≥Better Grades
Menu

Calculus I · Unit 2B · lesson

Pythagorean Related Rates

Concept

Learning objectives

Solve ladder, separation, and distance problems using the Pythagorean theorem.

Distances Connected by a Right Triangle

Explanation

Before the formulas

In Pythagorean Related Rates, distinguish a quantity from its rate and a distance from a component of that distance. Many errors come from assigning one symbol to two different lengths or from confusing the speed of an object with the speed of a shadow tip or line of sight.

Use a variable table if the diagram is crowded: quantity, meaning, units, known value, known rate. The table makes substitutions deliberate and prevents a snapshot value from being mistaken for a constant throughout the motion.

The wall, ground, and ladder form a right triangle. The fixed ladder length gives x 2+y 2=L 2, while dx/dt and dy/dt describe the horizontal and vertical endpoint speeds.
Read this graph as text

Sliding ladder geometry. The wall, ground, and ladder form a right triangle. The fixed ladder length gives x 2+y 2=L 2 , while dx/dt and dy/dt describe the horizontal and vertical endpoint speeds. As the foot moves away from the wall, x increases. Because the ladder length stays fixed, the top must move downward, so y decreases. Differentiating x 2+y 2=L 2 gives 2x dx/dt+2y dy/dt=0 , which already predicts opposite signs.

Every relationship in sliding ladder geometry is identified with written labels plus distinct solid, dashed, dotted, double, marker, or pattern cues; color is never the only carrier of meaning.

Why it matters: The diagram should support variable selection and sign reasoning before any numbers are used. It must show that L is constant while x and y vary. Arrows make the direction conventions explicit.

Visual study

The wall, ground, and ladder form a right triangle. The fixed ladder length gives x 2+y 2=L 2, while dx/dt and dy/dt describe the horizontal and vertical endpoint speeds.

Explanation

Right triangles connect rates along different directions

Whenever two changing perpendicular distances and a connecting distance appear, the Pythagorean theorem is the natural relationship. Differentiating x2+y2=z2x^2+y^2=z^2 gives 2xx˙+2yy˙=2zz˙2x\dot x+2y\dot y=2z\dot z, which weights each rate by the current geometry.

The snapshot lengths matter because the same horizontal speed can produce different changes in the diagonal depending on the triangle's shape at that moment.

Right-triangle models appear whenever two perpendicular distances determine a third: ladders, aircraft separation, radar range, and objects moving on perpendicular roads. The Pythagorean theorem supplies the constraint, while signs on the rates record the directions of motion.

A length can be positive while its derivative is negative. For example, the height of a ladder remains positive even as it decreases. Confusing a quantity with its rate is one of the quickest ways to produce a physically impossible answer.

Guided walkthrough

Sliding ladder

A 1313-foot ladder leans against a wall. The bottom slides away from the wall at 22 ft/s. How fast is the top sliding down when the bottom is 55 feet from the wall?

Answer reveal

Worked solution

Write a real attempt before opening the supplied answer.

Worked example

Two vehicles moving on perpendicular roads

If one vehicle is xx miles east of an intersection and another is yy miles north, their separation zz satisfies

z2=x2+y2.z^2=x^2+y^2.

Differentiating gives

zdzdt=xdxdt+ydydt.z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}.

Signs of dx/dtdx/dt and dy/dtdy/dt must reflect whether each vehicle moves toward or away from the intersection.

Common mistake

The constant hypotenuse in a ladder problem has derivative zero. Writing 2L dL/dt2L\,dL/dt is not wrong, but dL/dt=0dL/dt=0. More often, students mistakenly insert a nonzero rate for a length the problem says is fixed.

Modeling lab

Two aircraft tracked from a control station

Aircraft A is 3030 km east of a station and moving east at 600600 km/h. Aircraft B is 4040 km north and moving south at 500500 km/h. Their separation zz satisfies

z2=x2+y2.z^2=x^2+y^2.

At the instant described, z=50z=50, x′=600x'=600, and y′=−500y'=-500. Hence

50z′=30(600)+40(−500)=−2000,50z'=30(600)+40(-500)=-2000,

so

z′=−40 km/h.z'=-40\text{ km/h}.

Despite both aircraft moving rapidly, their separation is decreasing only at 4040 km/h because the two directional effects nearly cancel.

After the explanation

Use the section idea

Reading lens

Freeze the geometry at one instant, but differentiate the relationship while every changing quantity is still a function of time.

Mental model

The picture supplies a constraint; implicit differentiation transmits known rates through that constraint to the unknown rate.

Decision

Draw and label first, write one relationship, differentiate with time, then substitute the snapshot measurements and rates.

Common trap

Substituting numerical dimensions before differentiating and thereby erasing the very change the problem asks about.

Check yourself

Does your final rate have the predicted sign, the correct units, and a magnitude compatible with the diagram?

Interactive checkapp-aircraft-01

Aircraft coordinates are (30,0)(30,0) and (0,40)(0,40) km with rates x′=600x'=600, y′=−500y'=-500. Find separation rate.

Your work stays on this device. No account or AI grader is used.

Show hint

Use zz′=xx′+yy′zz'=xx'+yy' and z=50z=50.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Source & rights

Original instruction with traceable references.

BetterGrades-original composition declared by source handoff; owner provenance review required before public release

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.