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Arc length of parametric curves

Learning goals

  • Finds the coordinate derivatives needed for a parametric curve's arc length.
  • Sets up an arc-length integral for a parametric curve.
  • Uses the curve's parameter interval as arc-length integration bounds.
  • Evaluates definite arc-length integrals.

Click problems to add them to your worksheet.

Arc length of parametric curves 1
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

\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\)
Arc length of parametric curves 2
A runner follows a quarter-circle track from A to B. Which labeled route, c or d, shows the run? Is that route longer or shorter than the other route?
Figure for problem 555998

Hints

- Trace each labeled route between A and B. - Which route matches the runner's track? - Compare the track with the straight connection between its endpoints.

Solution

1. Route c follows the quarter-circle track between A and B. 2. Route d goes straight between the endpoints. 3. A quarter-circle arc is longer than its chord, so route c is longer.

Answer

Route c; it is longer than route d.
Arc length of parametric curves 3
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: \(25\)
Arc length of parametric curves 4
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: \(t\sqrt{5}\) Arc-length integral: \(\int_{1}^{4}t\sqrt{5}\,dt\) Length: \(\frac{15\sqrt{5}}{2}\)
Arc length of parametric curves 5
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

\(\frac{14}{3}\)
Arc length of parametric curves 6
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

\(12\)
Arc length of parametric curves 7
For \(x=3\cos t\), \(y=3\sin t\) on \(0\le t\le\frac{\pi}{2}\), 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: \(\frac{3\pi}{2}\)
Arc length of parametric curves 8
For \(x=5\cos t\), \(y=5\sin t\) on \(\frac{\pi}{6}\le t\le\frac{5\pi}{6}\), 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: \(\frac{10\pi}{3}\)
Arc length of parametric curves 9
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

\(\int_{0}^{\frac{\pi}{2}}2\sqrt{3\sin^{2}(t)+1}\,dt\)
Arc length of parametric curves 10
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

\(\int_{0}^{\pi}3\sqrt{3\sin^{2}(t)+1}\,dt\)
Arc length of parametric curves 11
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

\(\int_{0}^{1}\sqrt{(2t+\cos(t))^{2}+(3t^{2}+\sin(t))^{2}}\,dt\)
Arc length of parametric curves 12
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

\(\int_{0}^{\pi}\sqrt{(1-\sin(t))^{2}+(\cos(t)-1)^{2}}\,dt\)
Arc length of parametric curves 13
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

\(\int_{0}^{2}\sqrt{2}e^{t}\,dt\)
Arc length of parametric curves 14
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

\(\int_{1}^{2}\frac{\sqrt{16t^{8}+1}}{t}\,dt\)
Arc length of parametric curves 15
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

\(16\)
Arc length of parametric curves 16
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

\(8\)
Arc length of parametric curves 17
Use the midpoint rule to estimate the arc length from \(t=0\) to \(t=4\) using the four measured speeds.
Figure for problem 545438

Hints

- Each marked time is halfway through a subinterval. - What width does each midpoint speed represent? - Add the speeds and multiply by their common interval width.

Solution

1. The four subintervals each have width \(1\). 2. The speed values at their midpoints are \(2\), \(3\), \(5\), and \(4\). 3. The midpoint estimate is \(L\approx1(2+3+5+4)=14\).

Answer

\(L\approx14\)
Arc length of parametric curves 18
A parametric path has constant rates \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=-2\) for \(0\le t\le3\). Find the path length.

Hints

- Find speed from the two coordinate rates. - Constant speed makes the length calculation simple over the parameter interval.

Solution

1. The speed is \(\sqrt{1^2+(-2)^2}=\sqrt{5}\). 2. The length is \(\int_{0}^{3}\sqrt{5}\,dt=3\sqrt{5}\).

Answer

\(3\sqrt{5}\) units.
Arc length of parametric curves 19
The marked arc is the upper half of a circle. Find its length and compare it with the straight distance between its endpoints.
Figure for problem 562872

Hints

- A half-circle has half the circumference of a complete circle. - The shortest path between the endpoints is the diameter.

Solution

1. The half-circumference is \(\frac{1}{2}\cdot2\pi\cdot2=2\pi\). 2. The endpoints are \(4\) units apart, so the curved path is longer than the straight segment.

Answer

Arc length \(2\pi\) units; endpoint distance \(4\) units, so the arc is longer.
Arc length of parametric curves 20
A curve has speed \(v(t)=2t\) for \(0\le t\le2\). Find its length on this interval.

Hints

- Speed is already supplied, so no component derivatives are needed. - Integrate that nonnegative rate across the full interval.

Solution

1. Arc length is the integral of speed over the given interval. 2. \(L=\int_{0}^{2}2t\,dt=[t^2]_{0}^{2}=4\).

Answer

\(4\) units.
Arc length of parametric curves 21
The path is \(x=t^2\), \(y=0\) for \(-1\le t\le1\). Find its total length even though it starts and ends at the same point.

Hints

- The path reaches the origin and then retraces its steps. - Why must the speed use an absolute value here?

Solution

1. The speed is \(\sqrt{(2t)^2}=2|t|\). 2. Split at zero: \(L=\int_{-1}^{0}-2t\,dt+\int_{0}^{1}2t\,dt=1+1=2\).

Answer

\(2\) units.
Arc length of parametric curves 22
A moving point has velocity components at three times. Estimate the total distance traveled over \(0\le t\le2\) using the trapezoidal rule with step size \(1\). <table><tr><th>\(t\) (s)</th><th>\(v_x\) (m/s)</th><th>\(v_y\) (m/s)</th></tr><tr><td>\(0\)</td><td>\(3\)</td><td>\(4\)</td></tr><tr><td>\(1\)</td><td>\(0\)</td><td>\(3\)</td></tr><tr><td>\(2\)</td><td>\(-3\)</td><td>\(4\)</td></tr></table>

Hints

- What single number measures the speed of a point moving in two directions at once? - Find a speed at each listed time before applying the numerical rule. - The spacing between the measurements is one second.

Solution

1. Distance accumulates the magnitude of velocity, not the signed sum of its components. 2. The three speeds are \(\sqrt{3^2+4^2}=5\), \(\sqrt{0^2+3^2}=3\), and \(\sqrt{(-3)^2+4^2}=5\) meters per second. 3. The trapezoidal estimate is \(L\approx\frac{1}{2}(5+2\cdot3+5)=8\,\text{m}\).

Answer

\(L\approx8\,\text{m}\).
Arc length of parametric curves 23
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

\(\frac{2}{3}\left(5\sqrt{5}-1\right)\)
Arc length of parametric curves 24
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

\(5\sqrt{5}-1\)
Arc length of parametric curves 25
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

\(\frac{10\sqrt{10}-1}{3}\)
Arc length of parametric curves 26
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

\(L\approx4.105\)
Arc length of parametric curves 27
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

\(L\approx4.254\)
Arc length of parametric curves 28
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

\(L\approx2.0097\)
Arc length of parametric curves 29
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

\(L\approx8.000\)
Arc length of parametric curves 30
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

\(L\approx8.109\)
Arc length of parametric curves 31
A curve is given by \(x=kt\) and \(y=2kt\) for \(0\le t\le3\), where \(k>0\). Its arc length is \(15\sqrt{5}\). 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: \(k\sqrt{5}\) Arc-length equation: \(\int_0^3 k\sqrt{5}\,dt=15\sqrt{5}\) \(k=5\)
Arc length of parametric curves 32
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: \(\int_0^a5\,dt=50\) \(a=10\)
Arc length of parametric curves 33
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 \(4\pi\). 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: \(\int_0^{\frac{\pi}{3}}a\,dt=4\pi\) \(a=12\)
Arc length of parametric curves 34
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.
Arc length of parametric curves 35
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.
Arc length of parametric curves 36
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 \(\sqrt{13}\).
Arc length of parametric curves 37
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 \(\pi\).
Arc length of parametric curves 38
For \(x=t^2\), \(y=\frac{t^3}{3}-t\) on \(0\le t\le2\), Tomáš uses \(\int_0^2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)\,\text{d}t\). 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 \(\int_0^2(t^2+1)\,\text{d}t=\frac{14}{3}\).

Answer

Tomáš added the component rates instead of finding their vector magnitude. The correct exact length is \(\frac{14}{3}\).
Arc length of parametric curves 39
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) \(L\approx5.916\)
Arc length of parametric curves 40
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) \(L\approx2.343\)
Arc length of parametric curves 41
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: \(4\pi\)
Arc length of parametric curves 42
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

\(3\)
Arc length of parametric curves 43
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

\(\frac{\pi^2}{2}\)
Arc length of parametric curves 44
The path \(x=\sin t\), \(y=0\) is traced for \(0\le t\le2\pi\). Kojo integrates \(\cos t\) and concludes that the arc length is \(0\). Explain the error and find the correct total length.

Hints

- Compare what the proposed 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 \(\int_0^{2\pi}|\cos t|\,\text{d}t=4\).

Answer

Kojo used signed horizontal change instead of speed. The correct total length is \(4\).
Arc length of parametric curves 45
A path has two stages. For \(0\le t\le1\), \(x=t\) and \(y=0\). For \(1<t\le1+\frac{\pi}{2}\), let \(u=t-1\) and use \(x=1+\sin u\), \(y=1-\cos u\). Find the speed and arc-length integral for each stage and the exact total length.
Figure for problem 545421

Hints

- Find the speed separately on each part of the path. - The second stage uses \(u=t-1\); how does this affect differentiation with respect to \(t\)? - Add the two positive lengths after integrating over the correct intervals.

Solution

1. On the first stage, the coordinate rates are \(1\) and \(0\), so the speed is \(1\) and \(L_{1}=\int_{0}^{1}1\,dt=1\). 2. On the second stage, \(\frac{dx}{dt}=\cos u\) and \(\frac{dy}{dt}=\sin u\); thus the speed is also \(1\). 3. The second length is \(L_{2}=\int_{1}^{1+\pi/2}1\,dt=\frac{\pi}{2}\). 4. Adding gives \(L=1+\frac{\pi}{2}\).

Answer

Stage 1: speed \(1\), \(L_{1}=\int_{0}^{1}1\,dt=1\). Stage 2: speed \(1\), \(L_{2}=\int_{1}^{1+\pi/2}1\,dt=\frac{\pi}{2}\). Total length: \(1+\frac{\pi}{2}\).
Arc length of parametric curves 46
Use Simpson's rule with four equal subintervals to estimate the arc length from \(t=0\) to \(t=4\) using the recorded speeds.
Figure for problem 545423

Hints

- Read the speed at each marked time. - Check the common width of the four subintervals. - Give the first and last speeds the endpoint weights in Simpson's rule.

Solution

1. The subinterval width is \(h=1\). 2. The graph gives speeds \(1\), \(2\), \(4\), \(3\), and \(5\) at consecutive integer times. 3. Simpson's rule gives \(L\approx\frac{1}{3}[1+4(2)+2(4)+4(3)+5]=\frac{34}{3}\).

Answer

\(L\approx\frac{34}{3}\)
Arc length of parametric curves 47
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

\(\frac{14}{3}\)
Arc length of parametric curves 48
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

\(21\)
Arc length of parametric curves 49
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

- Find the two coordinate rates, then form their combined speed. - Does the permitted parameter interval determine the sign of the half-angle sine? - Integrate the speed up to the unknown endpoint and use the given length.

Solution

1. Differentiate: \(x'=1-\cos(t)\) and \(y'=\sin(t)\). 2. The speed is \(\sqrt{(1-\cos(t))^2+\sin^2(t)}=2\sin\left(\frac{t}{2}\right)\), because \(0\le t<2\pi\). 3. The arc length is \(L=\int_{0}^{a}2\sin\left(\frac{t}{2}\right)\,dt=4\left(1-\cos\left(\frac{a}{2}\right)\right)\). 4. Set \(L=2\). Then \(\cos\left(\frac{a}{2}\right)=\frac{1}{2}\), and \(0<a<2\pi\) gives \(a=\frac{2\pi}{3}\).

Answer

\(a=\frac{2\pi}{3}\)
Arc length of parametric curves 50
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 \(3.000\,\text{m}\).
Arc length of parametric curves 51
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, \(L=\int_0^1(1+t^4)\,dt=\frac{6}{5}\).

Answer

\(\frac{6}{5}\)
Arc length of parametric curves 52
Path A is \(x=2t\), \(y=0\), and Path B is \(x=1-\cos(\pi t)\), \(y=\sin(\pi t)\), both for \(0\le t\le1\). Find the speed and arc-length integral for each path. How much longer is B than A?
Figure for problem 545433

Hints

- Find the length from each path's component rates, not from the common endpoints. - One path is straight and the other is curved; compare their speeds over the stated interval. - Subtract the shorter total length from the longer.

Solution

1. For A, the rates are \(2\) and \(0\), so the speed is \(2\) and \(L_{A}=\int_{0}^{1}2\,dt=2\). 2. For B, the rates are \(\pi\sin(\pi t)\) and \(\pi\cos(\pi t)\). Its speed is \(\pi\), so \(L_{B}=\int_{0}^{1}\pi\,dt=\pi\). 3. The difference is \(L_{B}-L_{A}=\pi-2\).

Answer

A: speed \(2\), \(L_{A}=\int_{0}^{1}2\,dt=2\). B: speed \(\pi\), \(L_{B}=\int_{0}^{1}\pi\,dt=\pi\). B is longer by \(\pi-2\).
Arc length of parametric curves 53
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

\(18\)
Arc length of parametric curves 54
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

\(4\)
Arc length of parametric curves 55
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.
Arc length of parametric curves 56
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: \(\pi r+2r=10\) \(r=\frac{10}{\pi+2}\)
Arc length of parametric curves 57
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

\(8\)
Arc length of parametric curves 58
For \(x=3\cos t\), \(y=3\sin t\) on \(0\le t\le\frac{\pi}{3}\), find the speed, arc-length integral, endpoint distance, and how much longer the arc is than the straight chord.
Figure for problem 545602

Hints

- Find the speed from the component derivatives. - The dashed chord connects the endpoints directly, while the arc follows the curved path. - Subtract the straight-line length from the arc length.

Solution

1. The coordinate derivatives are \(-3\sin t\) and \(3\cos t\), so the speed is \(3\). 2. The arc length is \(L=\int_{0}^{\pi/3}3\,dt=\pi\). 3. The endpoints are \((3,0)\) and \(\left(\frac{3}{2},\frac{3\sqrt{3}}{2}\right)\), whose distance is \(3\). 4. The arc is longer by \(\pi-3\).

Answer

Speed: \(3\). Arc-length integral: \(\int_{0}^{\pi/3}3\,dt=\pi\). Chord length: \(3\). Difference: \(\pi-3\).
Arc length of parametric curves 59
The astroid has the parameterization \(x=4\cos^{3}t\), \(y=4\sin^{3}t\). Find the exact length of its first-quadrant arc, \(0\le t\le\frac{\pi}{2}\), and the length of the entire curve.
Figure for problem 545603

Hints

- Differentiate each cubed trigonometric coordinate using the chain rule. - On the first-quadrant interval, both sine and cosine are nonnegative. - How many matching arcs make up the full curve?

Solution

1. Differentiate: \(x'=-12\cos^{2}t\sin t\) and \(y'=12\sin^{2}t\cos t\). 2. On the first-quadrant interval, the speed is \(\sqrt{(x')^{2}+(y')^{2}}=12\sin t\cos t\). 3. The quarter length is \(\int_{0}^{\pi/2}12\sin t\cos t\,dt=6\). 4. Symmetry gives a total length of \(4\cdot6=24\).

Answer

First-quadrant length: \(6\). Total length: \(24\).
Arc length of parametric curves 60
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

\(a=3\)
Arc length of parametric curves 61
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

\(5\sqrt{5}\)
Arc length of parametric curves 62
Find the exact arc length of the curve \(x=t^{3}\), \(y=t^{2}\) for \(-1\le t\le1\).
Figure for problem 545413

Hints

- Find the speed from both coordinate rates. - Be careful with the absolute value when taking a square root across negative parameter values. - Compare the speed at \(t\) with the speed at \(-t\) before integrating.

Solution

1. The component derivatives are \(x'=3t^{2}\) and \(y'=2t\), so the speed is \(|t|\sqrt{9t^{2}+4}\). 2. By symmetry, \(L=2\int_{0}^{1}t\sqrt{9t^{2}+4}\,dt\). 3. With \(u=9t^{2}+4\), the integral from \(0\) to \(1\) equals \(\frac{13\sqrt{13}-8}{27}\). 4. Thus \(L=\frac{2(13\sqrt{13}-8)}{27}\).

Answer

\(\frac{2(13\sqrt{13}-8)}{27}\)
Arc length of parametric curves 63
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

\(1\)
Arc length of parametric curves 64
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

\(\frac{7}{3}\)
Arc length of parametric curves 65
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

\(\frac{80\sqrt{10}-13\sqrt{13}}{27}\)
Arc length of parametric curves 66
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

\(\frac{3\sqrt{5}}{2}\)
Arc length of parametric curves 67
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

\(\frac{3\sqrt{3}-1}{6}\)
Arc length of parametric curves 68
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

\(v=\frac{7}{2}\)
Arc length of parametric curves 69
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 \(\sqrt{(1+t^2)^2(\cos^2(t)+\sin^2(t))}=1+t^2\) on the interval. 3. Therefore, \(L=\int_0^1(1+t^2)\,dt=\frac{4}{3}\).

Answer

\(\frac{4}{3}\)
Arc length of parametric curves 70
Find the exact length of the closed parametric curve \(x=2\cos t-\cos(2t)\), \(y=2\sin t-\sin(2t)\) for \(0\le t\le2\pi\).
Figure for problem 545604

Hints

- Simplify the sum of squares of the component rates before taking a square root. - Use a half-angle identity and consider the sign of the resulting sine on the interval. - Integrate the nonnegative speed over one full tracing.

Solution

1. Differentiate both coordinates and simplify: \((x')^{2}+(y')^{2}=8-8\cos t=16\sin^{2}\left(\frac{t}{2}\right)\). 2. On \(0\le t\le2\pi\), \(\sin\left(\frac{t}{2}\right)\ge0\), so the speed is \(4\sin\left(\frac{t}{2}\right)\). 3. The length is \(L=\int_{0}^{2\pi}4\sin\left(\frac{t}{2}\right)\,dt=16\).

Answer

\(16\)
Arc length of parametric curves 71
The curve \(x=\cos^{2}t\), \(y=\sin^{2}t\) traces part of the line \(x+y=1\). For \(0\le t\le\pi\), find the total distance traveled and compare it with the length of the segment in the first quadrant.
Figure for problem 545605

Hints

- A square root of a squared expression gives an absolute value. Could the expression inside change sign? - Track the curve as \(t\) runs through the full interval, not only the first half. - The segment's geometric length depends only on its endpoints.

Solution

1. Differentiate both components; the speed is \(\sqrt{2}|\sin(2t)|\). 2. The total distance is \(L=\sqrt{2}\int_{0}^{\pi}|\sin(2t)|\,dt=2\sqrt{2}\). 3. The segment joining \((1,0)\) and \((0,1)\) has length \(\sqrt{2}\). 4. The parameterization traces the segment twice, so the travel distance is twice its geometric length.

Answer

Total distance traveled: \(2\sqrt{2}\). Geometric segment length: \(\sqrt{2}\).

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