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Arc length

Learning goals

  • Finds the derivative needed for a function graph's arc length.
  • Sets up an arc-length integral for a function graph.
  • Uses the graph's input interval as arc-length integration bounds.
  • Evaluates definite arc-length integrals.

Click problems to add them to your worksheet.

Arc length 1
A differentiable curve satisfies \(f'(x)=3\) for \(2\le x\le5\). Write, but do not evaluate, the arc-length integral for this part of the curve.

Hints

- Arc length depends on the derivative, not directly on the function's height. - Use the endpoints of the stated x-interval as the limits. - Substitute the given slope into the arc-length density without evaluating the integral.

Solution

1. Arc length for \(y=f(x)\) on \([a,b]\) is \(\int_a^b\sqrt{1+[f'(x)]^2}\,\mathrm{d}x\). 2. Substituting the given derivative and interval gives \(\int_2^5\sqrt{1+3^2}\,\mathrm{d}x\).

Answer

\(\int_2^5\sqrt{1+3^2}\,\mathrm{d}x\)
Arc length 2
Find the length of the straight segment between the two endpoints shown in the graph.
Figure for problem 562826

Hints

- Read the horizontal and vertical changes from the grid. - These changes are the legs of a right triangle whose hypotenuse is the segment.

Solution

1. The endpoints are \((0,0)\) and \((3,4)\), so the horizontal change is \(3\) and vertical change is \(4\). 2. By the Pythagorean theorem, the length is \(\sqrt{3^2+4^2}=5\).

Answer

\(5\) length units.
Arc length 3
At one point on a smooth curve, the tangent slope is \(3\). Replace a very short part of the curve with a straight segment whose horizontal change is \(\frac{1}{10}\) unit and whose slope is \(3\). Find the length of this straight-line approximation exactly.

Hints

- Use slope to estimate the vertical change over the small horizontal change. - Treat these two changes as perpendicular legs of a right triangle.

Solution

1. The approximating segment has vertical change \(3\cdot\frac{1}{10}=\frac{3}{10}\). 2. The chord length is \(\sqrt{\left(\frac{1}{10}\right)^2+\left(\frac{3}{10}\right)^2}=\frac{\sqrt{10}}{10}\) unit.

Answer

\(\frac{\sqrt{10}}{10}\) unit.
Arc length 4
Is the curved path longer than, shorter than, or equal in length to \(\sqrt{2}\) units? Explain without using an integral.
Figure for problem 562829

Hints

- Compare the path with the shortest route between its endpoints. - A straight-line distance can be found from the grid without following the curve.

Solution

1. The endpoints are \((0,0)\) and \((1,1)\). 2. The straight segment between them has length \(\sqrt{1^2+1^2}=\sqrt2\). 3. The curved path is not straight, so its length is greater than \(\sqrt2\).

Answer

Greater than \(\sqrt2\). A straight segment is the shortest path between the endpoints, and this path is curved.
Arc length 5
Use the arc-length integral to find the length of \(y=-\frac{3}{4}x+5\) on \([0,8]\), then confirm it using endpoint distance.

Hints

- A constant slope makes the arc-length integrand constant. - What are the two endpoint coordinates on the stated interval? - Use the changes in x and y as the legs of a right triangle for the geometric check.

Solution

1. Since \(y'(x)=-\frac{3}{4}\), \(L=\int_0^8\sqrt{1+\frac{9}{16}}\,\mathrm{d}x=\int_0^8\frac{5}{4}\,\mathrm{d}x=10\). 2. The endpoints are \((0,5)\) and \((8,-1)\), whose distance is \(\sqrt{8^2+(-6)^2}=10\).

Answer

\(10\) units
Arc length 6
The length of \(y=f(x)\) on \([a,b]\) is \(S\). Find the length of \(y=-f(x)\) on \([a,b]\).

Hints

- A reflection across the x-axis reverses every slope sign. - Check which part of the arc-length integrand removes the sign.

Solution

1. The reflected graph has derivative \(-f'(x)\). 2. Squaring gives \([-f'(x)]^2=[f'(x)]^2\). 3. Its arc-length integrand is unchanged, so the length is \(S\).

Answer

\(S\)
Arc length 7
Ilze claims the length of \(y=\sin x\) on \([0,\pi]\) is \(\int_0^\pi\sqrt{1+\cos x}\,\text{d}x\). Identify the error and give the correct integral. A decimal value is not required.

Hints

- Recall the complete arc-length formula before comparing integrands. - Check every operation applied to the derivative in the formula. - Use the fact that a valid arc-length density must be at least \(1\).

Solution

1. Since \(y'(x)=\cos x\), the derivative must be squared in the arc-length density. 2. The correct integral is \(\int_0^\pi\sqrt{1+\cos^2x}\,\text{d}x\). 3. Ilze's integrand can equal \(0\), but a valid arc-length density is always at least \(1\).

Answer

Ilze forgot to square the derivative. The correct setup is \(\int_0^\pi\sqrt{1+\cos^2x}\,\text{d}x\).
Arc length 8
A differentiable curve satisfies \(f'(x)=\sqrt{e^{2x}-1}\) for \(0\le x\le\ln3\). Find the exact arc length of its graph on this interval.

Hints

- Substitute the given derivative into the length density before trying to integrate. - Simplify the expression under the square root completely. - Use the domain to choose the correct sign of the square root.

Solution

1. The arc-length density is \(\sqrt{1+[f'(x)]^2}=\sqrt{e^{2x}}=e^x\). 2. The length is \(L=\int_0^{\ln3}e^x\,\mathrm{d}x\). 3. Evaluating gives \(L=3-1=2\).

Answer

\(2\) units
Arc length 9
Paolo claims that replacing \(f(x)\) by \(g(x)=f(x)+2x\) preserves arc length on the same interval because it only “tilts” the graph. Test the claim using \(f(x)=0\) on \([0,3]\). Find both arc lengths and state a conclusion.

Hints

- A single counterexample is enough to disprove a universal claim. - Compare how the transformation changes the derivative, not just the function values. - Use the same interval for both length calculations.

Solution

1. For \(f(x)=0\), \(f'(x)=0\), so \(L_f=\int_0^3 1\,\mathrm{d}x=3\). 2. Here \(g(x)=2x\), so \(g'(x)=2\) and \(L_g=\int_0^3\sqrt{1+4}\,\mathrm{d}x=3\sqrt5\). 3. Since \(3\sqrt5\ne3\), adding a nonconstant linear term does not generally preserve arc length.

Answer

The claim is false. The two lengths are \(L_f=3\) and \(L_g=3\sqrt5\).
Arc length 10
The graph shows a piecewise-linear path through four points. Find its total length.
Figure for problem 549561

Hints

- Treat each straight segment separately. - Use the horizontal and vertical changes between consecutive vertices. - Add segment lengths, not signed changes in height.

Solution

1. The segment lengths are \(\sqrt{3^2+4^2}=5\), \(\sqrt{4^2+0^2}=4\), and \(\sqrt{3^2+(-4)^2}=5\). 2. Add the nonoverlapping segment lengths: \(5+4+5=14\).

Answer

\(14\) units
Arc length 11
A piecewise-linear curve has slope \(0\) for \(0<x<1\), \(\sqrt3\) for \(1<x<3\), and \(-2\sqrt2\) for \(3<x<4\). Find its arc length on \([0,4]\).

Hints

- Treat each straight piece using its own slope. - For a line segment with slope \(m\), what is the arc-length density? - Account for the different horizontal interval lengths before adding the pieces.

Solution

1. The arc-length densities on the three pieces are \(\sqrt{1+0^2}=1\), \(\sqrt{1+(\sqrt3)^2}=2\), and \(\sqrt{1+(-2\sqrt2)^2}=3\). 2. Multiply each density by its horizontal interval length: \(1(1)+2(2)+3(1)\). 3. The total arc length is \(8\) units.

Answer

\(8\) units
Arc length 12
A smooth curve joins \((1,2)\) to \((7,10)\). What is the least possible arc length, and when is it attained?

Hints

- Compare any path with the direct route between its endpoints. - Use the distance formula for the endpoints. - Identify the path for which equality in the distance comparison is attained.

Solution

1. Any curve joining the points has length at least their straight-line distance. 2. The distance is \(\sqrt{(7-1)^2+(10-2)^2}=\sqrt{36+64}=10\). 3. Equality occurs for the line segment joining the points.

Answer

The least possible length is \(10\) units, attained by the line segment.
Arc length 13
Suppose \(f'(x)\ne0\) for all \(x\in[a,b]\). Compare the arc lengths of \(y=f(x)\) and \(y=3f(x)\) on \([a,b]\).

Hints

- Differentiate the vertically scaled function. - Compare the two arc-length densities at the same value of \(x\). - Use the nonzero-slope condition to decide whether the inequality is strict.

Solution

1. Their arc-length densities are \(\sqrt{1+[f'(x)]^2}\) and \(\sqrt{1+9[f'(x)]^2}\). 2. Because \([f'(x)]^2>0\), the second density is strictly larger at every point. 3. Therefore, \(y=3f(x)\) has strictly greater arc length.

Answer

The graph of \(y=3f(x)\) has greater arc length.
Arc length 14
A differentiable function \(f\) is strictly increasing on \([a,b]\), and its graph has length \(9.4\). Find the length of the graph of \(f^{-1}\) on \([f(a),f(b)]\), and justify your answer.

Hints

- Relate a function and its inverse geometrically. - The inverse graph is obtained by reflecting across \(y=x\). - Consider whether reflection changes distances along a curve.

Solution

1. The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y=x\). 2. Reflection preserves distances and therefore preserves arc length. 3. The inverse graph has length \(9.4\).

Answer

\(9.4\) units
Arc length 15
Find the length of \(y=|x-2|+1\) on \([0,5]\) using arc-length integrals split at the corner.

Hints

- Where does the absolute-value expression change its linear rule? - Write one arc-length integral for each straight piece. - A corner does not prevent adding the lengths of the two smooth pieces.

Solution

1. On \([0,2]\), \(y'(x)=-1\); on \([2,5]\), \(y'(x)=1\). 2. Therefore, \(L=\int_0^2\sqrt{1+(-1)^2}\,\mathrm{d}x+\int_2^5\sqrt{1+1^2}\,\mathrm{d}x\). 3. Thus, \(L=2\sqrt2+3\sqrt2=5\sqrt2\) units.

Answer

\(5\sqrt2\) units
Arc length 16
The displayed path has a curved portion followed by a line segment. Find the total length of the path.
Figure for problem 549575

Hints

- Treat the curved and straight portions separately. - Recognize the fraction of a full circle represented by the first portion. - Use endpoint distance only for the straight portion.

Solution

1. The quarter-circle portion has length \(\frac14(2\pi\cdot2)=\pi\). 2. The line segment has length \(\sqrt{(5-2)^2+(4-0)^2}=5\). 3. The total length is \(\pi+5\).

Answer

\(\pi+5\) units
Arc length 17
Mekdes says the length of \(y=x^2\) on \([0,1]\) is \(\int_0^1x^2\,\mathrm{d}x\). The graph shows the quantity represented by Mekdes's integral. Explain what it measures and write the correct arc-length integral.
Figure for problem 549580

Hints

- Identify the geometric quantity represented by integrating the function value. - Differentiate the function before forming the arc-length density. - Arc length uses \(\sqrt{1+[y'(x)]^2}\), not the graph's height.

Solution

1. The shaded quantity \(\int_0^1x^2\,\mathrm{d}x\) is the area between the graph and the x-axis. 2. Since \(y'(x)=2x\), the arc-length density is \(\sqrt{1+4x^2}\). 3. The correct length is \(\int_0^1\sqrt{1+4x^2}\,\mathrm{d}x\).

Answer

Mekdes's integral measures area. The correct arc-length integral is \(\int_0^1\sqrt{1+4x^2}\,\mathrm{d}x\).
Arc length 18
Which integral represents the length of \(x=y^3-2y\) from \(y=-1\) to \(y=2\)? A. \(\int_{-1}^{2}\sqrt{1+(3y^2-2)^2}\,\text{d}y\) B. \(\int_{-1}^{2}\sqrt{1+(y^3-2y)^2}\,\text{d}y\) C. \(\int_{-1}^{2}[1+(3y^2-2)^2] \,\text{d}y\) D. \(\int_{-1}^{2}\sqrt{1+(3x^2-2)^2}\,\text{d}x\) Justify your choice.

Hints

- Match the derivative and differential to the way the curve is defined. - Differentiate \(x\) with respect to \(y\). - The function value itself is not the slope in the arc-length density.

Solution

1. The curve is expressed as \(x\) in terms of \(y\), so differentiate with respect to \(y\): \(\frac{\text{d}x}{\text{d}y}=3y^2-2\). 2. The length density is \(\sqrt{1+(3y^2-2)^2}\). 3. The limits and differential are in \(y\), so choice A is correct.

Answer

A. Because \(\frac{\text{d}x}{\text{d}y}=3y^2-2\), the arc-length integral is \(\int_{-1}^{2}\sqrt{1+(3y^2-2)^2}\,\text{d}y\).
Arc length 19
A smooth curve has arc length \(10\) on \([0,6]\) and arc length \(4\) on \([1,5]\). Find the combined length of the portions on \([0,1]\) and \([5,6]\). Can the two individual outer lengths be determined?

Hints

- Decompose the full interval into three adjacent pieces. - Use what is known about the middle piece. - Decide whether the remaining information gives one unknown total or two separate values.

Solution

1. Arc length is additive over adjacent intervals. 2. The two outer portions together have length \(10-4=6\). 3. No information separates that total between \([0,1]\) and \([5,6]\), so their individual lengths cannot be determined.

Answer

The combined outer length is \(6\) units. The two individual outer lengths cannot be determined from the given information.
Arc length 20
A curve satisfies \(|f'(x)|=\sqrt{15}\) throughout an interval of width \(w\). Its arc length is \(28\). Find \(w\).

Hints

- A constant slope magnitude gives a constant arc-length density. - Use \(\sqrt{1+[f'(x)]^2}\) to find the length gained per horizontal unit. - Relate total length to interval width through that constant density.

Solution

1. The arc-length density is \(\sqrt{1+[f'(x)]^2}=\sqrt{1+15}=4\). 2. Because the density is constant, \(L=4w\). 3. Set \(4w=28\), giving \(w=7\).

Answer

\(w=7\)
Arc length 21
Calculate the broken-line length in each panel. Which is the better estimate of the curve's length, and why?
Figure for problem 562830

Hints

- Use the endpoint differences to find each chord's length. - A broken-line length adds all its straight segments. - Think about which path more closely follows the bend of the curve.

Solution

1. In a), the chord from \((0,0)\) to \((2,4)\) has length \(\sqrt{2^2+4^2}=2\sqrt{5}\). 2. In b), the chords run from \((0,0)\) to \((1,1)\) and from \((1,1)\) to \((2,4)\), with lengths \(\sqrt{2}\) and \(\sqrt{10}\). 3. The second estimate is \(\sqrt{2}+\sqrt{10}\approx4.576\), versus \(2\sqrt{5}\approx4.472\). Following the curve more closely makes the two-chord estimate better.

Answer

a) \(2\sqrt{5}\approx4.472\) units. b) \(\sqrt{2}+\sqrt{10}\approx4.576\) units. The two-chord estimate is better because its shorter chords follow the curve more closely.
Arc length 22
The curve \(y=x^2\) runs from \(x=0\) to \(x=2\). Over a very short horizontal step of size \(\Delta x>0\), its slope is approximately \(2x\). Use a right triangle to derive a formula for the short segment's length, then write an integral for the whole curve length. Do not evaluate it.

Hints

- The horizontal and vertical changes form the legs of a short right triangle. - Use the slope to express the vertical change in terms of \(\Delta x\). - Adding very short lengths leads to a definite integral.

Solution

1. The short vertical change is approximately \(\Delta y=(2x)\Delta x\). 2. By the Pythagorean theorem, the short chord length is \(\sqrt{(\Delta x)^2+(2x\Delta x)^2}=\sqrt{1+4x^2}\,\Delta x\). 3. Adding the short lengths as \(\Delta x\) approaches zero gives \(L=\int_0^2\sqrt{1+4x^2}\,\mathrm{d}x\).

Answer

A short chord has length approximately \(\sqrt{(\Delta x)^2+(2x\Delta x)^2}=\sqrt{1+4x^2}\,\Delta x\). \(L=\int_0^2\sqrt{1+4x^2}\,\mathrm{d}x\).
Arc length 23
For the curve \(x=\frac{y^2}{2}\) from \(y=0\) to \(y=2\), set up an integral for arc length using \(y\) as the variable. Explain why the short horizontal change is about \(y\,\Delta y\). Do not evaluate the integral.
Figure for problem 562832

Hints

- For a vertical step, start from the derivative of \(x\) with respect to \(y\). - Use horizontal and vertical changes as the legs of a short right triangle. - The bounds are the lowest and highest \(y\)-values.

Solution

1. Differentiating \(x=\frac{y^2}{2}\) gives \(\frac{\mathrm{d}x}{\mathrm{d}y}=y\), so a small vertical step \(\Delta y\) changes \(x\) by about \(y\,\Delta y\). 2. The short length is about \(\sqrt{(y\Delta y)^2+(\Delta y)^2}=\sqrt{1+y^2}\,\Delta y\). 3. The total length is \(L=\int_0^2\sqrt{1+y^2}\,\mathrm{d}y\).

Answer

\(\Delta x\approx y\,\Delta y\), because \(\frac{\mathrm{d}x}{\mathrm{d}y}=y\). \(L=\int_0^2\sqrt{1+y^2}\,\mathrm{d}y\).
Arc length 24
Estimate the length of \(y=x^2\) from \(x=0\) to \(x=1\) by joining the points at \(x=0\), \(x=\frac12\), and \(x=1\) with straight segments. Give the estimated length to three decimal places. Then write, but do not evaluate, the integral for the curve's exact length. Which is longer, the curve or the broken-line estimate?

Hints

- Find the three exact point coordinates before measuring chord lengths. - For each chord, use its horizontal and vertical changes. - The arc-length integrand uses the derivative, not the original function value.

Solution

1. The points are \((0,0)\), \((\frac12,\frac14)\), and \((1,1)\). 2. The two chords have lengths \(\frac{\sqrt5}{4}\) and \(\frac{\sqrt{13}}{4}\). Their sum is \(\frac{\sqrt5+\sqrt{13}}4\approx1.460\). 3. Since \(y'=2x\), the exact curve length is \(L=\int_0^1\sqrt{1+4x^2}\,\mathrm{d}x\). 4. A curved path is longer than the chords that cut across its bends.

Answer

Broken-line estimate: \(\frac{\sqrt5+\sqrt{13}}4\approx1.460\) units. Exact-length setup: \(L=\int_0^1\sqrt{1+4x^2}\,\mathrm{d}x\). The curve is longer.
Arc length 25
Find the exact length of the curve \(y=\frac{2}{3}x^{3/2}\) from \(x=0\) to \(x=3\) using an arc-length integral. Show your integral and evaluate it.

Hints

- Begin by differentiating the function rather than integrating it. - The arc-length integrand contains the square of the derivative. - Simplify the expression inside the square root before finding an antiderivative.

Solution

1. The derivative is \(y'=\sqrt{x}\). 2. The arc-length integral is \(L=\int_0^3\sqrt{1+(\sqrt{x})^2}\,\mathrm{d}x=\int_0^3\sqrt{1+x}\,\mathrm{d}x\). 3. Integrating gives \(L=\left[\frac{2}{3}(1+x)^{3/2}\right]_0^3=\frac{2}{3}(8-1)=\frac{14}{3}\) length units.

Answer

\(L=\int_0^3\sqrt{1+x}\,\mathrm{d}x=\frac{14}{3}\) length units.
Arc length 26
Find the exact length of the curve \(x=\frac{1}{3}(y+2)^{3/2}\) for \(-2\le y\le2\).

Hints

- Use the arc-length formula for \(x\) as a function of \(y\). - Differentiate with respect to the variable used in the interval. - Simplify the expression inside the square root before integrating.

Solution

1. Differentiate with respect to \(y\): \(\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{2}\sqrt{y+2}\). 2. The arc-length integrand is \(\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}=\frac{1}{2}\sqrt{y+6}\). 3. Therefore, \(L=\frac{1}{2}\int_{-2}^{2}\sqrt{y+6}\,\mathrm{d}y=\frac{16\sqrt2-8}{3}\) units.

Answer

\(\frac{16\sqrt2-8}{3}\) units
Arc length 27
The graph of \(y=kx+1\) on \([0,6]\) has length \(10\). Use the arc-length integral to find all possible values of \(k\), then verify the result using the distance between the endpoints.

Hints

- The required method begins with the arc-length integral, even though the graph is a line. - How does a constant slope simplify the integrand? - After solving for \(k\), use the endpoint changes only as a verification.

Solution

1. Since \(y'(x)=k\), the arc-length equation is \(10=\int_0^6\sqrt{1+k^2}\,\mathrm{d}x=6\sqrt{1+k^2}\). 2. Thus \(1+k^2=\frac{25}{9}\), so \(k^2=\frac{16}{9}\) and \(k=\frac{4}{3}\) or \(k=-\frac{4}{3}\). 3. For either value, the horizontal change is \(6\) and the magnitude of the vertical change is \(|6k|=8\). The endpoint distance is \(\sqrt{6^2+8^2}=10\), confirming the result.

Answer

\(\int_0^6\sqrt{1+k^2}\,\mathrm{d}x=10\), so \(k=\frac{4}{3}\) or \(k=-\frac{4}{3}\).
Arc length 28
A differentiable curve satisfies \(f'(x)=\sqrt{x^2-1}\) for \(x\ge1\). Find \(b>1\) if the length of the curve from \(x=1\) to \(x=b\) is \(4\).

Hints

- Substitute the given derivative into the arc-length formula. - Use the domain restriction when simplifying \(\sqrt{x^2}\). - Keep the unknown endpoint in the definite integral.

Solution

1. Since \(f'(x)=\sqrt{x^2-1}\), the arc-length integrand is \(\sqrt{1+[f'(x)]^2}=\sqrt{x^2}=x\) for \(x\ge1\). 2. Thus, the length is \(\int_1^b x\,\mathrm{d}x=\frac{b^2-1}{2}\). 3. Set \(\frac{b^2-1}{2}=4\). Since \(b>1\), \(b=3\).

Answer

\(b=3\)
Arc length 29
Find the exact length of \(y=\frac{x^2}{4}-\frac{1}{2}\ln x\) on \([1,3]\).

Hints

- Differentiate before expanding the arc-length expression. - Look for a perfect-square identity inside \(1+[y'(x)]^2\). - Use that \(x>0\) on the interval when removing the square root.

Solution

1. Differentiate: \(y'(x)=\frac{x}{2}-\frac{1}{2x}\). 2. On \([1,3]\), \(\sqrt{1+[y'(x)]^2}=\sqrt{\left(\frac{x}{2}+\frac{1}{2x}\right)^2}=\frac{x}{2}+\frac{1}{2x}\). 3. Therefore, the length is \(\int_1^3\left(\frac{x}{2}+\frac{1}{2x}\right)\,\mathrm{d}x=2+\frac{1}{2}\ln3\) units.

Answer

\(2+\frac{1}{2}\ln3\) units
Arc length 30
Find the exact length of \(y=\frac13(x^2+2)^{3/2}\) on \([0,2]\).

Hints

- Apply the chain rule carefully to the power \(\frac32\). - Simplify \(1+[y'(x)]^2\) before finding an antiderivative. - Check the sign of the expression obtained after taking the square root.

Solution

1. Differentiate: \(y'(x)=x\sqrt{x^2+2}\). 2. The length density simplifies because \(1+[y'(x)]^2=1+x^2(x^2+2)=(x^2+1)^2\). 3. Since \(x^2+1>0\), the length is \(\int_0^2(x^2+1)\,\mathrm{d}x=\frac{14}{3}\).

Answer

\(\frac{14}{3}\) units
Arc length 31
Consider the quarter-circle arc \(x=\sqrt{36-y^2}\) for \(0\le y\le6\). a) Use geometry to find its length. b) Set up, but do not evaluate, the corresponding arc-length integral with respect to \(y\), writing it as an improper integral.

Hints

- What circle and radius are encoded by the equation? - For the integral, treat \(x\) as a function of \(y\). - What happens to the derivative as \(y\) approaches the endpoint \(6\)?

Solution

1. The curve is one quarter of a circle of radius \(6\), so its geometric length is \(\frac14\cdot2\pi(6)=3\pi\). 2. Differentiate with respect to \(y\): \(\frac{\mathrm{d}x}{\mathrm{d}y}=-\frac{y}{\sqrt{36-y^2}}\). 3. The length density is \(\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}=\frac{6}{\sqrt{36-y^2}}\). 4. Because the integrand is unbounded at \(y=6\), the setup is \(\lim_{b\to6^-}\int_0^b\frac{6}{\sqrt{36-y^2}}\,\mathrm{d}y\).

Answer

a) \(3\pi\) units b) \(\lim_{b\to6^-}\int_0^b\frac{6}{\sqrt{36-y^2}}\,\mathrm{d}y\)
Arc length 32
For \(b>0\), the curve \(y=\frac23(x+1)^{3/2}\) is considered on \([0,b]\). Its arc length is \(\frac23(27-2\sqrt2)\). Find \(b\).

Hints

- Express the curve's length as a function of the unknown endpoint. - Simplify the quantity inside the square root before integrating. - After equating the two length expressions, isolate the power containing the endpoint.

Solution

1. The derivative is \(y'(x)=\sqrt{x+1}\), so the arc-length integrand is \(\sqrt{1+[y'(x)]^2}=\sqrt{x+2}\). 2. The length from \(0\) to \(b\) is \(L=\int_0^b\sqrt{x+2}\,\mathrm{d}x=\frac23[(b+2)^{3/2}-2\sqrt2]\). 3. Equating this to \(\frac23(27-2\sqrt2)\) gives \((b+2)^{3/2}=27\). 4. Therefore, \(b+2=9\), so \(b=7\).

Answer

\(b=7\)
Arc length 33
Find the exact length of \(y=-\ln(\cos x)\) on \([0,\frac{\pi}{3}]\). You may use \(\int\sec x\,\text{d}x=\ln|\sec x+\tan x|+C\).

Hints

- Differentiate the logarithmic function first. - Use a trigonometric identity to simplify the length density. - Check that the simplified function is positive on the interval before applying the supplied antiderivative.

Solution

1. Differentiate: \(y'(x)=\tan x\). 2. On the interval, \(\sqrt{1+[y'(x)]^2}=\sqrt{1+\tan^2x}=\sec x\). 3. Therefore, \(L=\int_0^{\pi/3}\sec x\,\text{d}x=[\ln(\sec x+\tan x)]_0^{\pi/3}=\ln(2+\sqrt3)\).

Answer

\(\ln(2+\sqrt3)\) units
Arc length 34
Let \(S(x)\) denote arc length accumulated along \(y=f(x)\) from \(x=0\) to \(x\). Suppose \(f'(x)=kx\) with \(k>0\), and \(S'(1)=\sqrt{10}\). Find \(k\).

Hints

- Interpret the derivative of accumulated length as the local arc-length density. - Substitute the given formula for \(f'(x)\) before evaluating at \(x=1\). - Use the positive parameter condition after squaring.

Solution

1. The arc-length accumulation rate is \(S'(x)=\sqrt{1+[f'(x)]^2}=\sqrt{1+k^2x^2}\). 2. At \(x=1\), \(\sqrt{1+k^2}=\sqrt{10}\). 3. Thus, \(k^2=9\), and the condition \(k>0\) gives \(k=3\).

Answer

\(k=3\)
Arc length 35
Define \(S(x)=\int_2^x\sqrt{1+9t^4}\,\mathrm{d}t\). If \(S\) is the arc-length function of a curve \(y=f(x)\), find all possible values of \(f'(x)\).

Hints

- Differentiate the accumulation function using the Fundamental Theorem of Calculus. - Set the result equal to the standard arc-length density. - Arc length determines slope magnitude but not slope direction.

Solution

1. By the Fundamental Theorem of Calculus, \(S'(x)=\sqrt{1+9x^4}\). 2. Arc-length density also satisfies \(S'(x)=\sqrt{1+[f'(x)]^2}\). 3. Therefore, \([f'(x)]^2=9x^4\), so \(f'(x)=3x^2\) or \(f'(x)=-3x^2\).

Answer

\(f'(x)=\pm3x^2\)
Arc length 36
A line \(y=f(x)\) satisfies \(f(0)=1\). Its accumulated arc length from \(0\) to \(x\) is \(S(x)=5x\) for \(0\le x\le3\). Find the two possible equations of the line.

Hints

- A line has a constant length gained per horizontal unit. - Relate that constant gain to the line's slope. - The given point determines the vertical intercept after the possible slopes are found.

Solution

1. Let the constant slope be \(m\). The length density of the line is \(\sqrt{1+m^2}\). 2. Since \(S'(x)=5\), solve \(\sqrt{1+m^2}=5\). 3. This gives \(m^2=24\), so \(m=\pm2\sqrt6\). 4. Using \(f(0)=1\), the two lines are \(f(x)=1+2\sqrt6x\) and \(f(x)=1-2\sqrt6x\).

Answer

\(f(x)=1+2\sqrt6x\) or \(f(x)=1-2\sqrt6x\)
Arc length 37
The graph of \(y=f(x)\) has arc length \(L\) on \([0,2]\). Define \(g(x)=2f\left(\frac{x}{2}\right)\) on \([0,4]\). Find the arc length of \(g\) in terms of \(L\).

Hints

- Determine how the transformation changes both horizontal and vertical distances. - Differentiate the transformed function carefully. - A change of variable can put the new length integral on the original interval.

Solution

1. Differentiate \(g(x)=2f\left(\frac{x}{2}\right)\): \(g'(x)=f'\left(\frac{x}{2}\right)\). 2. The length of \(g\) is \(\int_0^4\sqrt{1+\left[f'(x/2)\right]^2}\,\mathrm{d}x\). 3. With \(u=\frac{x}{2}\), \(\mathrm{d}x=2\,\mathrm{d}u\), and the limits become \(0\) and \(2\). 4. The length is \(2\int_0^2\sqrt{1+[f'(u)]^2}\,\mathrm{d}u=2L\).

Answer

\(2L\)
Arc length 38
Consider the right semicircle \(x=\sqrt{9-y^2}\), \(-3\le y\le3\). Set up its arc-length integral in terms of \(y\) and evaluate it.

Hints

- Treat \(x\) as a function of \(y\). - Simplify the arc-length density before integrating. - What happens to the derivative at the two endpoint heights?

Solution

1. Differentiate with respect to \(y\): \(\frac{\mathrm{d}x}{\mathrm{d}y}=-\frac{y}{\sqrt{9-y^2}}\). 2. The arc-length density simplifies to \(\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}=\frac{3}{\sqrt{9-y^2}}\). 3. The endpoints are improper, so \(L=\lim_{a\to-3^+,\,b\to3^-}\int_a^b\frac{3}{\sqrt{9-y^2}}\,\mathrm{d}y=3\pi\).

Answer

\(3\pi\) units
Arc length 39
A differentiable function satisfies \(2\le |f'(x)|\le3\) on \([1,5]\). Give the tightest arc-length bounds that follow directly from this information.

Hints

- Convert the slope-magnitude bounds into bounds on the arc-length density. - The square-root function preserves the order of nonnegative quantities. - Integrate the resulting constant bounds over the interval.

Solution

1. The length density satisfies \(\sqrt5\le\sqrt{1+[f'(x)]^2}\le\sqrt{10}\). 2. The interval length is \(5-1=4\). 3. Integrating the constant bounds gives \(4\sqrt5\le L\le4\sqrt{10}\).

Answer

\(4\sqrt5\le L\le4\sqrt{10}\)
Arc length 40
Show that the arc length of any differentiable graph \(y=f(x)\) on \([a,b]\) is at least \(b-a\). State when equality occurs.

Hints

- Compare the arc-length density with the constant \(1\). - Integrate the pointwise inequality over the full interval. - Determine when the pointwise inequality becomes equality everywhere.

Solution

1. Since \([f'(x)]^2\ge0\), \(\sqrt{1+[f'(x)]^2}\ge1\). 2. Therefore, \(L=\int_a^b\sqrt{1+[f'(x)]^2}\,\mathrm{d}x\ge\int_a^b1\,\mathrm{d}x=b-a\). 3. Equality occurs exactly when \(f'(x)=0\) throughout the interval, so the graph is horizontal.

Answer

\(L\ge b-a\), with equality exactly for a horizontal graph.
Arc length 41
For \(b>0\), the arc length of \(y=x^2\) from \(x=0\) to \(x=b\) is \(3\). Find \(b\) to three decimal places.

Hints

- Write the length as a function of the unknown endpoint. - Check whether that function increases as the endpoint moves right. - Use a numerical root-finding process after setting up the equation.

Solution

1. The length condition is \(\int_0^b\sqrt{1+4x^2}\,\mathrm{d}x=3\). 2. The left side is strictly increasing for \(b>0\), so there is one solution. 3. Numerical solution gives \(b\approx1.554045\). 4. To three decimal places, \(b\approx1.554\).

Answer

\(b\approx1.554\)
Arc length 42
A road profile is modeled by \(h(x)=0.002x^2\), where both \(x\) and \(h\) are measured in meters, for \(0\le x\le100\). Write an integral for the road-surface length and state the units. Then approximate the length to the nearest tenth of a meter.

Hints

- Differentiate the height profile to obtain rise per unit horizontal distance. - How does that slope enter the arc-length formula? - The integration variable supplies the final unit of length.

Solution

1. Differentiate: \(h'(x)=0.004x\), which is a dimensionless rise-per-run slope. 2. The road-surface length is \(L=\int_0^{100}\sqrt{1+(0.004x)^2}\,\mathrm{d}x\). 3. Numerical evaluation gives \(L\approx102.6\,\text{m}\).

Answer

\(L=\int_0^{100}\sqrt{1+(0.004x)^2}\,\mathrm{d}x\), and \(L\approx102.6\,\text{m}\).
Arc length 43
A suspension cable is modeled by \(y=0.05x^2\) for \(-10\le x\le10\), where both coordinates are measured in meters. For exact evaluation, you may use the antiderivative \(\int\sqrt{1+u^2}\,\text{d}u=\frac12\left(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\right)+C\). a) Write an integral for the cable's length. b) Find the exact length and round it to the nearest tenth of a meter. c) Interpret the result in context.

Hints

- Differentiate the height model before using the arc-length formula. - What symmetry does the model have on the stated interval? - The final unit is length, not area.

Solution

1. Differentiate: \(y'(x)=0.1x=\frac{x}{10}\). Therefore, \(L=\int_{-10}^{10}\sqrt{1+\left(\frac{x}{10}\right)^2}\,\text{d}x\). 2. By symmetry and the substitution \(u=\frac{x}{10}\), \(L=20\int_0^1\sqrt{1+u^2}\,\text{d}u\). 3. Using the supplied antiderivative gives \(L=10\left(\sqrt2+\ln(1+\sqrt2)\right)\,\text{m}\approx23.0\,\text{m}\). 4. This is the physical length of cable over the modeled horizontal span.

Answer

a) \(\int_{-10}^{10}\sqrt{1+\left(\frac{x}{10}\right)^2}\,\text{d}x\) b) \(10\left(\sqrt2+\ln(1+\sqrt2)\right)\,\text{m}\approx23.0\,\text{m}\) c) The cable is approximately \(23.0\,\text{m}\) long over the modeled span.
Arc length 44
An inspection robot follows the centerline \(y=f(x)\) of a pipe, where \(x\) and \(y\) are measured in meters. Engineers estimate the following slopes. <table><tr><td>\(x\)</td><td>\(0\)</td><td>\(1\)</td><td>\(3\)</td><td>\(5\)</td></tr><tr><td>\(f'(x)\)</td><td>\(0\)</td><td>\(\frac34\)</td><td>\(0\)</td><td>\(\frac43\)</td></tr></table> a) Find the corresponding values of \(g(x)=\sqrt{1+[f'(x)]^2}\). b) Apply the trapezoidal rule to \(g\) on the unequal subintervals to estimate the pipe length. c) State the unit and interpret the estimate.

Hints

- Convert each measured slope into an arc-length density before using the trapezoidal rule. - Use the actual width of each unequal subinterval. - The accumulated quantity is a physical length in meters.

Solution

1. The arc-length-density values are \(g(0)=1\), \(g(1)=\frac54\), \(g(3)=1\), and \(g(5)=\frac53\). 2. On \([0,1]\), the trapezoidal contribution is \(\frac12(1)\left(1+\frac54\right)=\frac98\). 3. On \([1,3]\), the contribution is \(\frac12(2)\left(\frac54+1\right)=\frac94\). 4. On \([3,5]\), the contribution is \(\frac12(2)\left(1+\frac53\right)=\frac83\). 5. The estimated length is \(\frac98+\frac94+\frac83=\frac{145}{24}\approx6.04\,\text{m}\).

Answer

a) \(1,\ \frac54,\ 1,\ \frac53\) b) \(\frac{145}{24}\approx6.04\) c) The estimated pipe-centerline length is \(6.04\,\text{m}\).
Arc length 45
For \(k>0\), the curve \(y=kx^2\) on \([0,1]\) has arc length \(2\). Find \(k\) to three decimal places.

Hints

- Express the curve's length in terms of the unknown scale factor. - Increasing \(k\) increases the slope magnitudes and therefore the length. - Use numerical equation solving after setting the length equal to \(2\).

Solution

1. The derivative is \(y'(x)=2kx\), so the length is \(L(k)=\int_0^1\sqrt{1+4k^2x^2}\,\mathrm{d}x\). 2. Set \(\int_0^1\sqrt{1+4k^2x^2}\,\mathrm{d}x=2\). 3. Numerical solution gives \(k\approx1.635\).

Answer

\(k\approx1.635\)
Arc length 46
A curve \(y=f(x)\) has a continuous derivative on \([0,4]\). Compare its length with the length of \(y=f(2x)\) on \([0,2]\). Prove the relationship by writing both integrals.

Hints

- Differentiate \(f(2x)\) with the chain rule. - Use a substitution so both length integrals run over the same interval. - Compare the two positive integrands point by point.

Solution

1. The original length is \(L_1=\int_0^4\sqrt{1+[f'(u)]^2}\,\text{d}u\). 2. For \(g(x)=f(2x)\), the chain rule gives \(g'(x)=2f'(2x)\). Thus, \(L_2=\int_0^2\sqrt{1+4[f'(2x)]^2}\,\text{d}x=\frac12\int_0^4\sqrt{1+4[f'(u)]^2}\,\text{d}u\). 3. For every real \(p\), \(\frac12\sqrt{1+4p^2}<\sqrt{1+p^2}\), because squaring both positive sides gives \(\frac14+p^2<1+p^2\). 4. Therefore, \(L_2<L_1\).

Answer

\(L_1=\int_0^4\sqrt{1+[f'(u)]^2}\,\text{d}u\) and \(L_2=\frac12\int_0^4\sqrt{1+4[f'(u)]^2}\,\text{d}u\), so \(L_2<L_1\).
Arc length 47
Prove that the arc length \(L\) of \(y=f(x)\) on \([a,b]\) satisfies \(L\le(b-a)+\int_a^b|f'(x)|\,\mathrm{d}x\).

Hints

- First establish a pointwise upper bound for the arc-length density. - Both sides of the proposed pointwise inequality are nonnegative, so squaring is valid. - Integrate the pointwise inequality over \([a,b]\).

Solution

1. For any real \(u\), \(\sqrt{1+u^2}\le1+|u|\), because both sides are nonnegative and \((1+|u|)^2=1+u^2+2|u|\ge1+u^2\). 2. Apply this inequality with \(u=f'(x)\). 3. Integrating gives \(L\le\int_a^b[1+|f'(x)|] \,\mathrm{d}x=(b-a)+\int_a^b|f'(x)|\,\mathrm{d}x\).

Answer

\(L\le(b-a)+\int_a^b|f'(x)|\,\mathrm{d}x\)

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