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Arc length

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55017212
The graph shows the horizontal segment \(y=3\) on \([1,7]\). Use an arc-length integral to find its length.
Figure for problem 550172

Hints

- Find the slope of the horizontal segment. - Substitute that slope into the arc-length density. - The integral should agree with the segment's horizontal length.

Solution

1. The derivative is \(y'(x)=0\). 2. Therefore, \(L=\int_1^7\sqrt{1+[y'(x)]^2}\,\mathrm{d}x=\int_1^7 1\,\mathrm{d}x=6\).

Answer

\(6\) units
54954912
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.
Figure for problem 549549

Hints

- A constant slope makes the arc-length integrand constant. - 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
54955412
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

\(S\)
54955612
A student claims the length of \(y=\sin x\) on \([0,\pi]\) is \(\int_0^\pi\sqrt{1+\cos x}\,\mathrm{d}x\). 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 \(\int_0^\pi\sqrt{1+\cos^2x}\,\mathrm{d}x\). 3. The student's integrand can equal \(0\), but a valid arc-length density is always at least \(1\).

Answer

The correct setup is \(\int_0^\pi\sqrt{1+\cos^2x}\,\mathrm{d}x\).
54955712
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
54955912
A student 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 \(L_g=3\sqrt5\).
54956112
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
54956312
The graph shows a piecewise-linear curve. Its slope is \(0\) for \(0<x<1\), \(\sqrt3\) for \(1<x<3\), and \(-2\sqrt2\) for \(3<x<4\). Find its arc length on \([0,4]\).
Figure for problem 549563

Hints

- Treat each straight piece using its own slope. - For a line segment with slope \(m\), the arc-length density is \(\sqrt{1+m^2}\). - Multiply each density by that piece's horizontal interval length before adding.

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
54956912
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.
54957012
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.
54957312
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
54957412
The graph shows \(y=|x-2|+1\) on \([0,5]\). Find its length using arc-length integrals split at the corner.
Figure for problem 549574

Hints

- Locate where the derivative changes definition. - 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
54957512
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
54958012
A student 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 the student'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

The student's integral measures area. The correct arc-length integral is \(\int_0^1\sqrt{1+4x^2}\,\mathrm{d}x\).
54958212
Which integral represents the length of \(x=y^3-2y\) from \(y=-1\) to \(y=2\)? A. \(\int_{-1}^{2}\sqrt{1+(3y^2-2)^2}\,\mathrm{d}y\) B. \(\int_{-1}^{2}\sqrt{1+(y^3-2y)^2}\,\mathrm{d}y\) C. \(\int_{-1}^{2}[1+(3y^2-2)^2] \,\mathrm{d}y\) D. \(\int_{-1}^{2}\sqrt{1+(3x^2-2)^2}\,\mathrm{d}x\) 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 \(y\): \(\frac{\mathrm{d}x}{\mathrm{d}y}=3y^2-2\). 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

Choice A.
54958312
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.
54958412
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

\(w=7\)
54954812
Find the exact length of the curve \(x=\frac{1}{3}(y+2)^{3/2}\) for \(-2\le y\le2\). The figure shows the curve segment.
Figure for problem 549548

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
54955112
The graph of \(y=kx+1\) on \([0,6]\) has length \(10\). Find all possible values of \(k\).

Hints

- A line's slope is constant across the interval. - Both upward and downward slopes can produce the same length. - Use the interval width as a constant factor in the arc-length integral.

Solution

1. The derivative is \(y'(x)=k\), so the length is \(L=6\sqrt{1+k^2}\). 2. Set \(6\sqrt{1+k^2}=10\), giving \(k^2=\frac{16}{9}\). 3. Therefore, \(k=\frac{4}{3}\) or \(k=-\frac{4}{3}\).

Answer

\(k=\frac{4}{3}\) or \(k=-\frac{4}{3}\)
54955212
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

\(b=3\)
54955312
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
54955512
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
54955812
The graph shows 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.
Figure for problem 549558

Hints

- Recognize the radius and the fraction of the full circle shown. - For the integral, treat \(x\) as a function of \(y\). - A vertical tangent at the endpoint requires a limit.

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) \(\lim_{b\to6^-}\int_0^b\frac{6}{\sqrt{36-y^2}}\,\mathrm{d}y\)
54956012
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

\(b=7\)
54956212
Find the exact length of \(y=-\ln(\cos x)\) on \([0,\frac{\pi}{3}]\).

Hints

- Differentiate the logarithmic function first. - Use a trigonometric identity to simplify the length density. - Check the sign of the simplified function on the interval.

Solution

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

Answer

\(\ln(2+\sqrt3)\) units
54956412
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

\(k=3\)
54956512
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

\(f'(x)=\pm3x^2\)
54956612
Let \(S(x)=\int_1^{x^2}\sqrt{1+[f'(t)]^2}\,\mathrm{d}t\) for \(x>0\). Find \(S'(x)\).

Hints

- View the expression as an accumulation with a moving endpoint. - The endpoint is itself a function of \(x\). - Account for both the value of the length density and the rate at which the endpoint moves.

Solution

1. The integrand evaluated at the upper limit is \(\sqrt{1+[f'(x^2)]^2}\). 2. The upper limit \(x^2\) has derivative \(2x\). 3. Therefore, \(S'(x)=2x\sqrt{1+[f'(x^2)]^2}\).

Answer

\(S'(x)=2x\sqrt{1+[f'(x^2)]^2}\)
54956712
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 \(f(x)=1-2\sqrt6x\)
54957112
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

\(2L\)
54957212
The graph shows 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.
Figure for problem 549572

Hints

- Treat \(x\) as a function of \(y\). - Simplify the arc-length density before integrating. - Handle the vertical-tangent endpoints with limits.

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
54957612
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

\(4\sqrt5\le L\le4\sqrt{10}\)
54957712
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.
54957912
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

\(b\approx1.554\)
54958112
The graph shows a road profile 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.
Figure for problem 549581

Hints

- Differentiate the height profile to obtain rise per unit horizontal distance. - Substitute the slope into the arc-length formula. - The integration variable supplies the final unit of meters.

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

\(\approx102.6\,\text{m}\)
54958612
A suspension cable is modeled by \(y=0.05x^2\) for \(-10\le x\le10\), where both coordinates are measured in meters. 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.
Figure for problem 549586

Hints

- Differentiate the height model before using the arc-length formula. - Use symmetry to reduce the integral to half the interval. - The final unit is length, not area.

Solution

1. Differentiate: \(y'(x)=0.1x=\frac{x}{10}\). Therefore, \(L=\int_{-10}^{10}\sqrt{1+\left(\frac{x}{10}\right)^2}\,\mathrm{d}x\). 2. By symmetry and the substitution \(u=\frac{x}{10}\), \(L=20\int_0^1\sqrt{1+u^2}\,\mathrm{d}u\). 3. Using \(\int\sqrt{1+u^2}\,\mathrm{d}u=\frac12\left(u\sqrt{1+u^2}+\ln\left(u+\sqrt{1+u^2}\right)\right)\), \(L=10\left(\sqrt2+\ln(1+\sqrt2)\right)\approx23.0\,\text{m}\). 4. This is the physical length of cable over the modeled horizontal span.

Answer

a) \(\int_{-10}^{10}\sqrt{1+\left(\frac{x}{10}\right)^2}\,\mathrm{d}x\) b) \(10\left(\sqrt2+\ln(1+\sqrt2)\right)\approx23.0\,\text{m}\) c) The cable is approximately \(23.0\,\text{m}\) long over the modeled span.
54958712
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 \(6.04\,\text{m}\).
54955012
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

\(k\approx1.635\)
54956812
A curve \(y=f(x)\) is defined 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 \(L_1=\int_0^4\sqrt{1+[f'(u)]^2}\,\mathrm{d}u\). 2. For \(g(x)=f(2x)\), the chain rule gives \(g'(x)=2f'(2x)\). Thus, \(L_2=\int_0^2\sqrt{1+4[f'(2x)]^2}\,\mathrm{d}x=\frac12\int_0^4\sqrt{1+4[f'(u)]^2}\,\mathrm{d}u\). 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

The graph of \(y=f(2x)\) on \([0,2]\) is shorter: \(L_2<L_1\).
54957812
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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