LEARN · EXPLAIN · REVISE

Rates and tangent meaning

Read the idea, work independently, then explain what changed.

高二選擇性必修第二册(A版).pdf · 5.1 · PDF 64 / printed page 59

TOPIC 01

Rates and tangent meaning

Derive tangent slope from shrinking secants, then connect slope, point and local approximation.

What you will be able to explain

Derive tangent slope from shrinking secants, then connect slope, point and local approximation.
Distinguish average rate, instantaneous rate, height, normal and inverse rate.

From secant to tangent

A journey’s average speed does not describe its rate at each instant.

The secant measures a finite input change; the tangent slope is its finite common limit.

f\prime(a)=\lim_{h\to0}[f(a+h)-f(a)]/h

The secant measures a finite input change; the tangent slope is its finite common limit.
$[(a+h)^2-a^2]/h=2a+h$
The secant measures a finite input change; the tangent slope is its finite common limit.
$\lim_{h\to0}(2a+h)=2a$

When it applies

h≠0 before taking the limit; f must be defined near a.

Counterexample: |x| is continuous at zero but has slopes −1 and 1 from opposite sides.

Table of secant slopes, curve with a contact point, tangent equation and approximation error.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: For y=x², predict how secant slope changes as h approaches zero from either side.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

Secant slope 2.5 tends to 2. The tangent error is h²=0.25.

Numerical data for the diagram; printable
Increment	Secant slope	Tangent error
0.5	2.5	0.25
0.25	2.25	0.0625
0.05	2.05	0.0025

Explain: Compare 2a+h with tangent slope 2a; keep the h=0 secant exception visible.

Transfer: Relate the vertical gap h² to the accuracy of the local approximation.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

For f(x)=x² at x=1, calculate the secant slope when h=1/2.

f(x)=x^2;\ h=1/2

For the secant quotient h≠0; take the limit only after simplifying.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the two function values before dividing by h.

Hint 2

Start from this intermediate relation.

m_h=[(1+h)^2-1]/h=2+h

Worked solution

Subtract the two function values before dividing by h.
Establish and simplify this mathematical relation.
$m_h=[(1+h)^2-1]/h=2+h$
Establish and simplify this mathematical relation.
$m_{1/2}=5/2$
The denominator is the input change, not the final input.

The requested value is 2.5.

Checks and common pitfalls: The denominator is the input change, not the final input.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Find the tangent line to y=x² at x=2.

y=x^2;\ a=2

First step provided — complete the reasoning:Find the slope and the point independently.

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Find the slope and the point independently.

Hint 2

Start from this intermediate relation.

f'(2)=4;\ f(2)=4

Worked solution

Find the slope and the point independently.
Establish and simplify this mathematical relation.
$f'(2)=4;\ f(2)=4$
Establish and simplify this mathematical relation.
$y-4=4(x-2)$
Slope alone does not determine a line.

y=4x−4.

Checks and common pitfalls: Slope alone does not determine a line.

Reasoning checklist · self / teacher assessment

Find the slope and the point independently.
Give the intermediate relations and explain the conclusion with its restrictions.
Slope alone does not determine a line.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

A sensor records s=t² and t>0. At t=2, find dt/ds and explain why it is the reciprocal of ds/dt.

s=t^2;\ t=2

Use the branch t>0, so s=t² is invertible and its derivative is nonzero at t=2.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Restrict to an invertible branch and differentiate t=√s.

Hint 2

Start from this intermediate relation.

ds/dt=2t=4

Worked solution

Restrict to an invertible branch and differentiate t=√s.
Establish and simplify this mathematical relation.
$ds/dt=2t=4$
Establish and simplify this mathematical relation.
$dt/ds=1/(2\sqrt{s})=1/4$
A reciprocal derivative needs a nonzero derivative and a local inverse.

The requested value is 0.25.

Checks and common pitfalls: A reciprocal derivative needs a nonzero derivative and a local inverse.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find f′(−2) for f(x)=x².

f(x)=x^2;\ a=-2

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Take the limit of the difference quotient.

Hint 2

Start from this intermediate relation.

[(-2+h)^2-4]/h=-4+h

Worked solution

Take the limit of the difference quotient.
Establish and simplify this mathematical relation.
$[(-2+h)^2-4]/h=-4+h$
Establish and simplify this mathematical relation.
$f'(-2)=-4$
A negative slope is a direction, not a negative distance.

The requested value is -4.

Checks and common pitfalls: A negative slope is a direction, not a negative distance.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Position is s(t)=t². Find the average velocity from t=1 to t=3.

s(t)=t^2;\ 1\le t\le3

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Divide the position change by the time change.

Hint 2

Start from this intermediate relation.

v_{avg}=(9-1)/(3-1)

Worked solution

Divide the position change by the time change.
Establish and simplify this mathematical relation.
$v_{avg}=(9-1)/(3-1)$
Establish and simplify this mathematical relation.
$v_{avg}=4$
Average velocity over an interval is not automatically the velocity at its first endpoint.

The requested value is 4.

Checks and common pitfalls: Average velocity over an interval is not automatically the velocity at its first endpoint.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

For f(x)=x³ at x=2, find the tangent slope rather than the height.

f(x)=x^3;\ a=2

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Differentiate before substituting.

Hint 2

Start from this intermediate relation.

f'(x)=3x^2

Worked solution

Differentiate before substituting.
Establish and simplify this mathematical relation.
$f'(x)=3x^2$
Establish and simplify this mathematical relation.
$f'(2)=12;\ f(2)=8$
Function value and derivative value answer different questions.

The requested value is 12.

Checks and common pitfalls: Function value and derivative value answer different questions.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Does |x| have a finite derivative at zero? Justify with one-sided quotients.

f(x)=|x|;\ a=0

A derivative requires equal finite one-sided limits; an extremum itself does not require differentiability.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Keep the sign of h when removing the absolute value.

Hint 2

Start from this intermediate relation.

|h|/h=-1\ (h<0)

Worked solution

Keep the sign of h when removing the absolute value.
Establish and simplify this mathematical relation.
$|h|/h=-1\ (h<0)$
Establish and simplify this mathematical relation.
$|h|/h=1\ (h>0)$
Continuity does not supply a common two-sided derivative.

No: the one-sided slopes are −1 and 1.

Checks and common pitfalls: Continuity does not supply a common two-sided derivative.

Reasoning checklist · self / teacher assessment

Keep the sign of h when removing the absolute value.
Give the intermediate relations and explain the conclusion with its restrictions.
Continuity does not supply a common two-sided derivative.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

For y=x² at x=1, find the slope of the normal.

y=x^2;\ a=1

For a finite normal slope the tangent slope must be nonzero.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Use the perpendicular-line slope relation after finding the tangent slope.

Hint 2

Start from this intermediate relation.

m_T=f'(1)=2

Worked solution

Use the perpendicular-line slope relation after finding the tangent slope.
Establish and simplify this mathematical relation.
$m_T=f'(1)=2$
Establish and simplify this mathematical relation.
$m_N=-1/m_T=-1/2$
Taking only the negative does not make the slopes perpendicular.

The requested value is -0.5.

Checks and common pitfalls: Taking only the negative does not make the slopes perpendicular.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find all tangents to y=x² that pass through (0,−4).

y=x^2;\ P=(0,-4)

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Let a denote the unknown contact point, not the external point.

Hint 2

Start from this intermediate relation.

y=2ax-a^2;\ -4=-a^2

Worked solution

Let a denote the unknown contact point, not the external point.
Establish and simplify this mathematical relation.
$y=2ax-a^2;\ -4=-a^2$
Establish and simplify this mathematical relation.
$a=\pm2;\ y=\pm4x-4$
The given point need not lie on the curve.

y=4x−4 and y=−4x−4.

Checks and common pitfalls: The given point need not lie on the curve.

Reasoning checklist · self / teacher assessment

Let a denote the unknown contact point, not the external point.
Give the intermediate relations and explain the conclusion with its restrictions.
The given point need not lie on the curve.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Describe the tangent and normal to y=x² at zero.

y=x^2;\ a=0

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Treat a vertical normal with its equation, not a finite slope.

Hint 2

Start from this intermediate relation.

f'(0)=0

Worked solution

Treat a vertical normal with its equation, not a finite slope.
Establish and simplify this mathematical relation.
$f'(0)=0$
Establish and simplify this mathematical relation.
$T:y=0;\ N:x=0$
The expression −1/0 is undefined; the normal is still a valid vertical line.

Tangent y=0; normal x=0.

Checks and common pitfalls: The expression −1/0 is undefined; the normal is still a valid vertical line.

Reasoning checklist · self / teacher assessment

Treat a vertical normal with its equation, not a finite slope.
Give the intermediate relations and explain the conclusion with its restrictions.
The expression −1/0 is undefined; the normal is still a valid vertical line.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

For f(2)=7 and f′(2)=−3, use a tangent approximation for f(2.04).

f(2)=7;\ f'(2)=-3

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Use the small input increment, retaining the derivative sign.

Hint 2

Start from this intermediate relation.

\Delta x=0.04

Worked solution

Use the small input increment, retaining the derivative sign.
Establish and simplify this mathematical relation.
$\Delta x=0.04$
Establish and simplify this mathematical relation.
$f(2.04)\approx7-3(0.04)=6.88$
The approximation is local and is not an exact value without further information.

The requested value is 6.88.

Checks and common pitfalls: The approximation is local and is not an exact value without further information.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Which values of c make y=x+c tangent to y=x²? Explain uniqueness.

x^2=x+c

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Translate tangency into a repeated intersection root.

Hint 2

Start from this intermediate relation.

x^2-x-c=0;\ \Delta=1+4c

Worked solution

Translate tangency into a repeated intersection root.
Establish and simplify this mathematical relation.
$x^2-x-c=0;\ \Delta=1+4c$
Establish and simplify this mathematical relation.
$\Delta=0\implies c=-1/4;\ x=1/2$
A line can cross a curve while sharing some nearby slope; the contact condition is essential.

c=−1/4, with contact at x=1/2.

Checks and common pitfalls: A line can cross a curve while sharing some nearby slope; the contact condition is essential.

Reasoning checklist · self / teacher assessment

Translate tangency into a repeated intersection root.
Give the intermediate relations and explain the conclusion with its restrictions.
A line can cross a curve while sharing some nearby slope; the contact condition is essential.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

For y=x², prove the tangent approximation at a never overestimates f(a+h), and find the error at h=−0.2.

f(x)=x^2;\ h=-0.2

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the tangent prediction from the exact function value.

Hint 2

Start from this intermediate relation.

(a+h)^2-[a^2+2ah]=h^2\ge0

Worked solution

Subtract the tangent prediction from the exact function value.
Establish and simplify this mathematical relation.
$(a+h)^2-[a^2+2ah]=h^2\ge0$
Establish and simplify this mathematical relation.
$h=-0.2\implies h^2=0.04$
The error is nonnegative for both signs of h; its first-order term cancels.

The requested value is 0.04.

Checks and common pitfalls: The error is nonnegative for both signs of h; its first-order term cancels.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Predict	Predict slopes of secants to x² at a=1 from both sides.
8	Build the idea	Sample response: quotient 2+h tends to 2; h cannot initially be zero.
10	Guided comparison	Sample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.
12	Independent practice and discussion	Contrast the corner task with normal and local-error tasks; require a reason before calculation.
5	Exit and next step	Exit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.

Question sequence

Predict slopes of secants to x² at a=1 from both sides.
Sample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.

Board development

From secant to tangent
$f\prime(a)=\lim_{h\to0}[f(a+h)-f(a)]/h$

Expected correct and incorrect responses

Correct sample: 6
Anticipated wrong response: 9 — Substitutes the function height for its derivative.
Anticipated wrong response: Derivative is 0. — Applies a differentiable interior-minimum theorem to a corner.
Confusing f(a) with f′(a).
Treating a corner as a horizontal tangent.

Follow-up questions

Follow-up: Differentiate the function, then evaluate its derivative. Then explain why 9 fails.
Follow-up: Compare left and right difference quotients. Then explain why Derivative is 0. fails.
Relate the vertical gap h² to the accuracy of the local approximation.

Remediation

Sample response: quotient 2+h tends to 2; h cannot initially be zero.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.

Assessment checklist and sample responses

Representation and conditions
For h≠0, [(1+h)²−1]/h=2+h; its limit is 2.
Connected mathematical reasoning
At (2,4), f′(2)=4, so y−4=4(x−2).
Interpretation and boundary check
The exact minus predicted value is h²≥0, even for negative h.