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Parametric shape models

Use parametric equations to model fixed shapes and profiles. Find widths, heights, depth, mean gradients and physical restrictions for trigonometric, logarithmic and exponential models.

Before you startParametric elimination, trigonometric ranges, logarithms and straight-line gradients

01 / A parameter can label a place instead of a time

A fixed shape does not imply a moving object.

Suppose a profile has x = 12t, y = 3 + 2 cos(πt), 0 ≤ t ≤ 1. Both coordinates are in metres and t is dimensionless. Increasing t moves our selected point across a stationary shape; the model says nothing about speed.

Read the coordinate meanings and physical domain before calculating dimensions.

The model compares three original profiles and their mean gradients from the left endpoint. The permanent notes explain the calculations without requiring the animation.

Measure a profileExplore

ProfilePosition along profile

x = 12t; y = 3 + 2 cos(πt), 0 ≤ t ≤ 1.

t = 0 gives (0, 5) m.

Horizontal change 0 m; vertical change 0 m.

Choose a different point for a mean gradient.

Total width 12 m; profile heights range from 1 m to 5 m.

The parameter is dimensionless and labels a place on a fixed profile; it is not time. The gold chord measures the mean gradient from the start, not a tangent or the length of the curved profile. Axis scales differ.

02 / Find width from the horizontal coordinates

Parameter length and physical width are different quantities.

Find the width and endpoint heights of x = 12t, y = 3 + 2 cos(πt), 0 ≤ t ≤ 1.Worked example

At t = 0: (0, 5) m

Use cos 0 = 1.

At t = 1: (12, 1) m

Use cos π = −1.

Width = 12 − 0 = 12 m

Horizontal distance between the ends.

The right endpoint is 4 m lower

Its height is 1 m; the vertical drop is 5 − 1 = 4 m.

A width of 12 m does not mean the curved profile is 12 m long. Arc length is a separate quantity.

Watch a chord measure the mean gradient

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Use the actual parameter interval for heights

The full trigonometric range may not be attained on a restricted piece.

Find the height range on the cosine profile.Worked example

0 ≤ πt ≤ π

Transform the parameter interval.

−1 ≤ cos(πt) ≤ 1

Both extremes occur in this interval.

1 ≤ y ≤ 5 m

Multiply by 2 and add 3.

On only 0 ≤ t ≤ 1/2, cosine decreases from 1 to 0, so the height range is 3 ≤ y ≤ 5 m. Do not include the right-hand minimum of the full profile when it lies outside the selected piece.

04 / Find where a profile reaches a given height

Solve for the parameter, then convert to the horizontal position.

Where is the cosine profile at height 4 m?Worked example

3 + 2 cos(πt) = 4

Use the stated height.

cos(πt) = 1/2

Rearrange.

πt = π/3, so t = 1/3

This is the only solution for πt in [0, π].

x = 4 m

The position is (4, 4) m.

If a question asks for a location, the parameter value alone is not the final answer. A height outside the attained range is impossible within the model.

05 / Measure mean gradient between two positions

Use vertical change divided by horizontal change.

Find the mean gradient from t = 0 to t = 1 on the cosine profile.Worked example

Endpoints (0, 5) and (12, 1)

Find both coordinate pairs.

Mean gradient = (1 − 5)/(12 − 0) = −1/3

The negative sign describes descent as x increases.

Angle of descent = arctan(1/3) ≈ 18.43°

Use the positive drop/run ratio for the stated descent angle.

Mean gradient = Δy/Δx, when Δx ≠ 0.

The chord’s gradient is an average over that interval, not the tangent gradient at every point. Reversing the endpoints changes both signs and leaves the gradient unchanged.

06 / Compare different portions of a profile

The same whole-profile average can hide steeper sections.

Compare the first quarter and the middle half of the cosine profile.Worked example

At t = 1/4: (3, 3 + √2)

Use cos(π/4) = √2/2.

First-quarter mean gradient = (√2 − 2)/3 ≈ −0.195

Compare with the starting point (0, 5).

At t = 3/4: (9, 3 − √2)

Use cos(3π/4) = −√2/2.

Middle-half mean gradient = −2√2/6 = −√2/3 ≈ −0.471

This section has a larger magnitude of average descent.

Two identical parameter values give the same point here and hence a zero horizontal change. They do not define a chord gradient.

07 / Define the reference level for depth

Height above a datum and depth below a reference are different coordinates.

If depth is measured below the horizontal level y = 5 m, express depth along the cosine profile.Worked example

Depth d = 5 − y

Subtract the height from the reference level.

d = 2 − 2 cos(πt)

Substitute the profile.

0 ≤ d ≤ 4 m

The greatest depth occurs at the right endpoint.

This is a change of reference for measurement. It does not claim that the original height y is negative or that the shape moves downward in time.

08 / Respect logarithm domains as well as physical bounds

An algebraically defined extension need not belong to the modelled shape.

For x = 10t, y = 2 + ln(1 + 3t), 0 ≤ t ≤ 1, find a Cartesian rule and the height range.Worked example

t = x/10

Eliminate the parameter.

y = 2 + ln(1 + 3x/10)

Substitute.

0 ≤ x ≤ 10 m; 2 ≤ y ≤ 2 + ln 4 m

The logarithm increases across this interval.

The logarithm alone requires t > −1/3

The physical domain [0, 1] is stricter.

The logarithm’s input is dimensionless: t is dimensionless here. Do not take the logarithm of a length without an appropriate scale or ratio.

09 / Read limiting behaviour without inventing an endpoint

A positive exponential can approach zero without reaching it.

For x = 8t, y = 6e^−t, 0 ≤ t ≤ 2, find the width, minimum height and half-height position.Worked example

Width = 16 m; endpoint heights 6 and 6e⁻² m

The exponential decreases with t.

Minimum height = 6e⁻² ≈ 0.812 m

This minimum is attained at the included endpoint t = 2.

6e^−t = 3 gives t = ln 2

The half-height parameter lies in [0, 2].

x = 8 ln 2 ≈ 5.545 m

The half-height position is (8 ln 2, 3).

If the mathematical curve were extended, y would tend towards zero as t increases. It is never zero at finite t, and the stated physical model ends at t = 2.

10 / Track horizontal and vertical scale separately

Changing one coordinate scale changes gradients.

Doubling every x-coordinate doubles the width while keeping the heights. The same vertical change is spread over twice the horizontal distance, so each corresponding mean gradient is halved.

Apply a scale change to the cosine profile.Worked example

Original: x = 12t, y = 3 + 2 cos(πt)

Whole-profile mean gradient −1/3.

Horizontal stretch: x = 24t, same y

Mean gradient becomes −1/6.

x = 5t, y = 8e^−t, now with 0 ≤ t < 2. Is 8e⁻² an attained minimum?

Hint

The endpoint t = 2 is excluded.

Worked solution

No. Heights approach 8e⁻² from above but do not attain it. The height range is 8e⁻² < y ≤ 8.

12 / Recap

Keep geometry, units and parameter restrictions together.

A parameter may label a fixed spatial position rather than time.
Use coordinate differences for widths, drops and mean gradients.
Restrict ranges and level solutions to the actual modelled interval.
Define depth relative to a stated reference height.
Keep logarithm inputs valid and distinguish limits from attained endpoints.
Record how horizontal and vertical scaling affect the geometry.

Section 1 of 12 · A parameter can label a place instead of a time