Unit 2 · Lesson 2 of 5

Curve length adds tiny diagonal segments.

A road or cable follows a curve rather than a straight horizontal line. Its length must include both horizontal and vertical change.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its arc length.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase upper x bound b? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Read the model

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Length contribution per dx

Upper x bound b1 m

arc length

L=\frac23[(1+1)^{3/2}-1]=1.21895\,\mathrm m

A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself.

Assumptions, units, and limits

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

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

From the situation to an equation

The Pythagorean relation for a small segment gives $ds=sqrt$ (1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Keep the reasoning visible

ds=\sqrt{dx^2+dy^2}

A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself.

What makes this model useful?

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Integrating the curve height gives area, not length. Arc length is at least the endpoint straight-line distance.

A worked example

Calculate the starting scenario

Use upper x bound b = 1 m.
$L=\frac23[(1+1)^{3/2}-1]=1.21895\,\mathrm m$
A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself. The checked quantity is arc length.

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 arc length using upper x bound b 1 m. State the meaning of its sign and units.

Checked quantity: arc length · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The Pythagorean relation for a small segment gives ds=sqrt(1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

Explore the visual solution

Read the model

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Length contribution per dx

Upper x bound b1 m

arc length

L=\frac23[(1+1)^{3/2}-1]=1.21895\,\mathrm m

A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself.

Assumptions, units, and limits

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

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.

The Pythagorean relation for a small segment gives $ds=sqrt$ (1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Calculate and explain

Apply the idea to a new scenario

For y=(2/3)x^(3/2), find arc length from $x=0$ to $x=2$ m.

Checked quantity: arc length · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The Pythagorean relation for a small segment gives ds=sqrt(1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

Explore the visual solution

Read the model

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Length contribution per dx

Upper x bound b2 m

arc length

L=\frac23[(1+2)^{3/2}-1]=2.79743\,\mathrm m

A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself.

Assumptions, units, and limits

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

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.

The Pythagorean relation for a small segment gives $ds=sqrt$ (1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-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 model

y=\frac23x^{3/2},\quad L=\int_0^b\sqrt{1+(y')^2}\,dx=\frac23[(1+b)^{3/2}-1]

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

Length contribution per dx

Upper x bound b1 m

arc length

L=\frac23[(1+1)^{3/2}-1]=1.21895\,\mathrm m

A tiny segment has length sqrt( $dx^2+dy^2$ ). Adding segment lengths gives the integral, rather than integrating y itself.

Assumptions, units, and limits

Coordinates are numeric meter values: $y=\frac23(1\,\mathrm m)[x/(1\,\mathrm m)]^{3/2}$ . The arc-length density includes horizontal and vertical movement.

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

The reasoning. The Pythagorean relation for a small segment gives $ds=sqrt$ (1+(y′)²) dx. Integrate this length density over the domain. With a different parameter, use the magnitude of the vector derivative.

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.

Completion is saved on this browser. This is separate from account practice progress.