riaology
Unit 2 · Lesson 11 of 11

The same inflow can raise a cone's level at different rates.

A conical tank widens as it fills. Change its depth and flow to see why level rise slows even with a constant inlet.

Change the input. Follow the mathematics.

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

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

A model you can changeMove one control at a time.
Conical tank, water depth 2 meters, filling at 0.5 cubic meters per minute2 mWider section → slower level rise
Depth h (m) versus Level-rise rate (m/min). Level rise rate. Selected level-rise rate: 0.15915 m/min. The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.001.2731.52.5463Level-rise rate (m/min)Depth h (m)
Level rise rate

level-rise ratedhdt=4(0.5)π(2)2=0.159155 m/min\frac{dh}{dt}=\frac{4(0.5)}{\pi(2)^2}=0.159155\,\mathrm{m/min}

The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.

Assumptions, units, and limits

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

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

01

From the situation to an equation

Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

r=h/2,V=π12h3,dVdt=π4h2dhdtr=h/2,\quad\mathcal V=\frac\pi{12}h^3,\quad\frac{d\mathcal V}{dt}=\frac\pi4h^2\frac{dh}{dt}
02

Keep the reasoning visible

r=h2,V=π3r2h=π12h3r=\frac h2,\quad\mathcal V=\frac\pi3r^2h=\frac\pi{12}h^3

The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.

03

What makes this model useful?

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

The derivative relates instantaneous rates at the stated depth. The cone's changing cross-section makes the level rate nonconstant.

A worked example

Calculate the starting scenario

  1. Use water depth = 2 m; inlet volume flow = 0.5 m³/min.
  2. dhdt=4(0.5)π(2)2=0.159155 m/min\frac{dh}{dt}=\frac{4(0.5)}{\pi(2)^2}=0.159155\,\mathrm{m/min}
  3. The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly. The checked quantity is level-rise rate.
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 level-rise rate using water depth 2 m, inlet volume flow 0.5 m³/min. State the meaning of its sign and units.

Checked quantity: level-rise rate · m/min. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

Explore the visual solution
Conical tank, water depth 2 meters, filling at 0.5 cubic meters per minute2 mWider section → slower level rise
Depth h (m) versus Level-rise rate (m/min). Level rise rate. Selected level-rise rate: 0.15915 m/min. The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.001.2731.52.5463Level-rise rate (m/min)Depth h (m)
Level rise rate

level-rise ratedhdt=4(0.5)π(2)2=0.159155 m/min\frac{dh}{dt}=\frac{4(0.5)}{\pi(2)^2}=0.159155\,\mathrm{m/min}

The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.

Assumptions, units, and limits

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

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.

Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

r=h/2,V=π12h3,dVdt=π4h2dhdtr=h/2,\quad\mathcal V=\frac\pi{12}h^3,\quad\frac{d\mathcal V}{dt}=\frac\pi4h^2\frac{dh}{dt}
Calculate and explain

Apply the idea to a new scenario

At depth 1 m and inflow 0.25 m³/min, find level-rise rate.

Checked quantity: level-rise rate · m/min. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

Explore the visual solution
Conical tank, water depth 1 meters, filling at 0.25 cubic meters per minute1 mWider section → slower level rise
Depth h (m) versus Level-rise rate (m/min). Level rise rate. Selected level-rise rate: 0.31831 m/min. The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.000.6371.51.2733Level-rise rate (m/min)Depth h (m)
Level rise rate

level-rise ratedhdt=4(0.25)π(1)2=0.31831 m/min\frac{dh}{dt}=\frac{4(0.25)}{\pi(1)^2}=0.31831\,\mathrm{m/min}

The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.

Assumptions, units, and limits

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

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.

Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

r=h/2,V=π12h3,dVdt=π4h2dhdtr=h/2,\quad\mathcal V=\frac\pi{12}h^3,\quad\frac{d\mathcal V}{dt}=\frac\pi4h^2\frac{dh}{dt}
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
Conical tank, water depth 2 meters, filling at 0.5 cubic meters per minute2 mWider section → slower level rise
Depth h (m) versus Level-rise rate (m/min). Level rise rate. Selected level-rise rate: 0.15915 m/min. The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.001.2731.52.5463Level-rise rate (m/min)Depth h (m)
Level rise rate

level-rise ratedhdt=4(0.5)π(2)2=0.159155 m/min\frac{dh}{dt}=\frac{4(0.5)}{\pi(2)^2}=0.159155\,\mathrm{m/min}

The cone widens with height. The same volume per minute spreads over more area, so the level rises more slowly.

Assumptions, units, and limits

An inverted cone has height 4 m and top radius 2 m. Prescribed inflow, no outflow, no evaporation.

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

The reasoning. Use similar triangles to express radius in terms of depth. Substitute this relation into volume before differentiating with respect to time. Then insert the instantaneous depth and volume rate; treating radius as fixed would model another tank.

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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