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Arc length of parametric curves

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55599712
For a parametric curve \(x=x(t)\), \(y=y(t)\) on \(a\le t\le b\), write the integrand used to accumulate arc length.

Hints

- Think of a very short piece of the curve and its horizontal and vertical changes. - What familiar distance relationship combines two perpendicular component changes? - The integrand should represent distance traveled per unit change in the parameter.

Solution

1. An infinitesimal change in position has horizontal component \(dx\) and vertical component \(dy\), so its length is \(ds=\sqrt{dx^2+dy^2}\). 2. Dividing by \(dt\) gives \(\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\). 3. This is the arc-length integrand.

Answer

\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\)
55599812
The figure shows a parametric path from \(A\) to \(B\). The dashed segment joins the same endpoints directly. Which geometric quantity is the arc length: the curved path or the dashed segment? Which one is longer in this figure?
Figure for problem 555998

Hints

- Distinguish distance traveled along a path from the direct change in position between endpoints. - Which visible object follows every point of the motion? - A straight segment gives the shortest distance between two distinct points.

Solution

1. Arc length measures distance along the traced curve, so it is the length of the curved path from \(A\) to \(B\). 2. The dashed segment is the straight-line displacement between the endpoints. 3. Here the curved path is a quarter of a unit circle, with length \(\frac{\pi}{2}\), while the dashed segment has length \(\sqrt{2}\). Since \(\frac{\pi}{2}>\sqrt{2}\), the arc length is longer.

Answer

The arc length is the curved path, and it is longer than the dashed endpoint-to-endpoint segment.
53918212
Using the parametric arc-length formula for \(x=3t\), \(y=4t\) on \(0\le t\le5\), report the speed \(\sqrt{(dx/dt)^2+(dy/dt)^2}\), write the definite arc-length integral, and evaluate it exactly.

Hints

- What quantity describes how fast the point is moving along the plane curve? - Use both component rates when determining that quantity. - Because the component rates are constant here, think about what that means for accumulation over 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

Speed: \(5\) Arc-length integral: \(\int_0^5 5\,dt\) Length: \(25\)
53918412
Using the parametric arc-length formula for \(x=t^2\), \(y=\frac{t^2}{2}\) on \(1\le t\le4\), report the speed, write the definite integral, and evaluate the exact arc length.

Hints

- Compare the two component rates: both have the same parameter factor. - When a square root simplifies to an absolute value, use the stated interval to decide its sign. - Check that your final quantity represents accumulated distance, not displacement between endpoints.

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

Speed: \(t\sqrt5\) Arc-length integral: \(\int_1^4 t\sqrt5\,dt\) Length: \(\frac{15\sqrt5}{2}\)
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

- Write the speed expression before expanding or simplifying it. - Inspect the expression under the square root for an algebraic pattern rather than approximating it. - Once the speed is simplified, verify that it is nonnegative throughout the interval.

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

- Form the speed and inspect the entire expression under the radical before doing numerical work. - Is the polynomial under the square root something that can be written as one square? - Check the sign of the simplified square root over the full parameter 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
For \(x=3\cos t\), \(y=3\sin t\) on \(0\le t\le\frac\pi2\), use the parametric arc-length formula: report the speed, the definite integral, and the exact arc length.

Hints

- Determine whether the point moves around this circle at a constant or variable speed. - Use both coordinate rates to justify that conclusion rather than relying only on the radius. - Relate the accumulated distance to the length of the 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

Speed: \(3\) Arc-length integral: \(\int_0^{\pi/2}3\,dt\) Length: \(\frac{3\pi}{2}\)
53919112
For \(x=5\cos t\), \(y=5\sin t\) on \(\frac\pi6\le t\le\frac{5\pi}{6}\), use the parametric arc-length formula: report the speed, the definite integral, and the exact arc length.

Hints

- Decide whether the magnitude of the derivative vector depends on \(t\). - A trigonometric identity can help you check the magnitude without evaluating particular angles. - Use the actual width of the stated parameter interval when accumulating distance.

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

Speed: \(5\) Arc-length integral: \(\int_{\pi/6}^{5\pi/6}5\,dt\) Length: \(\frac{10\pi}{3}\)
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

- The integral should accumulate the point's speed along the curve, not either component rate by itself. - Express that speed in terms of the parameter and simplify it only as far as useful. - Check that the limits match exactly the portion of the parameter interval named in the problem.

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

- Decide what single nonnegative quantity must be integrated to measure distance along a parametric curve. - Build that quantity from both component rates and simplify without changing its sign. - Leave the final result as a definite integral over the full stated interval.

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

- Identify the instantaneous quantity whose accumulation gives length along a parametric curve. - Build that quantity from both coordinate rates; keep each derivative's sign correct before squaring. - The problem asks only for a model, so stop once the correct definite integral is written.

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 carefully before thinking about the length model. - Length must use a nonnegative measure of the two-dimensional rate of motion. - Keep the original parameter bounds and leave the result unevaluated.

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

- Both coordinates are products, so obtain their component rates before forming the speed. - When the squared rates are added, look for cancellation and a familiar trigonometric identity. - The requested endpoint is the definite-integral setup, not its antiderivative.

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

- Start by finding the two component rates and remember the logarithm's domain. - Combine those rates into one nonnegative speed before simplifying fractions or radicals. - The requested answer is the definite-integral model itself; do not evaluate it.

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

\(\int_{1}^{2}\frac{\sqrt{16t^{8}+1}}{t}\,dt\)
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

\(8\)
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

- Form the speed from the two component rates before deciding how to integrate. - The stated parameter interval matters when simplifying a square root involving a factor of \(t^2\). - After simplifying the speed, look for an inner expression whose derivative is already present.

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

- Build the speed from the two component derivatives and simplify it before integrating. - Use the parameter interval when an absolute value appears during simplification. - Look for a substitution suggested by a repeated quadratic expression in the simplified speed.

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

- Begin with the component rates \(x'(t)\) and \(y'(t)\) and combine them into the speed. - Use the stated interval when simplifying any absolute-value factor that appears. - After simplifying the speed, look for an inner expression whose derivative is proportional to the remaining factor.

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

- Look for common factors in the two component rates before expanding the speed expression. - The sign of the parameter on the given interval matters when simplifying the square root. - Choose an integration strategy only after the speed has been reduced to its simplest useful form.

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 \(x=t+\sin(t)\), \(y=\sin(t)\) on \(0\le t\le\pi\). Round to 3 decimal places.

Hints

- Arc length depends on the magnitude of the two-component velocity, not on either coordinate change alone. - Decide whether the speed simplifies to an elementary antiderivative before choosing numerical evaluation. - Round only the final accumulated length.

Solution

1. The component rates are \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=\cos(t)\). 2. The speed is \(\sqrt{(1+\cos t)^2+\cos^2t}\). 3. Thus \(L=\int_0^\pi\sqrt{(1+\cos t)^2+\cos^2t}\,dt\). 4. Numerical evaluation gives \(L\approx4.105\).

Answer

\(L\approx4.105\)
53919512
Find the arc length of \(x=t^{2}\), \(y=\cos(t)\) on \(0\le t\le2\). Round to 3 decimal places.

Hints

- Differentiate both coordinates carefully, including the sign of the trigonometric derivative. - Use the two rates to determine the instantaneous speed along the path. - Keep guard digits through the numerical integration before rounding the final length.

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

- Determine the point's speed from the exponential and polynomial component rates. - Decide whether exact integration is realistic after simplifying the speed as far as possible. - Delay rounding until the requested four-decimal result is obtained.

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

- Simplify the speed expression before deciding whether numerical integration is necessary. - A trigonometric identity can turn \(2-2\cos(t)\) into a square-related expression; pay attention to sign over the full interval. - Keep the requested three-decimal format even if the simplified integral evaluates exactly.

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 domain of the logarithmic coordinate when simplifying the speed. - Decide how the two component rates combine to measure motion along the curve. - Use numerical integration only after obtaining a correct definite-integral model, and round at the end.

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 \(15\sqrt5\). Derive the parametric speed and an arc-length integral in terms of \(k\), then find \(k\).

Hints

- The unknown scale factor affects both component rates in the same way. - Express the path length in terms of \(k\) before using the given total length. - Use the condition \(k>0\) when interpreting the magnitude of the derivative vector.

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

Speed: \(k\sqrt5\) Arc-length equation: \(\int_0^3 k\sqrt5\,dt=15\sqrt5\) \(k=5\)
53920412
The curve \(x=3t\), \(y=4t\) is traced from \(t=0\) to \(t=a\), where \(a>0\), and has length \(50\). Write the parametric speed and the arc-length equation in \(a\), then solve for \(a\).

Hints

- Determine both component rates before deciding how distance accumulates as the upper parameter bound changes. - Express the accumulated length in terms of \(a\) before using the stated total. - Check that the recovered upper bound satisfies the required positivity condition.

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

Speed: \(5\) Arc-length equation: \(\int_0^a5\,dt=50\) \(a=10\)
53920512
A circular arc is parameterized by \(x=a\cos t\), \(y=a\sin t\) for \(0\le t\le\frac\pi3\), with \(a>0\). Its length is \(4\pi\). Derive the parametric speed and the arc-length equation, then find \(a\).

Hints

- Treat \(a\) as an unknown scale and determine how it affects the speed along the circular path. - Relate that speed to the angular parameter interval before using the given length. - The condition \(a>0\) determines how a magnitude involving \(a\) simplifies.

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

Speed: \(a\) Arc-length equation: \(\int_0^{\frac{\pi}{3}}a\,dt=4\pi\) \(a=12\)
53920712
Two curves are traced over \(0\le t\le2\). Curve A has \(x=3t\), \(y=4t\); Curve B has \(x=5t\), \(y=0\). For each curve, report its parametric speed and arc-length integral. Then give both lengths to three decimal places and compare them.

Hints

- Compare the actual speeds along the two parameterizations rather than comparing their coordinate formulas by appearance. - The two paths have the same parameter interval, so any difference in length must come from their speeds. - Report both computed lengths at the requested precision before making the comparison.

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

Curve A: speed \(5\), \(L_A=\int_0^2 5\,dt=10.000\) Curve B: speed \(5\), \(L_B=\int_0^2 5\,dt=10.000\) The lengths are equal.
53920812
Two curves are traced over \(0\le t\le1\). Curve A has \(x=t+\sin(t)\), \(y=t^2\). Curve B has \(x=2t+\sin(t)\), \(y=\frac{t^2}{2}\). Which curve is longer? Give both lengths to three decimal places.

Hints

- Treat the two paths independently; compare complete accumulated lengths rather than endpoint displacement. - Each path has its own two-component speed function. - Use the same numerical precision for both integrals before comparing.

Solution

1. Curve A has speed \(\sqrt{(1+\cos t)^2+4t^2}\), so \(L_A=\int_0^1\sqrt{(1+\cos t)^2+4t^2}\,dt\approx2.172\). 2. Curve B has speed \(\sqrt{(2+\cos t)^2+t^2}\), so \(L_B=\int_0^1\sqrt{(2+\cos t)^2+t^2}\,dt\approx2.902\). 3. Therefore, Curve B is longer.

Answer

\(L_A\approx2.172\), \(L_B\approx2.902\); 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

- Ask whether adding horizontal and vertical signed rates can represent distance traveled in the plane. - Think about what geometric quantity should replace that signed sum before integrating. - After correcting the integrand, inspect its algebraic form before doing the exact integration.

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 \(\frac{14}{3}\).
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

- Part a) asks for the exact accumulation model for distance along the parametric path. - Use both coordinate rates to determine what is accumulated over the parameter interval. - In part b), evaluate the same integral numerically and postpone rounding until the end.

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

- For part a), derive the speed carefully from the two trigonometric component rates. - Simplify the speed enough to obtain a clear exact integral, but do not replace it with a decimal yet. - Part b) uses that exact setup; keep guard digits before rounding to three decimal places.

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

- Compare what the student's integral measures with what arc length is supposed to measure. - Follow the point along the x-axis over the full interval and note whether its direction ever reverses. - A correct distance accumulation should not let motion in opposite directions cancel.

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 \(1<t\le1+\frac\pi2\), let \(u=t-1\) and define \(x=1+\sin u\), \(y=1-\cos u\). For each stage, report the parametric speed and its arc-length integral, then find the exact total 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. On the first stage, \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=0\), so the speed is \(1\) and \(L_1=\int_0^1 1\,dt=1\). 2. On the second stage, \(u=t-1\) gives \(\frac{dx}{dt}=\cos u\) and \(\frac{dy}{dt}=\sin u\), so the speed is again \(1\). 3. Therefore, \(L_2=\int_1^{1+\pi/2}1\,dt=\frac\pi2\). 4. The total length is \(1+\frac\pi2\).

Answer

Stage 1: speed \(1\), \(L_1=\int_0^1 1\,dt=1\) Stage 2: speed \(1\), \(L_2=\int_1^{1+\pi/2}1\,dt=\frac\pi2\) Total: \(1+\frac\pi2\)
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 \(x=t-\sin(t)\), \(y=1-\cos(t)\) is traced from \(t=0\) to \(t=a\), where \(0<a<2\pi\). Its arc length is \(2\). Find \(a\).

Hints

- Simplify the speed before introducing the unknown upper endpoint into the integral. - The sign of the half-angle expression is controlled by the stated range for \(a\). - Use the interval restriction to select the correct solution of the final trigonometric equation.

Solution

1. The component rates are \(1-\cos(t)\) and \(\sin(t)\). 2. On \(0\le t<2\pi\), the speed is \(\sqrt{(1-\cos t)^2+\sin^2t}=2\sin(\frac t2)\). 3. Thus \(L=\int_0^a2\sin(\frac t2)\,dt=4\left(1-\cos(\frac a2)\right)\). 4. Setting this equal to \(2\) gives \(\cos(\frac a2)=\frac12\). Since \(0<\frac a2<\pi\), \(a=\frac{2\pi}{3}\).

Answer

\(a=\frac{2\pi}{3}\)
54543112
A flexible trim piece must follow the path \(x=t+e^t-1\), \(y=t^2\) for \(0\le t\le1\), where coordinates are measured in meters. Installation requires an additional \(0.01\,\text{m}\) of trim. Is a \(3.00\,\text{m}\) piece long enough? Support your answer with a parametric arc-length calculation, rounding lengths to the nearest \(0.001\,\text{m}\).

Hints

- The installation decision depends on the length of the traced path, not on the distance between its endpoints. - Build the speed from both coordinate rates before using numerical integration. - Add the allowance after finding the curve length, then make the comparison at the requested precision.

Solution

1. The component rates are \(1+e^t\) and \(2t\), so the path length is \(L=\int_0^1\sqrt{(1+e^t)^2+4t^2}\,dt\). 2. Numerical evaluation gives \(L\approx2.923\,\text{m}\) to the nearest \(0.001\,\text{m}\). 3. Including the allowance requires approximately \(2.933\,\text{m}\). 4. Since \(2.933<3.000\), the piece is long enough by approximately \(0.067\,\text{m}\).

Answer

Yes. The required length is approximately \(2.933\,\text{m}\), which is about \(0.067\,\text{m}\) less than \(3.000\,\text{m}\).
54543212
Find the exact arc length of \(x=t-\frac{t^5}{5}\), \(y=\frac{2t^3}{3}\) for \(0\le t\le1\).

Hints

- Differentiate both polynomial coordinates before interpreting the radical. - Compare the expanded expression under the square root with a perfect square. - The interval determines the sign of the simplified speed.

Solution

1. The component rates are \(\frac{dx}{dt}=1-t^4\) and \(\frac{dy}{dt}=2t^2\). 2. The speed is \(\sqrt{(1-t^4)^2+4t^4}=\sqrt{(1+t^4)^2}=1+t^4\) on the interval. 3. Therefore, \(L=\int_0^1(1+t^4)\,dt=\frac65\).

Answer

\(\frac65\)
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 \(0\le t\le1\). Report the speed and arc-length integral for each path, then find both exact lengths and 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. For Path A, \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=0\), so the speed is \(2\) and \(L_A=\int_0^1 2\,dt=2\). 2. For Path B, \(\frac{dx}{dt}=\pi\sin(\pi t)\) and \(\frac{dy}{dt}=\pi\cos(\pi t)\), so the speed is \(\pi\) and \(L_B=\int_0^1\pi\,dt=\pi\). 3. Path B is longer by \(\pi-2\).

Answer

Path A: speed \(2\), \(L_A=\int_0^1 2\,dt=2\) Path B: speed \(\pi\), \(L_B=\int_0^1 \pi\,dt=\pi\) 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 \(10\), report the arc's parametric speed and arc-length integral, then 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. For the semicircular arc, \(\frac{dx}{dt}=-r\sin t\) and \(\frac{dy}{dt}=r\cos t\), so the parametric speed is \(r\). 2. Its arc length is \(\int_0^\pi r\,dt=\pi r\). The diameter contributes \(2r\). 3. The total-length equation is \(\pi r+2r=10\). 4. Therefore, \(r=\frac{10}{\pi+2}\).

Answer

Arc speed: \(r\) Arc-length integral: \(\int_0^\pi r\,dt=\pi r\) Total-length equation: \(\pi r+2r=10\) \(r=\frac{10}{\pi+2}\)
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 \(x=3\cos t\), \(y=3\sin t\) traced from \(t=0\) to \(t=\frac\pi3\), together with the chord joining its endpoints. Report the parametric speed and arc-length integral, then find the exact arc length, 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 component rates are \(\frac{dx}{dt}=-3\sin t\) and \(\frac{dy}{dt}=3\cos t\), so the speed is \(3\). 2. The arc-length integral is \(\int_0^{\frac{\pi}{3}}3\,dt=\pi\). 3. The endpoints are \((3,0)\) and \(\left(\frac32,\frac{3\sqrt3}{2}\right)\), whose distance is \(3\). 4. The arc is longer than the endpoint distance by \(\pi-3\).

Answer

Speed: \(3\) Arc-length integral: \(\int_0^{\frac{\pi}{3}}3\,dt=\pi\) 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

- Think of the accumulated arc length as a function of the unknown endpoint \(a\). - Use the given total length only after obtaining an exact expression for that accumulated length. - Check monotonicity and the condition \(a>0\) when choosing among algebraic candidates.

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 \(x=t^2\sin(t)+2t\cos(t)-\sin(t)\), \(y=-t^2\cos(t)+2t\sin(t)+\cos(t)\) for \(0\le t\le1\).

Hints

- Differentiate the two coordinate expressions before judging the apparent complexity of the curve. - Look for a common factor in the two component rates. - The trigonometric identity affects the speed, not the path's endpoint geometry.

Solution

1. Differentiating gives \(\frac{dx}{dt}=(1+t^2)\cos(t)\) and \(\frac{dy}{dt}=(1+t^2)\sin(t)\). 2. The speed is \(\sqrt{(1+t^2)^2(\cos^2t+\sin^2t)}=1+t^2\) on the interval. 3. Therefore, \(L=\int_0^1(1+t^2)\,dt=\frac43\).

Answer

\(\frac43\)
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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