riaology
Unit 1 · Lesson 4 of 7

A hanging cable links exponentials and arc length.

A uniformly weighted flexible cable hangs in a catenary rather than a parabola. Hyperbolic functions describe its ideal shape.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its cable slope.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase cable parameter a? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.71716 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.52103.0432Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

y=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1
02

Keep the reasoning visible

cosh⁡u=eu+e−u2,sinh⁡u=eu−e−u2\cosh u=\frac{e^u+e^{-u}}2,\quad\sinh u=\frac{e^u-e^{-u}}2

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

03

What makes this model useful?

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

Catenary and parabola can look similar near the bottom but obey different models. Do not exchange their equations over a wide span.

A worked example

Calculate the starting scenario

  1. Use cable parameter a = 1.5 m; horizontal x = 1 m.
  2. y′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158
  3. Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u. The checked quantity is cable slope.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the cable slope using cable parameter a 1.5 m, horizontal x 1 m. State the meaning of its sign and units.

Checked quantity: cable slope · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Explore the visual solution
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.71716 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.52103.0432Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

y=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1
Calculate and explain

Apply the idea to a new scenario

For cable parameter a=2a=2 m at x=1x=1 m, find the slope.

Checked quantity: cable slope · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Explore the visual solution
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.5211 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.54303.0862Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/2)=0.521095y'(1)=\sinh(1/2)=0.521095

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

y=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modely=acosh⁡(x/a),y′=sinh⁡(x/a),cosh⁡2u−sinh⁡2u=1y=a\cosh(x/a),\quad y'=\sinh(x/a),\quad\cosh^2u-\sinh^2u=1

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

x (m) versus Height (m). Cable. Selected cable slope: 0.71716 . Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because $\sqrt{1+\sinh^2u}=\cosh u$.0-21.52103.0432Height (m)x (m)
Cable

cable slopey′(1)=sinh⁡(1/1.5)=0.717158y'(1)=\sinh(1/1.5)=0.717158

Hyperbolic functions come from exponentials. Their identity simplifies catenary arc length because 1+sinh⁡2u=cosh⁡u\sqrt{1+\sinh^2u}=\cosh u.

Assumptions, units, and limits

Ideal catenary shape for a uniform flexible cable under gravity; height reference is a convention.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Hyperbolic sine and cosine are defined from opposite exponentials. Their squared difference is one, unlike the circular identity's sum. This relation simplifies the catenary's arc-length integrand to a hyperbolic cosine.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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