Answer Average rates: 4 on [1,3] and 8 on [3,5]. Local tangent readings should be about 2 at x=1 and 6 at x=3. An interval average and a local rate answer different questions.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw y=3x²−2, with secants on [0,2] and [2,4]. Calculate both average rates, then draw local tangent directions at x=0,2,4 before checking reference readings. Compare with the constant-rate line y=6x−2. Label the original heights separately from the rates; both functions have all real inputs.
Give me a hint
Calculate the curve’s endpoint heights first. A line has the same rate on every interval, while a curve can have different secants and local directions.
Compare my reasoning
- The quadratic gives heights −2,10,46 at x=0,2,4. Its average rates are [10−(−2)]/2=6 and (46−10)/2=18. Draw both secants with their interval labels.
- After drawing local directions, compare with reference rates 0,12,24 at x=0,2,4 respectively. The origin-input point is (0,−2), with a horizontal local direction; its height −2 is not its rate 0.
- The line y=6x−2 has constant rate 6, both locally and on every nonzero interval. Its rate happens to match the quadratic’s first interval average, but it does not match all of the quadratic’s local rates or its second average.
Quadratic average rates: 6 and 18. Local readings: about 0,12,24 at x=0,2,4. The comparison line has constant rate 6.
Look for these in your work
- Draw the original curve, both finite secants and the three local directions.
- Keep exact average calculations separate from approximate local readings.
- Explain the line’s constant rate without assigning it to the entire quadratic.
For the genuine cubic y=x³, calculate the average rates on [0,1], [1,2] and [2,3]. Draw the curve, the three secants and local directions at x=0,1,2 before opening the reference values. Explain why the horizontal direction at zero does not make the function constant. Use the all-real domain.
Give me a hint
Use the cubic heights at the interval endpoints, then compare the curve’s direction at each selected input. A horizontal local direction is only one pointwise observation.
Compare my reasoning
- The heights at x=0,1,2,3 are 0,1,8,27. Each interval has width 1, so the average rates are 1−0=1, 8−1=7 and 27−8=19. Draw each secant on the original cubic.
- After attempting the local directions, reference rates at x=0,1,2 are 0,3,12. Report freehand readings approximately. The local direction at the origin is horizontal, but the curve rises on either side and is not a constant line.
- Compare the two lists: the interval averages 1,7,19 describe three complete intervals; the local values 0,3,12 describe three chosen inputs. Neither the heights nor the averages can be copied as the general local-rate function.
Average rates: 1,7,19. Local readings at 0,1,2: about 0,3,12. The cubic changes despite its horizontal direction at zero.
Look for these in your work
- Calculate all three finite averages from the original cubic.
- Draw local directions before using the references.
- Explain why zero local rate at one input does not imply a constant function.
An invented cooling record is T=20−t² degrees Celsius for 0≤t≤4 minutes. Draw its permitted curve and the secants over [0,2] and [2,4]. Calculate the two average rates of temperature change, then draw/read local directions at t=1 and t=3 before checking reference values. State the signed rates in °C/min and retain the time domain.
Give me a hint
Use final temperature minus initial temperature over elapsed minutes. Cooling gives a negative signed temperature-change rate; it does not require a negative temperature.
Compare my reasoning
- The record gives T(0)=20, T(2)=16 and T(4)=4. The two average rates are (16−20)/2=−2°C/min and (4−16)/2=−6°C/min. Draw the secants only over their labelled intervals.
- After drawing the local directions, compare with reference rates −2°C/min at t=1 and −6°C/min at t=3. These particular local values match the corresponding interval averages, but they are readings at different inputs and do not give one constant rate for the whole record.
- The graph falls more steeply later in the permitted interval. Negative rates describe cooling while the temperatures here remain positive. Do not apply the invented record before t=0 or after t=4; the exact finite averages and approximate graphical local readings have different sources.
Average temperature-change rates: −2°C/min on [0,2], −6°C/min on [2,4]. Local readings: about −2°C/min at 1 min and −6°C/min at 3 min. Domain: 0≤t≤4 min.
Look for these in your work
- Show the restricted original curve, both secants and both local directions.
- Use signed °C/min and distinguish temperature from its rate.
- Keep the model interval and the average/local distinction even when values coincide.
Common mistakes
- Using an interval average as the rate at every point of that interval.
- Confusing the original output, the rate at one input and the entire rate function.
- Dropping a negative rate sign, omitting units, or extending a practical model beyond its stated interval.
What can you explain now?
A curve has a horizontal tangent at one input. Must the function be constant everywhere?
Compare with the explanation
No. Its local rate is zero there, but it can change elsewhere.
The cubic y=x³ has a horizontal direction at x=0 yet rises through that point. A statement about one input is not a statement about the whole graph.
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Make it stick
Define a finite average and a local rate, explain their graph meanings, and give a signed rate unit. State why one horizontal tangent does not make a function constant.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.