TC8, TC9 - Arc length and envelopes

TC8, TC9 - Arc length and envelopes

Set up and evaluate arc-length integrals in cartesian, parametric and polar form. Find candidate envelopes by differentiating a family equation with respect to its parameter, then verify the contacts, tangencies and permitted parameter range.

Length measures the curve rather than its projection

Slide a circle of radius 11 along the horizontal axis. Two straight lines touch every circle you draw. Can one algebraic calculation find both lines without plotting infinitely many circles?

We will return to those contacts after measuring individual curves. A horizontal input interval gives the width of an arc, not its length. Small horizontal and vertical displacements combine by Pythagoras.

You already know

A tangent gradient is a ratio of coordinate rates. CAS can differentiate and evaluate definite integrals, but its input still needs the correct domain and limits. A parameter interval may retrace part of a curve, so eliminating the parameter alone does not determine a travelled length.

For a small horizontal change dx\mathrm dx, the corresponding vertical change is approximately y′dxy'\mathrm dx. The local length factor is therefore 1+(y′)2\sqrt{1+(y')^2}. The examination supplies arc-length formulae; the task is to choose and apply them correctly.

Cartesian arc length

For a continuously differentiable graph y=f(x)y=f(x), traced once from x=ax=a to x=bx=b, a<ba<b,

S=∫ab1+(dydx)2 dxS=\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

Split the calculation if separate pieces or singular endpoints require it.

Not in the formula booklet; formula supplied in the question.

Integrating yy would calculate a signed area. The length integrand is non-negative and depends on the derivative.

Worked example 1: a length with an elementary antiderivative

Question. Find the length of y=23x3/2y=\frac23x^{3/2} for 0≤x≤30\le x\le3.

Plan. Differentiate, substitute into the length formula, and integrate the resulting square root.

y′=xdifferentiate on the stated domainS=∫031+x dxsquare the derivative=[23(1+x)3/2]03CAS or direct antiderivative=23(8−1)=143evaluate both endpoints\begin{aligned} y'&=\sqrt{x}&&\text{differentiate on the stated domain}\\ S&=\int_0^3\sqrt{1+x}\,\mathrm dx&&\text{square the derivative}\\ &=\left[\frac23(1+x)^{3/2}\right]_0^3 &&\text{CAS or direct antiderivative}\\ &=\frac23(8-1)=\frac{14}{3} &&\text{evaluate both endpoints} \end{aligned}

Why this step? The factor is 1+(y′)2\sqrt{1+(y')^2}, not 1+y′1+y' or 1+y′\sqrt{1+y'}.

Answer. 14/314/3 units.

Check. It exceeds the horizontal displacement 33, as a non-horizontal arc should.

Change only the upper input bound from 33 to 88.

Worked example 2: a longer part of the same graph

Question. Find the length for 0≤x≤80\le x\le8.

Plan. Retain the derivative and antiderivative, changing only the evaluation limit.

S=∫081+x dxnew arc interval=[23(1+x)3/2]08same antiderivative=23(27−1)=523evaluate at 8 and 0\begin{aligned} S&=\int_0^8\sqrt{1+x}\,\mathrm dx&&\text{new arc interval}\\ &=\left[\frac23(1+x)^{3/2}\right]_0^8 &&\text{same antiderivative}\\ &=\frac23(27-1)=\frac{52}{3} &&\text{evaluate at }8\text{ and }0 \end{aligned}

Why this step? Changing the endpoint changes the travelled arc, even though the curve and local length factor stay the same.

Answer. 52/352/3 units.

Check. The new arc contains the first one and has greater length. Its horizontal displacement is 8<52/38<52/3.

If CAS returns an unevaluated definite integral, a correct integral may already answer an instruction to “give an expression”. Use numerical evaluation when a numerical length is requested.

Watch out

Wrong: enter the original function into a numerical integration tool and call the result its arc length. Right: construct the length integrand first, then supply the correct limits.

Change only the upper endpoint once more.

Integrate the speed along a parameterisation

Two coordinate rates combine to give the speed of tracing a curve. This removes the need to express yy as a single-valued function of xx.

Parametric arc length

For a piecewise smooth parameterisation over α≤t≤β\alpha\le t\le\beta,

S=∫αβ(dxdt)2+(dydt)2 dtS=\int_\alpha^\beta \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+ \left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

The integral counts the length traced, including repeated traversal. Choose a once-traced interval when the geometric arc's length is required.

Not in the formula booklet; formula supplied in the question.

For x=3cos⁡t, y=bsin⁡tx=3\cos t,\ y=b\sin t, 0≤t≤π/20\le t\le\pi/2, the parameterisation traces the first-quadrant arc once. Its rates are −3sin⁡t-3\sin t, bcos⁡tb\cos t.

Worked example 3: equal coordinate scales

Question. Find the length when b=3b=3.

Plan. Square and add the rates, using the trigonometric identity before integrating.

v(t)=9sin⁡2t+9cos⁡2t=3constant tracing speedS=∫0π/23 dtonce-traced interval=3π2evaluate\begin{aligned} v(t)&=\sqrt{9\sin^2t+9\cos^2t}=3 &&\text{constant tracing speed}\\ S&=\int_0^{\pi/2}3\,\mathrm dt&&\text{once-traced interval}\\ &=\frac{3\pi}{2}&&\text{evaluate} \end{aligned}

Why this step? The sign of dx/dt\mathrm dx/\mathrm dt is negative, but length uses squared rates; the curve does not acquire negative length while moving left.

Answer. 3π/23\pi/2 units.

Check. The equation x2+y2=9x^2+y^2=9 identifies a radius-33 circle; this is its quarter circumference.

Change only bb from 33 to 22. The speed is no longer constant.

Worked example 4: unequal coordinate scales

Question. Set up the length integral for b=2b=2, and evaluate the length to 55 decimal places.

Plan. Keep both coordinate rates in the square root, then ask CAS for the definite integral numerically.

v(t)=9sin⁡2t+4cos⁡2tnew vertical rateS=∫0π/29sin⁡2t+4cos⁡2t dtexact integral expression≈3.9663598973numerical CAS evaluation\begin{aligned} v(t)&=\sqrt{9\sin^2t+4\cos^2t} &&\text{new vertical rate}\\ S&=\int_0^{\pi/2}\sqrt{9\sin^2t+4\cos^2t}\,\mathrm dt &&\text{exact integral expression}\\ &\approx3.9663598973&&\text{numerical CAS evaluation} \end{aligned}

Why this step? A square root of a sum of squares is not the sum of their square roots. Replacing the speed by 3sin⁡t+2cos⁡t3\sin t+2\cos t would give a different quantity.

Answer. The displayed integral; length 3.966363.96636 units to 55 decimal places.

Check. 2≤v(t)≤32\le v(t)\le3, so π≤S≤3π/2\pi\le S\le3\pi/2, consistent with the numerical value.

An exact integral expression is useful even when no elementary antiderivative is available. Record its limits and radicand before the decimal output.

Watch out

Wrong: integrate over 0≤t≤2π0\le t\le2\pi when only the first-quadrant arc was requested. Right: use the parameter interval that traces the requested arc once.

Change only the vertical scale again.

Keep the absolute value in a polar speed

Differentiating x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta, squaring the two rates and adding cancels their cross terms. This gives a convenient length formula directly in polar form.

Polar arc length

For a piecewise smooth polar tracing over α≤θ≤β\alpha\le\theta\le\beta,

S=∫αβr2+(drdθ)2 dθS=\int_\alpha^\beta \sqrt{r^2+\left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2} \,\mathrm d\theta

Choose the tracing interval carefully. Signed rr is squared here; length remains non-negative.

Not in the formula booklet; formula supplied in the question.

For r=1+cos⁡θr=1+\cos\theta, the local speed simplifies:

(1+cos⁡θ)2+sin⁡2θ=2+2cos⁡θ=2∣cos⁡(θ/2)∣\sqrt{(1+\cos\theta)^2+\sin^2\theta} =\sqrt{2+2\cos\theta} =2|\cos(\theta/2)|

The absolute value is essential beyond the upper half.

Worked example 5: one half of the curve

Question. Find the arc length for 0≤θ≤π0\le\theta\le\pi.

Plan. Use the simplified speed and its sign on this interval.

0≤θ/2≤π/2⟹cos⁡(θ/2)≥0resolve the absolute valueS=∫0π2cos⁡(θ/2) dθupper-half length=[4sin⁡(θ/2)]0π=4evaluate\begin{aligned} 0\le\theta/2\le\pi/2&\Longrightarrow\cos(\theta/2)\ge0 &&\text{resolve the absolute value}\\ S&=\int_0^\pi2\cos(\theta/2)\,\mathrm d\theta &&\text{upper-half length}\\ &=[4\sin(\theta/2)]_0^\pi=4 &&\text{evaluate} \end{aligned}

Why this step? Removing the absolute value is justified by the interval, not by a general rule that square roots cancel squares.

Answer. 44 units.

Check. The arc runs from (2,0)(2,0) to the pole along the upper half, with positive length.

Change only the requested angle interval to a full turn, 0≤θ≤2π0\le\theta\le2\pi.

Worked example 6: the complete once-traced curve

Question. Find the full curve's length.

Plan. Split at θ=π\theta=\pi, where cos⁡(θ/2)\cos(\theta/2) changes sign.

S=∫0π2cos⁡(θ/2) dθ−∫π2π2cos⁡(θ/2) dθnon-negative speed on both halves=4−(−4)=8sum the two positive lengths\begin{aligned} S&=\int_0^\pi2\cos(\theta/2)\,\mathrm d\theta -\int_\pi^{2\pi}2\cos(\theta/2)\,\mathrm d\theta &&\text{non-negative speed on both halves}\\ &=4-(-4)=8&&\text{sum the two positive lengths} \end{aligned}

Why this step? Using 2cos⁡(θ/2)2\cos(\theta/2) on the full interval would incorrectly cancel the upper and lower lengths and give zero.

Answer. 88 units.

Check. Reflection in the initial line gives two equal halves of length 44. The pole cusp does not make this curve's length infinite.

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