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Arc length of parametric curves

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53918212
Find the exact arc length of the parametric curve \(x=3t\), \(y=4t\) for \(0\le t\le5\).

Hints

- Differentiate both linear coordinate functions. - Use the magnitude of the derivative vector as the constant speed. - Multiply that speed by the length of the parameter interval.

Solution

1. The component derivatives are \(\frac{dx}{dt}=3\) and \(\frac{dy}{dt}=4\). 2. The arc-length integrand is \(\sqrt{3^{2}+4^{2}}=5\). 3. The definite integral is \(\int_{0}^{5}5\,dt=25\).

Answer

\(25\)
53918412
Find the exact arc length of the parametric curve \(x=t^{2}\), \(y=\frac{t^{2}}{2}\) for \(1\le t\le4\).

Hints

- Compute the two component derivatives and form their Euclidean magnitude. - Use \(t>0\) on the interval to remove any unnecessary absolute value. - Integrate the resulting constant multiple of \(t\).

Solution

1. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=t\). 2. Because \(t>0\) on the interval, the arc-length integrand simplifies to \(\sqrt{5}\,t\). 3. The definite integral is \(\int_{1}^{4}\sqrt{5}\,t\,dt=\frac{15\sqrt{5}}{2}\).

Answer

\(\frac{15\sqrt{5}}{2}\)
53918612
Find the exact arc length of the parametric curve \(x=\frac{t^{3}}{3}-t\), \(y=t^{2}\) for \(0\le t\le2\).

Hints

- Square the two component derivatives and look for a perfect-square trinomial. - Use the interval to select the nonnegative square root. - Integrate the simplified polynomial speed exactly.

Solution

1. The component derivatives are \(\frac{dx}{dt}=t^{2}-1\) and \(\frac{dy}{dt}=2t\). 2. The arc-length integrand simplifies to \(\sqrt{(t^{2}-1)^{2}+4t^{2}}=t^{2}+1\). 3. The definite integral is \(\int_{0}^{2}(t^{2}+1)\,dt=\frac{14}{3}\).

Answer

\(\frac{14}{3}\)
53918912
Find the exact arc length of the parametric curve \(x=-\frac{t^{3}}{3}+t\), \(y=t^{2}\) for \(0\le t\le3\).

Hints

- Square the component derivatives and simplify the resulting polynomial. - Recognize the expression under the radical as a perfect square. - Use the nonnegative square root and integrate over the full interval.

Solution

1. The component derivatives are \(\frac{dx}{dt}=1-t^{2}\) and \(\frac{dy}{dt}=2t\). 2. The arc-length integrand simplifies to \(\sqrt{(1-t^{2})^{2}+4t^{2}}=t^{2}+1\). 3. The definite integral is \(\int_{0}^{3}(t^{2}+1)\,dt=12\).

Answer

\(12\)
53919012
Find the exact length of the arc traced by \(x=3\cos(t)\), \(y=3\sin(t)\) as \(t\) runs from \(0\) to \(\frac{\pi}{2}\).

Hints

- Differentiate both coordinate functions to form the curve’s derivative vector. - Simplify the magnitude of that vector before integrating. - Use the full stated parameter interval.

Solution

1. The component derivatives are \(\frac{dx}{dt}=-3\sin(t)\) and \(\frac{dy}{dt}=3\cos(t)\), so the speed is \(3\). 2. Integrating over the parameter interval gives \(\int_{0}^{\frac{\pi}{2}}3\,dt=\frac{3\pi}{2}\).

Answer

\(\frac{3\pi}{2}\)
53919112
Find the exact length of the arc traced by \(x=5\cos(t)\), \(y=5\sin(t)\) as \(t\) runs from \(\frac{\pi}{6}\) to \(\frac{5\pi}{6}\).

Hints

- Differentiate both trigonometric coordinate functions. - Simplify the magnitude of the derivative vector using \(\sin^{2}(t)+\cos^{2}(t)=1\). - Use the full stated parameter interval.

Solution

1. The component derivatives are \(\frac{dx}{dt}=-5\sin(t)\) and \(\frac{dy}{dt}=5\cos(t)\), so the speed is \(5\). 2. Integrating over the parameter interval gives \(\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}5\,dt=\frac{10\pi}{3}\).

Answer

\(\frac{10\pi}{3}\)
53919212
Set up, but do not evaluate, a definite integral for the arc length of \(x=4\cos(t)\), \(y=2\sin(t)\) over \(0\le t\le\frac{\pi}{2}\).

Hints

- Differentiate the cosine and sine components separately. - Place the squared component derivatives under one square root. - Use the given bounds and leave the integral unevaluated.

Solution

1. Differentiate: \(\frac{dx}{dt}=-4\sin(t)\) and \(\frac{dy}{dt}=2\cos(t)\). 2. Combine the squared derivatives and take the nonnegative square root: \(\sqrt{16\sin^{2}(t)+4\cos^{2}(t)}=2\sqrt{3\sin^{2}(t)+1}\). 3. The required setup is \(\int_{0}^{\frac{\pi}{2}}2\sqrt{3\sin^{2}(t)+1}\,dt\).

Answer

\(\int_{0}^{\frac{\pi}{2}}2\sqrt{3\sin^{2}(t)+1}\,dt\)
53919312
Set up, but do not evaluate, a definite integral for the arc length of \(x=6\cos(t)\), \(y=3\sin(t)\) over \(0\le t\le\pi\).

Hints

- Differentiate both components before forming the speed. - Simplify the sum of squared component derivatives as far as possible. - Use the entire stated interval and do not evaluate the integral.

Solution

1. Differentiate: \(\frac{dx}{dt}=-6\sin(t)\) and \(\frac{dy}{dt}=3\cos(t)\). 2. Combine the squared derivatives and take the nonnegative square root: \(\sqrt{36\sin^{2}(t)+9\cos^{2}(t)}=3\sqrt{3\sin^{2}(t)+1}\). 3. The required setup is \(\int_{0}^{\pi}3\sqrt{3\sin^{2}(t)+1}\,dt\).

Answer

\(\int_{0}^{\pi}3\sqrt{3\sin^{2}(t)+1}\,dt\)
53919912
Write a definite integral that gives the length of the curve \(x=t^{2}+\sin(t)\), \(y=t^{3}-\cos(t)\) from \(t=0\) to \(t=1\). Do not evaluate it.

Hints

- Differentiate both mixed polynomial-trigonometric coordinate functions. - Put the squared component derivatives under one square root. - Use the stated parameter bounds and stop after writing the integral.

Solution

1. Differentiate the coordinate functions: \(\frac{dx}{dt}=2t+\cos(t)\) and \(\frac{dy}{dt}=3t^{2}+\sin(t)\). 2. Their combined magnitude is \(\sqrt{(2t+\cos(t))^{2}+(3t^{2}+\sin(t))^{2}}\). 3. The length is represented by \(\int_{0}^{1}\sqrt{(2t+\cos(t))^{2}+(3t^{2}+\sin(t))^{2}}\,dt\).

Answer

\(\int_{0}^{1}\sqrt{(2t+\cos(t))^{2}+(3t^{2}+\sin(t))^{2}}\,dt\)
53920012
Write a definite integral that gives the length of the curve \(x=t+\cos(t)\), \(y=-t+\sin(t)\) from \(t=0\) to \(t=\pi\). Do not evaluate it.

Hints

- Differentiate each coordinate, preserving the signs of the trigonometric derivatives. - Form the speed from the sum of the squared component derivatives. - Use \(0\) and \(\pi\) as bounds without evaluating.

Solution

1. Differentiate the coordinate functions: \(\frac{dx}{dt}=1-\sin(t)\) and \(\frac{dy}{dt}=\cos(t)-1\). 2. Their combined magnitude is \(\sqrt{(1-\sin(t))^{2}+(\cos(t)-1)^{2}}\). 3. The length is represented by \(\int_{0}^{\pi}\sqrt{(1-\sin(t))^{2}+(\cos(t)-1)^{2}}\,dt\).

Answer

\(\int_{0}^{\pi}\sqrt{(1-\sin(t))^{2}+(\cos(t)-1)^{2}}\,dt\)
53920112
Write a definite integral that gives the length of the curve \(x=e^{t}\cos(t)\), \(y=e^{t}\sin(t)\) from \(t=0\) to \(t=2\). Do not evaluate it.

Hints

- Apply the product rule to both exponential-trigonometric components. - Simplify the squared speed using \(\sin^{2}(t)+\cos^{2}(t)=1\). - Write the definite integral over the full interval and do not evaluate it.

Solution

1. Differentiate the coordinate functions: \(\frac{dx}{dt}=e^{t}(\cos(t)-\sin(t))\) and \(\frac{dy}{dt}=e^{t}(\sin(t)+\cos(t))\). 2. Their combined magnitude simplifies to \(\sqrt{2}e^{t}\). 3. The length is represented by \(\int_{0}^{2}\sqrt{2}e^{t}\,dt\).

Answer

\(\int_{0}^{2}\sqrt{2}e^{t}\,dt\)
53920212
Write a definite integral that gives the length of the curve \(x=t^{4}\), \(y=\ln(t)\) from \(t=1\) to \(t=2\). Do not evaluate it.

Hints

- Differentiate the fourth power and logarithm components. - Combine the squared derivatives over a common positive denominator. - Use \(t>0\) on the interval when simplifying the square root.

Solution

1. Differentiate the coordinate functions: \(\frac{dx}{dt}=4t^{3}\) and \(\frac{dy}{dt}=\frac{1}{t}\). 2. Because \(t>0\), their combined magnitude is \(\frac{\sqrt{16t^{8}+1}}{t}\). 3. The length is represented by \(\int_{1}^{2}\frac{\sqrt{16t^{8}+1}}{t}\,dt\).

Answer

\(\int_{1}^{2}\frac{\sqrt{16t^{8}+1}}{t}\,dt\)
54543812
The graph shows a curve’s measured speed at the midpoints of four one-second intervals from \(t=0\) to \(t=4\). Use the midpoint rule to estimate the arc length.
Figure for problem 545438

Hints

- Read the speed at each interval midpoint from the graph. - Each midpoint represents an interval of width \(1\). - Multiply the sum of the four midpoint speeds by the common width.

Solution

1. Each subinterval has width \(1\). 2. The midpoint estimate is \(1[2+3+5+4]\). 3. Therefore, the estimated arc length is \(14\).

Answer

\(L\approx14\)
53918312
Find the exact arc length of the parametric curve \(x=t^{2}\), \(y=\frac{2t^{3}}{3}\) for \(0\le t\le2\).

Hints

- Differentiate both power functions and factor the common nonnegative term from the speed. - Simplify the radical to an expression involving \(1+t^{2}\). - Use a substitution based on the expression inside that radical.

Solution

1. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=2t^{2}\). 2. Because \(t\ge0\), the arc-length integrand simplifies to \(2t\sqrt{t^{2}+1}\). 3. Therefore, \(L=\int_{0}^{2}2t\sqrt{t^{2}+1}\,dt=\frac{2}{3}\left(5\sqrt{5}-1\right)\).

Answer

\(\frac{2}{3}\left(5\sqrt{5}-1\right)\)
53918512
Find the exact arc length of the parametric curve \(x=t^{3}\), \(y=\frac{3t^{2}}{2}\) for \(0\le t\le2\).

Hints

- Factor the common power of \(t\) from the speed expression. - Use the nonnegative interval to simplify the absolute value. - Apply a substitution involving \(t^{2}+1\).

Solution

1. The component derivatives are \(\frac{dx}{dt}=3t^{2}\) and \(\frac{dy}{dt}=3t\). 2. Because \(t\ge0\), the arc-length integrand simplifies to \(3t\sqrt{t^{2}+1}\). 3. The definite integral is \(\int_{0}^{2}3t\sqrt{t^{2}+1}\,dt=5\sqrt{5}-1\).

Answer

\(5\sqrt{5}-1\)
53918712
Find the exact arc length of the parametric curve \(x=\frac{t^{2}}{2}\), \(y=\frac{t^{3}}{3}\) for \(0\le t\le3\).

Hints

- Form the speed from the derivatives \(x'(t)\) and \(y'(t)\). - Factor the nonnegative parameter from the radical. - Use a substitution based on \(1+t^{2}\) and evaluate at both endpoints.

Solution

1. The component derivatives are \(\frac{dx}{dt}=t\) and \(\frac{dy}{dt}=t^{2}\). 2. Because \(t\ge0\), the arc-length integrand simplifies to \(t\sqrt{t^{2}+1}\). 3. The definite integral is \(\int_{0}^{3}t\sqrt{t^{2}+1}\,dt=\frac{10\sqrt{10}-1}{3}\).

Answer

\(\frac{10\sqrt{10}-1}{3}\)
53918812
Find the exact arc length of the parametric curve \(x=2t^{2}\), \(y=\frac{4t^{3}}{3}\) for \(0\le t\le2\).

Hints

- Factor the common numerical and parameter factors from the derivative-vector magnitude. - Use \(t\ge0\) to simplify the speed. - Apply a substitution involving \(1+t^{2}\) before evaluating the bounds.

Solution

1. The component derivatives are \(\frac{dx}{dt}=4t\) and \(\frac{dy}{dt}=4t^{2}\). 2. Because \(t\ge0\), the arc-length integrand simplifies to \(4t\sqrt{t^{2}+1}\). 3. The definite integral is \(\int_{0}^{2}4t\sqrt{t^{2}+1}\,dt=\frac{20\sqrt{5}-4}{3}\).

Answer

\(\frac{20\sqrt{5}-4}{3}\)
53919412
Find the arc length of \(x=t\), \(y=\sin(t)\) on \(0\le t\le\pi\). Round to 3 decimal places.

Hints

- Differentiate the linear and sine coordinate functions. - Set up the speed as the square root of the sum of squared component derivatives. - Evaluate numerically and round only the final result to three decimal places.

Solution

1. The component derivatives are \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=\cos(t)\). 2. The arc-length integral is \(L=\int_{0}^{\pi}\sqrt{1+\cos^{2}(t)}\,dt\). 3. Numerical evaluation gives \(L\approx3.820\).

Answer

\(L\approx3.820\)
53919512
Find the arc length of \(x=t^{2}\), \(y=\cos(t)\) on \(0\le t\le2\). Round to 3 decimal places.

Hints

- Use \(x'(t)=2t\) and differentiate the cosine coordinate with its negative sign. - Integrate the magnitude of the derivative vector over the stated interval. - Keep guard digits until rounding to three decimal places.

Solution

1. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=-\sin(t)\). 2. The arc-length integral is \(L=\int_{0}^{2}\sqrt{4t^{2}+\sin^{2}(t)}\,dt\). 3. Numerical evaluation gives \(L\approx4.254\).

Answer

\(L\approx4.254\)
53919612
Find the arc length of \(x=e^{t}\), \(y=t^{2}\) on \(0\le t\le1\). Round to 4 decimal places.

Hints

- Differentiate the exponential and quadratic components independently. - Use their squared derivatives to form the arc-length integrand. - Perform the numerical integration before rounding to four decimal places.

Solution

1. The component derivatives are \(\frac{dx}{dt}=e^{t}\) and \(\frac{dy}{dt}=2t\). 2. The arc-length integral is \(L=\int_{0}^{1}\sqrt{e^{2t}+4t^{2}}\,dt\). 3. Numerical evaluation gives \(L\approx2.0097\).

Answer

\(L\approx2.0097\)
53919712
Find the arc length of \(x=t-\sin(t)\), \(y=1-\cos(t)\) on \(0\le t\le2\pi\). Round to 3 decimal places.

Hints

- Differentiate both cycloid components and simplify the squared speed trigonometrically. - Use a half-angle identity and the sign of the sine factor on the interval. - Evaluate exactly before expressing the result to three decimal places.

Solution

1. The component derivatives are \(\frac{dx}{dt}=1-\cos(t)\) and \(\frac{dy}{dt}=\sin(t)\). 2. The arc-length integral is \(L=\int_{0}^{2\pi}\sqrt{2-2\cos(t)}\,dt\). 3. Numerical evaluation gives \(L\approx8.000\).

Answer

\(L\approx8.000\)
53919812
Find the arc length of \(x=\ln(t)\), \(y=t^{2}\) on \(1\le t\le3\). Round to 3 decimal places.

Hints

- Respect the positive domain when differentiating \(\ln(t)\). - Form the speed from \(1/t\) and the derivative of \(t^{2}\). - Numerically integrate first, then round to three decimal places.

Solution

1. The component derivatives are \(\frac{dx}{dt}=\frac{1}{t}\) and \(\frac{dy}{dt}=2t\). 2. Because \(t>0\), the arc-length integral is \(L=\int_{1}^{3}\frac{\sqrt{4t^{4}+1}}{t}\,dt\). 3. Numerical evaluation gives \(L\approx8.109\).

Answer

\(L\approx8.109\)
53920312
A curve is given by \(x=kt\) and \(y=2kt\) for \(0\le t\le3\), where \(k>0\). Its arc length is \(15\sqrt{5}\). Find \(k\).

Hints

- Differentiate the two linear coordinates and form the constant speed in terms of \(k\). - Multiply by the parameter-interval length to obtain the total arc length. - Use \(k>0\) when solving the resulting absolute-value equation.

Solution

1. The component derivatives are \(\frac{dx}{dt}=k\) and \(\frac{dy}{dt}=2k\), so the speed is \(\sqrt{5}\,k\) because \(k>0\). 2. The length is \(L=\int_{0}^{3}\sqrt{5}\,k\,dt=3\sqrt{5}\,k\). 3. Solving \(3\sqrt{5}\,k=15\sqrt{5}\) gives \(k=5\).

Answer

\(k=5\)
53920412
The curve \(x=3t\), \(y=4t\) is traced from \(t=0\) to \(t=a\), where \(a>0\). If its length is \(50\), find \(a\).

Hints

- Find the constant speed of the line parameterization. - Express the total length as speed times the interval length \(a\). - Apply \(a>0\) when solving the length equation.

Solution

1. The component derivatives are \(\frac{dx}{dt}=3\) and \(\frac{dy}{dt}=4\), so the speed is \(5\). 2. The length through parameter \(a\) is \(\int_{0}^{a}5\,dt=5a\). 3. Solving \(5a=50\) gives \(a=10\).

Answer

\(a=10\)
53920512
A circular arc is parameterized by \(x=a\cos(t)\), \(y=a\sin(t)\) for \(0\le t\le\frac{\pi}{3}\), with \(a>0\). Its length is \(4\pi\). Find \(a\).

Hints

- Show that the derivative vector has constant magnitude \(a\). - Multiply that speed by the angular parameter interval. - Use the positive-radius condition when isolating \(a\).

Solution

1. The component derivatives are \(\frac{dx}{dt}=-a\sin(t)\) and \(\frac{dy}{dt}=a\cos(t)\), so the speed is \(a\) because \(a>0\). 2. The length is \(\int_{0}^{\frac{\pi}{3}}a\,dt=\frac{\pi a}{3}\). 3. Solving \(\frac{\pi a}{3}=4\pi\) gives \(a=12\).

Answer

\(a=12\)
53920712
Two curves are traced over \(0\le t\le2\). Curve A has \(x=3t\), \(y=4t\); Curve B has \(x=5t\), \(y=0\). Which curve is longer? Give both lengths to three decimal places.

Hints

- Find the constant derivative-vector magnitude for each line. - Integrate each speed over the same parameter interval independently. - Compare the resulting lengths at the requested decimal precision.

Solution

1. Curve A has speed \(\sqrt{3^{2}+4^{2}}=5\), so \(L_A=\int_{0}^{2}5\,dt=10\approx10.000\). 2. Curve B has speed \(5\), so \(L_B=\int_{0}^{2}5\,dt=10\approx10.000\). 3. The curves have the same length.

Answer

\(L_A\approx10.000\), \(L_B\approx10.000\); the lengths are equal.
53920812
Two curves are traced over \(0\le t\le1\). Curve A has \(x=t\), \(y=t^{2}\); Curve B has \(x=2t\), \(y=\frac{t^{2}}{2}\). Which curve is longer? Give both lengths to three decimal places.

Hints

- Form a separate speed function for each curve. - Numerically evaluate both arc-length integrals over \([0,1]\). - Round consistently before stating which completed length is greater.

Solution

1. Curve A has length \(L_A=\int_{0}^{1}\sqrt{1+4t^{2}}\,dt\approx1.479\). 2. Curve B has length \(L_B=\int_{0}^{1}\sqrt{4+t^{2}}\,dt\approx2.080\). 3. Therefore, Curve B is longer.

Answer

\(L_A\approx1.479\), \(L_B\approx2.080\); Curve B is longer.
53920912
Curve A is \(x=2t\), \(y=3t\) for \(0\le t\le1\). Curve B is \(x=2t^{2}\), \(y=3t^{2}\) for \(0\le t\le1\). Verify by arc length that the two parameterizations trace paths of the same length.

Hints

- Compute the derivative-vector magnitude for each parameterization. - Notice that the second curve changes speed while tracing the same line segment. - Evaluate both integrals on their stated intervals and compare exact values.

Solution

1. Curve A has speed \(\sqrt{2^{2}+3^{2}}=\sqrt{13}\), so its length is \(L_A=\int_{0}^{1}\sqrt{13}\,dt=\sqrt{13}\). 2. Curve B has speed \(\sqrt{(4t)^{2}+(6t)^{2}}=2\sqrt{13}\,t\) on \([0,1]\), so its length is \(L_B=\int_{0}^{1}2\sqrt{13}\,t\,dt=\sqrt{13}\). 3. Since the values agree, both paths have length \(\sqrt{13}\).

Answer

Both lengths equal \(\sqrt{13}\).
53921012
Curve A is \(x=\cos(t)\), \(y=\sin(t)\) for \(0\le t\le\pi\). Curve B is \(x=\cos(2t)\), \(y=\sin(2t)\) for \(0\le t\le\frac{\pi}{2}\). Verify by arc length that the two parameterizations trace paths of the same length.

Hints

- Determine the constant speed for each circular parameterization. - Account for the factor from the inner angle in the second derivative vector. - Multiply each speed by its own parameter-interval length.

Solution

1. Curve A has speed \(1\), so its length is \(L_A=\int_{0}^{\pi}1\,dt=\pi\). 2. Curve B has speed \(2\), so its length is \(L_B=\int_{0}^{\frac{\pi}{2}}2\,dt=\pi\). 3. Since the values agree, both paths have length \(\pi\).

Answer

Both lengths equal \(\pi\).
53921112
For \(x=t^{2}\), \(y=\frac{t^{3}}{3}-t\) on \(0\le t\le2\), a student uses \(\int_{0}^{2}\left(\frac{dx}{dt}+\frac{dy}{dt}\right)dt\). Explain why this is not an arc-length integral and find the correct exact length.

Hints

- Arc length uses the Euclidean magnitude of the derivative vector, not a component sum. - Simplify the sum of squared component derivatives and look for a perfect square. - Use the nonnegative speed over the stated interval before integrating.

Solution

1. Arc length uses the magnitude of the component-rate vector, not the sum of its components. 2. The correct integrand is \(\sqrt{(2t)^{2}+(t^{2}-1)^{2}}=t^{2}+1\), which is nonnegative. 3. The exact length is \(\int_{0}^{2}(t^{2}+1)\,dt=\frac{14}{3}\).

Answer

The student added the component rates instead of finding their vector magnitude. The correct exact length is \(\frac{14}{3}\).
53921212
A parametric curve has the recorded speeds shown. Use the trapezoidal rule on the given intervals to estimate its arc length. <table><tr><th>\(t\)</th><th>Speed along the curve</th></tr><tr><td>\(0\)</td><td>\(5\)</td></tr><tr><td>\(1\)</td><td>\(4\)</td></tr><tr><td>\(2\)</td><td>\(6\)</td></tr><tr><td>\(3\)</td><td>\(7\)</td></tr></table>

Hints

- Interpret the table values as samples of the speed function. - Apply the trapezoidal rule to each unit-width subinterval. - Add all trapezoidal contributions to estimate the accumulated length.

Solution

1. Arc length is the accumulation of speed over time. 2. The trapezoidal estimate is \(\frac{1}{2}(5+4)+\frac{1}{2}(4+6)+\frac{1}{2}(6+7)=16\).

Answer

\(16\)
53921312
A parametric curve has the recorded speeds shown. Use the trapezoidal rule on the given intervals to estimate its arc length. <table><tr><th>\(t\)</th><th>Speed along the curve</th></tr><tr><td>\(0\)</td><td>\(2\)</td></tr><tr><td>\(0.5\)</td><td>\(3\)</td></tr><tr><td>\(1\)</td><td>\(5\)</td></tr><tr><td>\(1.5\)</td><td>\(4\)</td></tr><tr><td>\(2\)</td><td>\(6\)</td></tr></table>

Hints

- Determine the common spacing between consecutive parameter values. - Use the trapezoidal endpoint and interior weights on the speed samples. - Keep the spacing factor outside the weighted sum until the final step.

Solution

1. Arc length is the accumulation of speed over time. 2. With width \(0.5\), the trapezoidal estimate is \(0.5\left(\frac{2+3}{2}+\frac{3+5}{2}+\frac{5+4}{2}+\frac{4+6}{2}\right)=8\).

Answer

\(8\)
53921412
The path is \(x=2t\), \(y=t^{2}\) for \(0\le t\le2\). a) Write the exact arc-length integral. b) Approximate the length to three decimal places.

Hints

- Differentiate both coordinates and simplify the speed before writing the integral. - Use the exact integral as the answer to part a). - Numerically evaluate that same integral and round only for part b).

Solution

1. The component derivatives are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so the speed is \(2\sqrt{t^{2}+1}\). 2. The exact setup is \(\int_{0}^{2}2\sqrt{t^{2}+1}\,dt\). 3. Numerical evaluation gives \(L\approx5.916\).

Answer

a) \(\int_{0}^{2}2\sqrt{t^{2}+1}\,dt\) b) \(L\approx5.916\)
53921512
The path is \(x=t+\cos(t)\), \(y=\sin(t)\) for \(0\le t\le\pi\). a) Write the exact arc-length integral. b) Approximate the length to three decimal places.

Hints

- Differentiate the trigonometric components and simplify the squared speed. - Preserve the exact radical integral for part a). - Use numerical integration on that exact setup and round for part b).

Solution

1. The component derivatives are \(\frac{dx}{dt}=1-\sin(t)\) and \(\frac{dy}{dt}=\cos(t)\), so the speed is \(\sqrt{2-2\sin(t)}\). 2. The exact setup is \(\int_{0}^{\pi}\sqrt{2-2\sin(t)}\,dt\). 3. Numerical evaluation gives \(L\approx2.343\).

Answer

a) \(\int_{0}^{\pi}\sqrt{2-2\sin(t)}\,dt\) b) \(L\approx2.343\)
54541412
The path \(x=2\cos(t)\), \(y=2\sin(t)\) is traced for \(0\le t\le5\pi\). Find the total distance traveled and compare it with the geometric length of the circle.

Hints

- Determine the speed and the total parameter duration. - Separate distance traveled from the length of the underlying geometric curve. - Count how many complete and partial revolutions occur.

Solution

1. The speed is \(\sqrt{(-2\sin t)^2+(2\cos t)^2}=2\). 2. The total distance is \(\int_0^{5\pi}2\,dt=10\pi\). 3. The circle has radius \(2\), so its geometric circumference is \(4\pi\). 4. The parameter interval traces \(\frac{5}{2}\) revolutions, so the distance exceeds one circumference.

Answer

Total distance: \(10\pi\) Circle circumference: \(4\pi\)
54541812
A curve is defined by \(x(t)=\int_0^t\cos(s^2)\,ds\) and \(y(t)=\int_0^t\sin(s^2)\,ds\). Find its exact arc length for \(0\le t\le3\).

Hints

- Differentiate each accumulated coordinate using its upper limit. - Look for a trigonometric identity inside the speed. - Interpret the resulting constant speed over the interval.

Solution

1. The component derivatives are \(\frac{dx}{dt}=\cos(t^2)\) and \(\frac{dy}{dt}=\sin(t^2)\). 2. The speed is \(\sqrt{\cos^2(t^2)+\sin^2(t^2)}=1\). 3. Therefore, \(L=\int_0^3 1\,dt=3\).

Answer

\(3\)
54541912
Find the exact arc length of \(x=\cos(t)+t\sin(t)\), \(y=\sin(t)-t\cos(t)\) for \(0\le t\le\pi\).

Hints

- Product-rule terms cancel when each coordinate is differentiated. - Factor the parameter from the speed before using a trigonometric identity. - Use the interval to remove any absolute-value ambiguity.

Solution

1. The component derivatives simplify to \(\frac{dx}{dt}=t\cos(t)\) and \(\frac{dy}{dt}=t\sin(t)\). 2. The speed is \(\sqrt{t^2\cos^2(t)+t^2\sin^2(t)}=t\) on \([0,\pi]\). 3. Therefore, \(L=\int_0^\pi t\,dt=\frac{\pi^2}{2}\).

Answer

\(\frac{\pi^2}{2}\)
54542012
The path \(x=\sin(t)\), \(y=0\) is traced for \(0\le t\le2\pi\). A student integrates \(\cos(t)\) and concludes that the arc length is \(0\). Explain the error and find the correct total length.

Hints

- Arc length accumulates distance, which cannot be negative. - Identify where the path reverses direction along the x-axis. - Use the magnitude of the component rate across the full interval.

Solution

1. The horizontal component rate is \(\cos(t)\), but speed is its magnitude, \(|\cos(t)|\). 2. The path reverses direction whenever \(\cos(t)=0\), so signed changes cancel while distance does not. 3. The total length is \(\int_0^{2\pi}|\cos(t)|\,dt=4\).

Answer

The student used signed horizontal change instead of speed. The correct total length is \(4\).
54542112
The figure shows a two-stage path. For \(0\le t\le1\), \(x=t\) and \(y=0\). For \(1<t\le1+\frac{\pi}{2}\), let \(u=t-1\) and define \(x=1+\sin(u)\), \(y=1-\cos(u)\). Find the exact total arc length.
Figure for problem 545421

Hints

- Treat the two stages separately and confirm that they meet at the same point. - Compute the speed on each parameter interval. - Add the two positive lengths after evaluating them.

Solution

1. The first stage is a horizontal segment with speed \(1\), so its length is \(1\). 2. In the second stage, the component derivatives with respect to \(t\) are \(\cos(u)\) and \(\sin(u)\), so the speed is \(1\). 3. The second stage lasts for a parameter interval of length \(\frac{\pi}{2}\), so its length is \(\frac{\pi}{2}\). 4. The total length is \(1+\frac{\pi}{2}\).

Answer

\(1+\frac{\pi}{2}\)
54542312
The graph shows a parametric curve’s recorded speed at five equally spaced times. Use Simpson’s rule with four equal subintervals to estimate the arc length from \(t=0\) to \(t=4\).
Figure for problem 545423

Hints

- Read the five speed values from the plotted points at \(t=0,1,2,3,4\). - Confirm that four equal subintervals permit Simpson’s rule. - Apply the endpoint, four, and two weights to approximate the integral of speed.

Solution

1. The subinterval width is \(h=1\). 2. Simpson’s rule gives \(L\approx\frac{1}{3}[1+4(2)+2(4)+4(3)+5]\). 3. Therefore, \(L\approx\frac{34}{3}\).

Answer

\(L\approx\frac{34}{3}\approx11.333\)
54542612
A curve is defined by \(x(t)=\int_0^t(1-s^2)\,ds\) and \(y(t)=\int_0^t2s\,ds\). Find its exact arc length for \(0\le t\le2\).

Hints

- Differentiate each accumulated coordinate using the upper limit. - Expand or recognize a perfect square inside the speed. - Integrate the simplified nonnegative expression.

Solution

1. The component derivatives are \(1-t^2\) and \(2t\). 2. The speed is \(\sqrt{(1-t^2)^2+4t^2}=1+t^2\). 3. Therefore, \(L=\int_0^2(1+t^2)\,dt\). 4. The exact length is \(2+\frac{8}{3}=\frac{14}{3}\).

Answer

\(\frac{14}{3}\)
54542912
A curve \(\mathbf{r}(t)\) has arc length \(7\) for \(0\le t\le2\). A new curve is \(\mathbf{R}(t)=3\mathbf{r}(2-t)+\langle4,-1\rangle\) on the same parameter interval. Find the arc length of \(\mathbf{R}\).

Hints

- Consider separately the effects of reversing the parameter, scaling, and translating. - Only one of these transformations changes the magnitude of the tangent vector. - Apply the length-changing factor to the original arc length.

Solution

1. Replacing \(t\) by \(2-t\) reverses the tracing direction but does not change the path length. 2. Multiplying every coordinate by \(3\) multiplies every speed and every arc length by \(3\). 3. The translation \(\langle4,-1\rangle\) does not affect derivatives or length. 4. Therefore, the new arc length is \(3(7)=21\).

Answer

\(21\)
54543012
The curve \(x=\ln(t)\), \(y=2\ln(t)\) is traced from \(t=1\) to \(t=a\), where \(a>1\). Its arc length is \(2\sqrt{5}\). Find \(a\).

Hints

- Differentiate both logarithmic coordinates and simplify the speed. - Keep the upper endpoint symbolic when evaluating the length integral. - Solve the resulting logarithmic equation using the condition \(a>1\).

Solution

1. The component derivatives are \(\frac{1}{t}\) and \(\frac{2}{t}\), so the speed is \(\frac{\sqrt{5}}{t}\). 2. The length is \(\sqrt{5}\int_1^a\frac{1}{t}\,dt=\sqrt{5}\ln(a)\). 3. Setting this equal to \(2\sqrt{5}\) gives \(\ln(a)=2\). 4. Therefore, \(a=e^2\).

Answer

\(a=e^2\)
54543112
A flexible trim piece must follow the path \(x=t\), \(y=t^2\) for \(0\le t\le1\), where coordinates are measured in meters. Installation requires an additional \(0.01\,\text{m}\) of trim. Is a \(1.50\,\text{m}\) piece long enough? Support your answer with an arc-length calculation.

Hints

- Find the length of the curved path before adding the installation allowance. - Evaluate the arc-length integral with enough precision for the comparison. - Compare the total requirement, not just the curve length, with the available piece.

Solution

1. The path length is \(L=\int_0^1\sqrt{1+4t^2}\,dt\). 2. Numerical evaluation gives \(L\approx1.478943\,\text{m}\). 3. Including the allowance requires approximately \(1.488943\,\text{m}\). 4. Since \(1.488943<1.50\), the piece is long enough by approximately \(0.011057\,\text{m}\).

Answer

Yes. The required length is approximately \(1.488943\,\text{m}\), which is about \(0.011057\,\text{m}\) less than \(1.50\,\text{m}\).
54543212
Find the exact arc length of \(x=t\), \(y=-\ln(\cos(t))\) for \(0\le t\le\frac{\pi}{3}\).

Hints

- Differentiate the logarithmic coordinate carefully. - Use a trigonometric identity to simplify the speed. - Evaluate the standard secant antiderivative at exact angle values.

Solution

1. The component derivatives are \(1\) and \(\tan(t)\). 2. The speed is \(\sqrt{1+\tan^2(t)}=\sec(t)\) on the stated interval. 3. Therefore, \(L=\int_0^{\pi/3}\sec(t)\,dt\). 4. The exact length is \(\ln(\sec(\frac{\pi}{3})+\tan(\frac{\pi}{3}))=\ln(2+\sqrt{3})\).

Answer

\(\ln(2+\sqrt{3})\)
54543312
The figure shows two paths from \((0,0)\) to \((2,0)\). Path A is \(x=2t\), \(y=0\), and Path B is \(x=1-\cos(\pi t)\), \(y=\sin(\pi t)\), each for \(0\le t\le1\). Find both exact lengths and determine how much longer Path B is.
Figure for problem 545433

Hints

- Compute each path’s speed independently. - Confirm that both parameterizations have the same endpoints. - Compare total path lengths rather than endpoint displacement.

Solution

1. Path A has constant speed \(2\), so its length is \(2\). 2. Path B has speed \(\pi\sqrt{\sin^2(\pi t)+\cos^2(\pi t)}=\pi\), so its length is \(\pi\). 3. Path B is longer by \(\pi-2\).

Answer

Path A: \(2\) Path B: \(\pi\) Difference: \(\pi-2\)
54543412
A parametric curve has component derivatives \(\frac{dx}{dt}=3t^2-3\) and \(\frac{dy}{dt}=6t\). Find the exact arc length from \(t=-1\) to \(t=2\).

Hints

- The coordinate functions themselves are not needed when their derivatives are given. - Look for a perfect square after combining the squared component rates. - Integrate the nonnegative speed over the full interval.

Solution

1. The speed is \(\sqrt{(3t^2-3)^2+(6t)^2}\). 2. Factoring and simplifying gives speed \(3(t^2+1)\). 3. Therefore, \(L=\int_{-1}^{2}3(t^2+1)\,dt\). 4. The exact length is \(18\).

Answer

\(18\)
54543512
Find the exact arc length of \(x=t+\sin(t)\), \(y=1-\cos(t)\) for \(0\le t\le\pi\).

Hints

- Combine the squared component derivatives before simplifying. - Use a half-angle identity and check the sign on the interval. - Integrate the resulting speed exactly.

Solution

1. The component derivatives are \(1+\cos(t)\) and \(\sin(t)\). 2. The squared speed is \(2+2\cos(t)=4\cos^2\left(\frac{t}{2}\right)\). 3. On the stated interval, the speed is \(2\cos\left(\frac{t}{2}\right)\). 4. Therefore, \(L=\int_0^\pi2\cos\left(\frac{t}{2}\right)\,dt=4\).

Answer

\(4\)
54543612
Curve A is \(x=\cos(t)\), \(y=\sin(t)\) for \(0\le t\le2\pi\). Curve B is \(x=\cos(2t)\), \(y=\sin(2t)\) for \(0\le t\le2\pi\). Both trace the same unit circle. Find each total arc length and explain the difference.

Hints

- Compute the speed for each parameterization rather than relying only on the geometric curve. - Compare how much the angular input changes over the interval. - Arc length counts repeated tracing.

Solution

1. Curve A has speed \(1\), so its total length is \(2\pi\). 2. Curve B has speed \(2\), so its total length is \(4\pi\). 3. Curve A traces the circle once, while Curve B traces it twice on the same parameter interval.

Answer

Curve A: \(2\pi\) Curve B: \(4\pi\) Curve B is longer because it traces the circle twice.
54543712
A closed path consists of a semicircular arc of radius \(r\) together with its diameter. The arc is parameterized by \(x=r\cos(t)\), \(y=r\sin(t)\) for \(0\le t\le\pi\). If the total path length is \(10\), find \(r\).

Hints

- Separate the curved part from the straight part of the path. - Use the parameterization to identify the semicircle’s arc length. - Add the two lengths before solving for the radius.

Solution

1. The semicircular arc has length \(\pi r\). 2. The diameter segment has length \(2r\). 3. The total-length condition is \(\pi r+2r=10\). 4. Therefore, \(r=\frac{10}{\pi+2}\).

Answer

\(r=\frac{10}{\pi+2}\)
54543912
Find the exact arc length of \(x=3t-t^3\), \(y=3t^2\) for \(-1\le t\le1\).

Hints

- Factor the common constant from the squared component rates. - Look for a perfect square in the remaining polynomial. - Use symmetry or evaluate the simplified integral directly.

Solution

1. The component derivatives are \(3-3t^2\) and \(6t\). 2. The speed simplifies to \(3(1+t^2)\). 3. Therefore, \(L=\int_{-1}^{1}3(1+t^2)\,dt\). 4. The exact length is \(8\).

Answer

\(8\)
54560212
The figure shows the curve \(x=3\cos(t)\), \(y=3\sin(t)\) traced from \(t=0\) to \(t=\frac{\pi}{3}\), together with the chord joining its endpoints. Find the exact arc length, the straight-line endpoint distance, and the amount by which the arc is longer.
Figure for problem 545602

Hints

- Use the parameterization to find the distance traveled along the circle. - The endpoint distance is the chord, not the arc. - Subtract the two exact lengths only after finding each separately.

Solution

1. The speed is \(3\), so the arc length is \(3\left(\frac{\pi}{3}\right)=\pi\). 2. The central angle is \(\frac{\pi}{3}\), so the chord length is \(2(3)\sin\left(\frac{\pi}{6}\right)=3\). 3. The arc is longer by \(\pi-3\).

Answer

Arc length: \(\pi\) Endpoint distance: \(3\) Difference: \(\pi-3\)
54560312
The figure shows the full astroid whose first-quadrant arc is parameterized by \(x=4\cos^3(t)\), \(y=4\sin^3(t)\) for \(0\le t\le\frac{\pi}{2}\). Find the exact length of this quarter and the total length of the full astroid.
Figure for problem 545603

Hints

- Differentiate the cubed trigonometric coordinates carefully. - Use the first-quadrant signs when simplifying the speed. - Apply the curve’s symmetry only after finding one quarter’s length.

Solution

1. On the first-quadrant interval, the speed simplifies to \(12\sin(t)\cos(t)\). 2. The quarter length is \(\int_0^{\pi/2}12\sin(t)\cos(t)\,dt=6\). 3. The astroid has four congruent quarters, so its total length is \(4(6)=24\).

Answer

Quarter length: \(6\) Total length: \(24\)
53920612
The curve \(x=t^{2}\), \(y=\frac{2t^{3}}{3}\) is traced for \(0\le t\le a\), where \(a>0\). The arc length is \(\frac{20\sqrt{10}-2}{3}\). Find \(a\).

Hints

- Derive the exact arc-length expression with upper bound \(a\). - Set that increasing length expression equal to the given exact value. - Use \(a>0\) to choose the valid solution after simplifying radicals.

Solution

1. Because \(t\ge0\), the arc-length integrand is \(2t\sqrt{t^{2}+1}\). 2. The length through \(a\) is \(L(a)=\int_{0}^{a}2t\sqrt{t^{2}+1}\,dt=\frac{2}{3}\left((a^{2}+1)^{3/2}-1\right)\). 3. Setting this equal to \(\frac{20\sqrt{10}-2}{3}\) gives \((a^{2}+1)^{3/2}=10\sqrt{10}\), so \(a^{2}+1=10\). 4. Since \(a>0\), \(a=3\).

Answer

\(a=3\)
54541212
A path is parameterized by \(x=(t-1)^2\), \(y=2(t-1)^2\) for \(0\le t\le3\). The path reverses direction at \(t=1\). Find the exact total arc length traveled.

Hints

- Find the speed rather than the signed component rates. - Notice where the common derivative factor changes sign. - Split the interval at the reversal so distance remains nonnegative.

Solution

1. The component derivatives are \(\frac{dx}{dt}=2(t-1)\) and \(\frac{dy}{dt}=4(t-1)\). 2. The speed is \(2\sqrt{5}|t-1|\). 3. Split at the reversal: \(L=\int_0^1 2\sqrt{5}(1-t)\,dt+\int_1^3 2\sqrt{5}(t-1)\,dt\). 4. The two lengths are \(\sqrt{5}\) and \(4\sqrt{5}\), so the total is \(5\sqrt{5}\).

Answer

\(5\sqrt{5}\)
54541312
The figure shows the cusp-shaped curve \(x=t^3\), \(y=t^2\) for \(-1\le t\le1\). Find its exact arc length.
Figure for problem 545413

Hints

- Simplify the speed carefully near the cusp, where a factor changes sign. - Use symmetry to avoid integrating the absolute value across the whole interval. - A substitution turns the remaining radical integral into a power integral.

Solution

1. The component derivatives are \(3t^2\) and \(2t\), so the speed is \(|t|\sqrt{9t^2+4}\). 2. By symmetry, \(L=2\int_0^1 t\sqrt{9t^2+4}\,dt\). 3. Using \(u=9t^2+4\), the integral evaluates to \(\frac{13\sqrt{13}-8}{27}\). 4. Therefore, \(L=\frac{2(13\sqrt{13}-8)}{27}\).

Answer

\(\frac{2(13\sqrt{13}-8)}{27}\)
54541512
A cycloid is parameterized by \(x=t-\sin(t)\), \(y=1-\cos(t)\) for \(0\le t\le2\pi\). Find the parameter value \(a\) such that the arc length from \(t=0\) to \(t=a\) is \(2\).

Hints

- Simplify the speed using a half-angle identity. - Integrate only to an unknown upper endpoint. - Use the stated parameter interval to select the correct trigonometric solution.

Solution

1. On the interval, the speed simplifies to \(2\sin\left(\frac{t}{2}\right)\). 2. The length from \(0\) to \(a\) is \(4\left(1-\cos\left(\frac{a}{2}\right)\right)\). 3. Setting this equal to \(2\) gives \(\cos\left(\frac{a}{2}\right)=\frac{1}{2}\). 4. Since \(0\le a\le2\pi\), \(a=\frac{2\pi}{3}\).

Answer

\(a=\frac{2\pi}{3}\)
54541612
Find the exact arc length of \(x=\ln(\sec(t)+\tan(t))\), \(y=\sec(t)\) for \(0\le t\le\frac{\pi}{4}\).

Hints

- Differentiate the logarithmic coordinate using its standard simplification. - Factor the common trigonometric term inside the speed. - Use an identity to turn the radical into a single familiar function.

Solution

1. The component derivatives are \(\frac{dx}{dt}=\sec(t)\) and \(\frac{dy}{dt}=\sec(t)\tan(t)\). 2. The speed is \(\sqrt{\sec^2(t)+\sec^2(t)\tan^2(t)}=\sec^2(t)\). 3. Therefore, \(L=\int_0^{\pi/4}\sec^2(t)\,dt\). 4. The exact length is \(\tan\left(\frac{\pi}{4}\right)-\tan(0)=1\).

Answer

\(1\)
54541712
Find the exact arc length of \(x=\frac{t^2}{2}\), \(y=\frac{(t^2-1)^{3/2}}{3}\) for \(1\le t\le2\).

Hints

- Differentiate the radical power using the chain rule. - Factor the common power of the parameter inside the speed. - Use the interval to simplify the remaining square root correctly.

Solution

1. The component derivatives are \(\frac{dx}{dt}=t\) and \(\frac{dy}{dt}=t\sqrt{t^2-1}\). 2. The speed is \(\sqrt{t^2+t^2(t^2-1)}=t^2\) on the stated interval. 3. Therefore, \(L=\int_1^2 t^2\,dt\). 4. The exact length is \(\frac{8-1}{3}=\frac{7}{3}\).

Answer

\(\frac{7}{3}\)
54542212
The positive-parameter branch of a curve is \(x=t^2+1\), \(y=t^3\). Find the exact arc length of the portion whose x-coordinate runs from \(2\) to \(5\).

Hints

- Convert the coordinate bounds into parameter bounds using the branch restriction. - Simplify the speed before integrating. - A substitution based on the radical’s inner expression evaluates the integral.

Solution

1. On the positive branch, \(x=2\) gives \(t=1\), and \(x=5\) gives \(t=2\). 2. The speed is \(\sqrt{(2t)^2+(3t^2)^2}=t\sqrt{4+9t^2}\). 3. Thus, \(L=\int_1^2 t\sqrt{4+9t^2}\,dt\). 4. The exact length is \(\frac{80\sqrt{10}-13\sqrt{13}}{27}\).

Answer

\(\frac{80\sqrt{10}-13\sqrt{13}}{27}\)
54542412
Find the exact arc length of the logarithmic spiral parameterization \(x=e^{2t}\cos(t)\), \(y=e^{2t}\sin(t)\) for \(0\le t\le\ln(2)\).

Hints

- Product-rule terms combine cleanly when the squared component rates are added. - Factor the common exponential term from the speed. - Evaluate the exponential endpoint exactly before simplifying.

Solution

1. Differentiating both components and simplifying gives speed \(\sqrt{5}e^{2t}\). 2. Therefore, \(L=\int_0^{\ln(2)}\sqrt{5}e^{2t}\,dt\). 3. The integral is \(\frac{\sqrt{5}}{2}[e^{2t}]_0^{\ln(2)}\). 4. Since \(e^{2\ln(2)}=4\), the exact length is \(\frac{3\sqrt{5}}{2}\).

Answer

\(\frac{3\sqrt{5}}{2}\)
54542512
Find the exact arc length of \(x=\frac{t^3}{3}\), \(y=\frac{(t^2+1)^{3/2}}{3}\) for \(0\le t\le1\).

Hints

- Differentiate the radical power before forming the speed. - Factor the common parameter term inside the radical. - Use the inner quadratic expression as a substitution.

Solution

1. The component derivatives are \(t^2\) and \(t\sqrt{t^2+1}\). 2. The speed simplifies to \(t\sqrt{2t^2+1}\) on the stated interval. 3. Therefore, \(L=\int_0^1 t\sqrt{2t^2+1}\,dt\). 4. With \(u=2t^2+1\), the exact length is \(\frac{3\sqrt{3}-1}{6}\).

Answer

\(\frac{3\sqrt{3}-1}{6}\)
54542712
A curve’s speed is recorded at equally spaced times. Simpson’s rule gives an estimated arc length of \(15\) from \(t=0\) to \(t=4\). Find the missing speed \(v\). <table><tr><th>\(t\)</th><th>Speed</th></tr><tr><td>\(0\)</td><td>\(2\)</td></tr><tr><td>\(1\)</td><td>\(v\)</td></tr><tr><td>\(2\)</td><td>\(5\)</td></tr><tr><td>\(3\)</td><td>\(4\)</td></tr><tr><td>\(4\)</td><td>\(3\)</td></tr></table>

Hints

- Place the unknown speed in its correct Simpson’s-rule weight. - Set the resulting weighted sum equal to the given length estimate. - Solve the resulting linear equation.

Solution

1. With \(h=1\), Simpson’s estimate is \(\frac{1}{3}[2+4v+2(5)+4(4)+3]\). 2. Setting this equal to \(15\) gives \(31+4v=45\). 3. Therefore, \(v=\frac{7}{2}\).

Answer

\(v=\frac{7}{2}\)
54542812
Find the exact arc length of the rational parameterization \(x=\frac{1-t^2}{1+t^2}\), \(y=\frac{2t}{1+t^2}\) for \(0\le t\le1\).

Hints

- Combine the two rational derivatives before simplifying the radical. - Look for a common denominator and a perfect-square numerator. - Recognize the standard antiderivative of the simplified speed.

Solution

1. Differentiating and combining the squared component rates gives speed \(\frac{2}{1+t^2}\). 2. Therefore, \(L=\int_0^1\frac{2}{1+t^2}\,dt\). 3. The exact length is \(2[\arctan(t)]_0^1=\frac{\pi}{2}\).

Answer

\(\frac{\pi}{2}\)
54560412
The figure shows the closed curve \(x=2\cos(t)-\cos(2t)\), \(y=2\sin(t)-\sin(2t)\) for \(0\le t\le2\pi\). Find its exact length.
Figure for problem 545604

Hints

- Simplify the squared speed before taking its square root. - Use a half-angle identity and check the sign on the full interval. - Integrate the resulting nonnegative speed over one complete tracing.

Solution

1. The squared speed simplifies to \(8-8\cos(t)=16\sin^2\left(\frac{t}{2}\right)\). 2. On \([0,2\pi]\), \(\sin\left(\frac{t}{2}\right)\ge0\), so the speed is \(4\sin\left(\frac{t}{2}\right)\). 3. Therefore, \(L=\int_0^{2\pi}4\sin\left(\frac{t}{2}\right)\,dt=16\).

Answer

\(16\)
54560512
The figure shows the geometric path of \(x=\cos^2(t)\), \(y=\sin^2(t)\). For \(0\le t\le\pi\), find the total distance traveled and compare it with the geometric length of the line segment \(x+y=1\) in the first quadrant.
Figure for problem 545605

Hints

- Simplify the speed using double-angle expressions. - Track the sign changes over the full interval. - Compare the tracing behavior with the distance between the segment’s endpoints.

Solution

1. The speed is \(\sqrt{2}|\sin(2t)|\). 2. The total distance is \(\sqrt{2}\int_0^\pi|\sin(2t)|\,dt=2\sqrt{2}\). 3. The segment from \((1,0)\) to \((0,1)\) has length \(\sqrt{2}\). 4. The parameterization traces the segment twice, so the total distance is twice its geometric length.

Answer

Total distance: \(2\sqrt{2}\) Segment length: \(\sqrt{2}\)

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