riaology
Unit 2 · Lesson 6 of 11

A rate can pass through two transformations.

A sensor first rescales time-dependent input, then cubes the response. Follow each stage to understand its final rate.

Change the input. Follow the mathematics.

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

What will change if you increase time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Time tu=1+0.2tu=1+0.2ty=u3y=u^3

Multiply the rates through both stages.

Time (s) versus Response. Composite response. Selected response time derivative: 1.176 1/s. Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.0042.585ResponseTime (s)
Composite response

response time derivativey′(2)=3(1+0.2(2))2(0.2)=1.176 s−1y'(2)=3(1+0.2(2))^2(0.2)=1.176\,\mathrm{s^{-1}}

Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.

Assumptions, units, and limits

A dimensionless sensor response to a linearly changing dimensionless input.

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

01

From the situation to an equation

For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

y(t)=(1+0.2t)3,dydt=3(1+0.2t)2(0.2)y(t)=(1+0.2t)^3,\quad\frac{dy}{dt}=3(1+0.2t)^2(0.2)
02

Keep the reasoning visible

dydt=dydududt\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}

Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.

03

What makes this model useful?

A dimensionless sensor response to a linearly changing dimensionless input.

Differentiate the outer operation and then account for how its input changes. Treating the inner input as time misses a scale factor.

A worked example

Calculate the starting scenario

  1. Use time = 2 s.
  2. y′(2)=3(1+0.2(2))2(0.2)=1.176 s−1y'(2)=3(1+0.2(2))^2(0.2)=1.176\,\mathrm{s^{-1}}
  3. Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside. The checked quantity is response time derivative.
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 response time derivative using time 2 s. State the meaning of its sign and units.

Checked quantity: response time derivative · 1/s. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

Explore the visual solution
Time tu=1+0.2tu=1+0.2ty=u3y=u^3

Multiply the rates through both stages.

Time (s) versus Response. Composite response. Selected response time derivative: 1.176 1/s. Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.0042.585ResponseTime (s)
Composite response

response time derivativey′(2)=3(1+0.2(2))2(0.2)=1.176 s−1y'(2)=3(1+0.2(2))^2(0.2)=1.176\,\mathrm{s^{-1}}

Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.

Assumptions, units, and limits

A dimensionless sensor response to a linearly changing dimensionless input.

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.

For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

y(t)=(1+0.2t)3,dydt=3(1+0.2t)2(0.2)y(t)=(1+0.2t)^3,\quad\frac{dy}{dt}=3(1+0.2t)^2(0.2)
Calculate and explain

Apply the idea to a new scenario

At t=4t=4 s, find the rate of y=(1+0.21+0.2t)³.

Checked quantity: response time derivative · 1/s. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

Explore the visual solution
Time tu=1+0.2tu=1+0.2ty=u3y=u^3

Multiply the rates through both stages.

Time (s) versus Response. Composite response. Selected response time derivative: 1.944 1/s. Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.0042.585ResponseTime (s)
Composite response

response time derivativey′(4)=3(1+0.2(4))2(0.2)=1.944 s−1y'(4)=3(1+0.2(4))^2(0.2)=1.944\,\mathrm{s^{-1}}

Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.

Assumptions, units, and limits

A dimensionless sensor response to a linearly changing dimensionless input.

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.

For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

y(t)=(1+0.2t)3,dydt=3(1+0.2t)2(0.2)y(t)=(1+0.2t)^3,\quad\frac{dy}{dt}=3(1+0.2t)^2(0.2)
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
Time tu=1+0.2tu=1+0.2ty=u3y=u^3

Multiply the rates through both stages.

Time (s) versus Response. Composite response. Selected response time derivative: 1.176 1/s. Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.0042.585ResponseTime (s)
Composite response

response time derivativey′(2)=3(1+0.2(2))2(0.2)=1.176 s−1y'(2)=3(1+0.2(2))^2(0.2)=1.176\,\mathrm{s^{-1}}

Follow input → inner response → outer response. Multiply their local rates rather than differentiating only the outside.

Assumptions, units, and limits

A dimensionless sensor response to a linearly changing dimensionless input.

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

The reasoning. For a composition, multiply the outer derivative evaluated at the inner value by the inner derivative. The units of intermediate response cancel in the product of rates. A missing inner rate changes the physical meaning.

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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