How fast does the top of a sliding ladder move?
You will be able to: Use a fixed-length constraint to connect signed horizontal and vertical rates.
BC foundation: Unit 4 shares these contextual differentiation objectives with AB. Interpret signed rates with units, relate changing quantities before substituting an instant, and check the conditions behind approximations and limit methods. Parametric and polar motion come later.
How fast does the top of a sliding ladder move?
A 10-meter ladder leans against a vertical wall. Its foot is 6 meters from the wall and its top is 8 meters high. As the foot slides outward, the top moves downward.
A useful starting point: Why must changing dimensions stay variable until differentiation? →
Words and symbols before equations
- x(t)
- Horizontal foot distance from the wall, positive away from it.
- y(t)
- Top height above the ground, positive upward.
- Constraint
- An equation that holds throughout the modeled motion.
- Fixed length L
- The unchanging ladder length, here 10 m.
What this picture assumes
Original mathematical model; readouts are rounded. Rigid ladder L=10 m; vertical wall and level ground, equal spatial scales. x>0 and y>0. The ideal sliding-contact model is not continued to the ground endpoint.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x=6 m, y=8 m; x²+y²=100 m². Outward rate 2 m/s gives top rate -1.5 m/s (downward). The model excludes y=0.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The wall, ground and ladder form a right triangle: x²+y²=L²=100. Both x and y vary, but L stays fixed.
Differentiate with respect to t: 2x x′+2y y′=0. Solve y′=−(x/y)x′, valid while y>0.
With x=6, y=8 and x′=2 m/s, y′=−1.5 m/s. The minus sign indicates downward motion under the chosen positive-up convention.
As the ladder approaches the ground, the ideal formula predicts a very large downward rate for a fixed outward foot rate. The contact model has a restricted physical range and is not continued through y=0. A real ladder would not follow it without limit.
A worked example, step by step
At x=8 m, a 10 m ladder’s foot moves outward at 0.6 m/s. Find the top’s rate.
- Use the constraint to find y=√(100−64)=6 m.
- Differentiate before substituting: 2x x′+2y y′=0.
- Insert values: 16(0.6)+12y′=0.
- Thus y′=−0.8 m/s; the top moves downward at 0.8 m/s.
The foot and top generally do not have equal speeds. The fixed ladder length has derivative zero; the snapshot x and y values do not.
Why is L′ zero here?
Compare with an explanation
The model states that the ladder is rigid and its length stays fixed.
Predict. Change one thing. Explain.
Keep the outward foot rate fixed and increase x. Compare y and y′; explain why the top descends faster near the ground.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=6 m, y=8 m; x²+y²=100 m². Outward rate 2 m/s gives top rate -1.5 m/s (downward). The model excludes y=0.
Original mathematical model; readouts are rounded. Rigid ladder L=10 m; vertical wall and level ground, equal spatial scales. x>0 and y>0. The ideal sliding-contact model is not continued to the ground endpoint.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant rate relationship, tangent estimate, or limit argument and its conditions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 13 m ladder has its foot 5 m from the wall. If the top is moving down at 1 m/s, find the foot’s rate.
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Compare with the answer and four-point rubric
- 1 point: y=√(169−25)=12 m.
- 1 point: Use x x′+y y′=0.
- 1 point: Substitute 5x′+12(−1)=0.
- 1 point: x′=12/5=2.4 m/s outward.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which dimensions change?
The horizontal foot distance and vertical top height.
RECALL 2What fixes the sign of y′?
The positive-up convention and the direction of motion.
RECALL 3Why check the physical range?
The ideal constraint and derivative formula can fail at contact endpoints.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How fast does the top of a sliding ladder move?
- x²+y²=L² ⇒ x x′+y y′=0.
- y′=−(x/y)x′ for y>0.
Remember: The foot and top generally do not have equal speeds. The fixed ladder length has derivative zero; the snapshot x and y values do not.
Conditions: Original mathematical model; readouts are rounded. Rigid ladder L=10 m; vertical wall and level ground, equal spatial scales. x>0 and y>0. The ideal sliding-contact model is not continued to the ground endpoint.
Refresh Kid · AP Calculus BC Unit 4 · Objectives CHA-3.E · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.5, CHA-3.E. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 4 has seven official topics. Topic 4.7 assesses 0/0 and ∞/∞ quotient forms; other indeterminate forms are excluded from the core lessons. Focused lesson titles and questions are original Refresh Kid teaching material.
Derivative units and signs are interpreted in context. Speed is the magnitude of velocity; turning requires a sign change. Related-rate equations hold at nearby times and are differentiated before snapshot values are inserted. Cone and ladder models have explicit physical domains. Tangent approximations remain estimates; error direction requires behavior on the relevant interval. L’Hôpital’s rule requires an eligible quotient form, nearby differentiability, nonzero denominator derivative and an existing finite or infinite derivative-ratio limit. A failed derivative-ratio limit is inconclusive about the original quotient.
The Organic Chemistry Tutor video creators and relevant descriptions were checked; full videos were not reviewed. Khan Academy’s unit destination was checked; its JavaScript lesson content was not fully readable by the research tool. OpenStax Sections 3.4, 4.1, 4.2 and 4.8 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. Graph geometry is generated from the stated original equations; self-hosted Three.js retains its MIT license. The optional 3D model shows the circular water surface and axial cross-section of a tip-down conical tank. The water radius and height obey the same similar-triangle ratio in 2D and 3D. Camera rotation only changes the view; signed flow and height controls describe an instantaneous state. Complete labeled 2D geometry, rates and equations remain available without WebGL.
Independent teacher review and observation of students remain pending. Checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned to this lesson.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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