Free math worksheets

- Convert the time difference into a difference in travel distances. - Think about the largest possible difference between distances to two fixed points. - Use the sensor separation to determine the limiting delay.

1. A time difference of \(0.070\,\text{s}\) would require a difference in travel distances of \(340(0.070)=23.8\,\text{m}\). 2. For any point, the absolute difference of its distances from two fixed sensors cannot exceed the distance between the sensors. Here that maximum distance difference is \(20\,\text{m}\). 3. Because \(23.8>20\), no source location can produce the reported time difference. 4. The greatest possible time difference is \(\frac{20}{340}=\frac{1}{17}\,\text{s}\approx0.0588\,\text{s}\).

No. The reported delay would require a distance difference of \(23.8\,\text{m}\), which exceeds the \(20\,\text{m}\) separation of the sensors. The greatest possible arrival-time difference is \(\frac{1}{17}\,\text{s}\approx0.0588\,\text{s}\).

- Use half the cooker’s width as the horizontal coordinate of a rim point. - Relate that point to the focal parameter in a vertical parabola. - Compare the calculated focus height with the current receiver height.

1. Model the cross section by \(x^2=4py\), where the focus is \((0, p)\). 2. The rim is \(0.90\,\text{m}\) from the axis and \(0.225\,\text{m}\) above the vertex, so \((0.90)^2=4p(0.225)\). 3. Solving gives \(p=0.90\,\text{m}\). 4. The receiver is at height \(0.80\,\text{m}\), which is \(0.10\,\text{m}\) below the focus. It must be raised by \(0.10\,\text{m}\).

The focal distance is \(0.90\,\text{m}\). The receiver must be moved upward by \(0.10\,\text{m}\).

- Convert the full gallery dimensions to semi-axis lengths. - Use the relationship among an ellipse's semi-axes and focal distance. - Recall what quantity stays constant for every point on an ellipse.

1. The semi-major axis is \(a=15\), and the semi-minor axis is \(b=9\). 2. The focal distance satisfies \(c^2=a^2-b^2=225-81=144\), so \(c=12\). 3. The listening points should be placed at the foci \((-12, 0)\) and \((12, 0)\). 4. For every point on an ellipse, the sum of its distances to the two foci is \(2a\), which here is \(30\,\text{ft}\).

Place the listening points at \((-12, 0)\) and \((12, 0)\). The total two-segment path length is always \(30\,\text{ft}\).

- Use the closest branch point to determine the transverse semi-axis. - Convert eccentricity into focal distance, then find the remaining semi-axis. - Interpret asymptote slopes as long-range direction angles.

1. The vertex gives \(a=5\). The eccentricity gives \(c=ea=13\). 2. For a hyperbola, \(b^2=c^2-a^2=169-25=144\), so \(b=12\). 3. The standard equation is \(\frac{x^2}{25}-\frac{y^2}{144}=1\). The specified trajectory is the right branch, so \(x\ge5\). 4. The asymptotes are \(y=\pm\frac{b}{a}x=\pm\frac{12}{5}x\). 5. Their direction angles are \(\theta=\pm\arctan(\frac{12}{5})\approx\pm67.38^\circ\).

Trajectory: \(\frac{x^2}{25}-\frac{y^2}{144}=1\), restricted to the right branch \(x\ge5\) Asymptotic direction angles: \(\pm\arctan(\frac{12}{5})\approx\pm67.38^\circ\)

- Convert the full width and height into ellipse semi-axis lengths. - The cross section is half of a full ellipse. - Treat the hangar as a prism after finding the cross-sectional area.

1. The semiellipse has horizontal semi-axis \(a=30\,\text{ft}\) and vertical semi-axis \(b=24\,\text{ft}\). 2. Its cross-sectional area is half the area of a full ellipse: \(A=\frac{1}{2}\pi ab=\frac{1}{2}\pi(30)(24)=360\pi\,\text{ft}^2\). 3. Multiply by the hangar length: \(V=(360\pi)(150)=54{,}000\pi\,\text{ft}^3\). 4. Numerically, \(V\approx169{,}646\,\text{ft}^3\).

\(54{,}000\pi\,\text{ft}^3\approx169{,}646\,\text{ft}^3\)

- Interpret the stakes as the two foci. - Relate their separation and the fixed distance sum to \(c\) and \(a\). - Use the standard ellipse relationship to find the remaining semi-axis.

1. The stakes are the foci, so \(2c=20\) and \(c=10\). 2. The constant focal-distance sum is \(2a=26\), so \(a=13\). 3. Then \(b^2=a^2-c^2=169-100=69\), so \(b=\sqrt{69}\). 4. The major-axis length is \(2a=26\,\text{ft}\), and the minor-axis length is \(2b=2\sqrt{69}\,\text{ft}\). 5. The equation is \(\frac{x^2}{169}+\frac{y^2}{69}=1\).

Major-axis length: \(26\,\text{ft}\) Minor-axis length: \(2\sqrt{69}\,\text{ft}\) Equation: \(\frac{x^2}{169}+\frac{y^2}{69}=1\)

- Determine the focal distance from the two hyperbola denominators. - Locate the vertex on the branch used by the reflector. - Compare the two vertex-to-focus distances with the hyperbola’s constant difference.

1. The hyperbola has \(a=3\) and \(b=4\), so \(c^2=a^2+b^2=25\) and \(c=5\). 2. The foci are \((-5, 0)\) and \((5, 0)\), separated by \(10\) units. 3. The right-hand vertex is \((3, 0)\). 4. Its distance to the right focus is \(5-3=2\), and its distance to the left focus is \(3-(-5)=8\). 5. The difference \(8-2=6\) equals \(2a\), consistent with the hyperbola’s defining property.

Foci: \((-5, 0)\), \((5, 0)\) Focal separation: \(10\) Distances from the right vertex: \(8\) to the left focus and \(2\) to the right focus

- Separate the optical or acoustic reflection conditions from the distance-locus condition. - Recall which conic has one focus and a single axis-directed reflection property. - Compare the sum and difference versions of the two-focus definitions.

1. A parabola reflects rays parallel to its axis toward its focus, so design a) uses a parabola. 2. An ellipse reflects a ray leaving one focus toward the other focus, so design b) uses an ellipse. 3. A hyperbola is the locus of points for which the absolute difference of the distances to two foci is constant, so design c) uses a hyperbola.

a) Parabola — parallel incoming rays reflect toward the focus. b) Ellipse — a ray from one focus reflects toward the other focus. c) Hyperbola — the absolute difference of the distances to the two foci is constant.

- Identify which direction is unchanged by the tilt. - Determine the projection factor for lengths perpendicular to the tilt axis. - Use the two projected semi-axis lengths to write the ellipse and find its area.

1. Orthogonal projection preserves lengths parallel to the tilt axis, so the projected semi-axis along the x-axis is \(6\,\text{ft}\). 2. Lengths perpendicular to the tilt axis are multiplied by \(\cos 60^\circ=\frac{1}{2}\), so the other projected semi-axis is \(6\cdot\frac{1}{2}=3\,\text{ft}\). 3. The projected boundary is the ellipse \(\frac{x^2}{36}+\frac{y^2}{9}=1\). 4. Its area is \(\pi ab=\pi(6)(3)=18\pi\,\text{ft}^2\).

Projected boundary: \(\frac{x^2}{36}+\frac{y^2}{9}=1\) Projected area: \(18\pi\,\text{ft}^2\)

- Compare each plane-axis angle with the generator-axis angle. - Equality corresponds to a plane parallel to a generator. - Decide whether the plane cuts one nappe or both nappes.

1. A plane perpendicular to the axis produces a circle, so cut a) is a circle. 2. When the plane-axis angle is greater than the generator-axis angle but less than \(90^\circ\), the plane cuts one nappe in an ellipse. Thus, cut b) is an ellipse. 3. When the plane is parallel to a generator, the angles are equal and the section is a parabola. Thus, cut c) is a parabola. 4. When the plane-axis angle is less than the generator-axis angle, the plane cuts both nappes and produces a hyperbola. Thus, cut d) is a hyperbola.

a) circle b) ellipse c) parabola d) hyperbola

- Start with a parabola written in vertex form. - Use the nozzle location to determine the remaining coefficient. - Set the height equal to zero and select the physically relevant solution.

1. Use vertex form \(y=a(x-3)^2+5\). 2. Substitute the nozzle point \((0, 1)\): \(1=9a+5\), so \(a=-\frac{4}{9}\). 3. The path is \(y=-\frac{4}{9}(x-3)^2+5\). 4. Set \(y=0\) to find the pool-level intersections: \((x-3)^2=\frac{45}{4}\). 5. Thus, \(x=3\pm\frac{3\sqrt{5}}{2}\). The positive landing coordinate is \(3+\frac{3\sqrt{5}}{2}\).

Path: \(y=-\frac{4}{9}(x-3)^2+5\) Landing point: \(\left(3+\frac{3\sqrt{5}}{2}, 0\right)\), approximately \((6.35, 0)\)

- Use the wheel's center and radius in the standard equation of a circle. - Substitute the specified height for \(y\) before solving for the horizontal coordinate. - The two square roots represent positions on opposite sides of the vertical centerline.

1. The circle has center \((0,20)\) and radius \(18\), so its equation is \(x^2+(y-20)^2=324\). 2. At a height of \(29\,\text{ft}\), substitute \(y=29\): \(x^2+9^2=324\). 3. Thus, \(x^2=243\), so \(x=\pm9\sqrt{3}\). 4. Therefore, the rider is about \(15.6\,\text{ft}\) to either side of the centerline.

a) \(x^2+(y-20)^2=324\) b) \(x=\pm9\sqrt{3}\,\text{ft}\), approximately \(15.6\,\text{ft}\) to either side of the centerline

- Use the focal length to write the parabola's cross-sectional equation. - Evaluate the parabola at the aperture radius to find the depth. - Substitute the radius and depth into the supplied volume formula.

1. With focal length \(p=2\), the cross section is \(x^2=4py=8y\). 2. At the rim, \(x=3\), so the depth is \(d=y=\frac{9}{8}\,\text{m}\). 3. The receiver belongs at the focus, \((0, 2)\), which is \(2\,\text{m}\) above the vertex. 4. The volume is \(V=\frac{1}{2}\pi(3)^2\left(\frac{9}{8}\right)=\frac{81\pi}{16}\,\text{m}^3\).

Rim depth: \(\frac{9}{8}\,\text{m}\) Receiver location: \((0, 2)\) Volume: \(\frac{81\pi}{16}\,\text{m}^3\)

- One axis of the oblique cut remains the pipe's diameter. - Relate the elongated axis to its perpendicular projection. - Use the resulting semi-axis lengths in the ellipse area formula.

1. The minor axis is perpendicular to the direction of elongation and remains equal to the pipe diameter, so its length is \(8\,\text{in.}\). 2. Let \(L\) be the major-axis length. Its projection onto a plane perpendicular to the pipe axis equals the diameter: \(L\sin30^\circ=8\). 3. Therefore, \(L=16\,\text{in.}\). 4. The semi-axis lengths are \(a=8\,\text{in.}\) and \(b=4\,\text{in.}\). 5. The opening area is \(\pi ab=32\pi\,\text{in.}^2\).

Major axis: \(16\,\text{in.}\) Minor axis: \(8\,\text{in.}\) Area: \(32\pi\,\text{in.}^2\)

- Apply the supplied approximation separately to the two pairs of semi-axes. - Keep the two perimeter calculations organized before subtracting. - Round only after evaluating each full expression.

1. For the inner lane, \(C_{\text{in}}\approx\pi\left[3(60+40)-\sqrt{(180+40)(60+120)}\right]\). 2. This simplifies to \(C_{\text{in}}\approx60\pi(5-\sqrt{11})\approx317.31\,\text{m}\). 3. For the outer lane, \(C_{\text{out}}\approx\pi\left[3(64+44)-\sqrt{(192+44)(64+132)}\right]\). 4. This simplifies to \(C_{\text{out}}\approx4\pi(81-7\sqrt{59})\approx342.21\,\text{m}\). 5. The difference is approximately \(342.21-317.31=24.90\,\text{m}\).

Inner-lane lap: approximately \(317.31\,\text{m}\) Outer-lane lap: approximately \(342.21\,\text{m}\) Outer-lane increase: approximately \(24.90\,\text{m}\)

- Build a vertex-form equation from the arch's width and height. - Check the arch height at the doorway's horizontal edges. - Subtract the rectangular area from the supplied parabolic-segment area.

1. Put the arch endpoints at \((-12, 0)\) and \((12, 0)\), with vertex \((0, 9)\). Its equation is \(y=9-\frac{x^2}{16}\). 2. The doorway's upper corners have \(x=\pm4\). At either coordinate, the arch height is \(9-\frac{16}{16}=8\,\text{ft}\). 3. Since \(8\,\text{ft}>6\,\text{ft}\), the doorway fits. 4. The parabolic-segment area is \(\frac{2}{3}(24)(9)=144\,\text{ft}^2\). 5. The doorway area is \((8)(6)=48\,\text{ft}^2\). 6. The remaining façade area is \(144-48=96\,\text{ft}^2\).

The arch height at each upper doorway corner is \(8\,\text{ft}\), which is above the \(6\,\text{ft}\)-high doorway, so the doorway fits. Remaining façade area: \(96\,\text{ft}^2\)

- Locate both foci from the semi-axis lengths. - Use the constant sum of focal distances before substituting coordinates. - What line contains points equidistant from the two foci?

1. The focal distance is \(c=\sqrt{100-64}=6\), so the star is at \((6, 0)\). 2. The sum of the distances from an ellipse point to the two foci is \(2a=20\). 3. If the distance to the right focus is \(10\), then the distance to the left focus is also \(10\). Therefore the point lies on the perpendicular bisector of the focal segment, so \(x=0\). 4. Substituting \(x=0\) into the ellipse gives \(\frac{y^2}{64}=1\), so \(y=\pm8\).

The probe can be at \((0, 8)\) or \((0, -8)\), in millions of miles.

- Halve the focal separation and major-axis length to obtain \(c\) and \(a\). - Use the ellipse relation \(a^2=b^2+c^2\) to find the other semi-axis. - The defining focal-distance sum of an ellipse equals its full major-axis length.

1. Half the focal separation is \(c=15\,\text{cm}\), and half the major-axis length is \(a=25\,\text{cm}\). 2. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=625-225=400\) and \(b=20\,\text{cm}\). 3. The standard equation is \(\frac{x^2}{625}+\frac{y^2}{400}=1\). 4. The minor-axis length is \(2b=40\,\text{cm}\). 5. The sum of the focal distances is \(2a=50\,\text{cm}\).

a) \(\frac{x^2}{625}+\frac{y^2}{400}=1\) b) \(40\,\text{cm}\) c) \(50\,\text{cm}\)

- Use the positive y-value because only the upper half models the roof. - For a horizontal cross section, substitute the given height and solve for both x-coordinates. - The total width is the distance between the left and right intersection coordinates.

1. Substitute \(x=8\): \(\frac{64}{100}+\frac{y^2}{36}=1\). 2. Since the roof is the upper half, \(y=\frac{18}{5}=3.6\). This is the support height. 3. For the width at \(y=3\), substitute \(y=3\): \(\frac{x^2}{100}+\frac{9}{36}=1\). 4. Thus, \(\frac{x^2}{100}=\frac{3}{4}\), so \(x=\pm5\sqrt{3}\). 5. The total width is \(5\sqrt{3}-(-5\sqrt{3})=10\sqrt{3}\,\text{ft}\approx17.3\,\text{ft}\).

a) \(\frac{18}{5}\,\text{ft}=3.6\,\text{ft}\) b) \(10\sqrt{3}\,\text{ft}\approx17.3\,\text{ft}\)

- One diameter is unchanged by the projection because it lies along the tilt axis. - Relate the shortened diameter to the original diameter with a trigonometric projection factor. - Use the two projected semi-axis lengths for the ellipse area.

1. Under orthographic projection, the diameter parallel to the tilt axis keeps its full length. Therefore the original circle's diameter is \(12\,\text{cm}\), so its radius is \(6\,\text{cm}\). 2. If \(\alpha\) is the angle between the two planes, the perpendicular diameter is shortened by the factor \(\cos\alpha\). 3. Thus \(12\cos\alpha=7.2\), so \(\cos\alpha=0.6\). 4. Therefore \(\alpha=\arccos(0.6)\approx53.1^\circ\). 5. The projected ellipse has semi-axes \(6\,\text{cm}\) and \(3.6\,\text{cm}\), so its area is \(\pi(6)(3.6)=21.6\pi\,\text{cm}^2\).

Original radius: \(6\,\text{cm}\) Tilt angle: \(\arccos(0.6)\approx53.1^\circ\) Projected area: \(21.6\pi\,\text{cm}^2\)

- Convert the full span and full vertical axis into semi-axis lengths for the ellipse model. - Substitute the given horizontal position and select the nonnegative y-value for the arch. - At the specified height, find both symmetric x-coordinates before computing the width.

1. The horizontal semi-axis is \(a=15\), and the vertical semi-axis is \(b=5\). 2. The complete ellipse is \(\frac{x^2}{225}+\frac{y^2}{25}=1\). The arch is its upper half, so \(y\ge0\). 3. At \(x=9\), \(\frac{81}{225}+\frac{y^2}{25}=1\). 4. Therefore \(\frac{y^2}{25}=\frac{16}{25}\), so \(y=4\) on the upper half. 5. At height \(y=4\), the same equation gives \(\frac{x^2}{225}=\frac{9}{25}\), so \(x=\pm9\). 6. The horizontal clearance width is \(9-(-9)=18\,\text{ft}\).

a) \(\frac{x^2}{225}+\frac{y^2}{25}=1\), with \(y\ge0\) for the arch b) \(4\,\text{ft}\) c) \(18\,\text{ft}\)

- Interpret the fixed difference between distances to two fixed points as a conic condition. - Compare half the focal separation with half the stated distance difference. - Use which tower is farther away to select the correct branch before substituting the x-coordinate.

1. The towers are the foci, so \(c=6\). The constant difference of focal distances is \(2a=8\), so \(a=4\). 2. For a horizontal hyperbola, \(c^2=a^2+b^2\). Thus \(b^2=36-16=20\). 3. Because the strike is farther from the western tower, it lies on the eastern branch. The branch is part of \(\frac{x^2}{16}-\frac{y^2}{20}=1\) with \(x>0\). 4. Substituting \(x=10\) gives \(\frac{100}{16}-\frac{y^2}{20}=1\), so \(y^2=105\) and \(y=\pm\sqrt{105}\). 5. The two locations are \((10, \sqrt{105})\) and \((10, -\sqrt{105})\), approximately \((10, 10.25)\) and \((10, -10.25)\).

a) \(\frac{x^2}{16}-\frac{y^2}{20}=1\), restricted to the branch \(x>0\). b) \((10, \sqrt{105})\) and \((10, -\sqrt{105})\), approximately \((10, 10.25)\) and \((10, -10.25)\), in miles.

- Relate the two extreme distances from one focus to the semi-major axis and focal distance. - Add and subtract those two relationships. - Use the ellipse relationship among the two semi-axes and focal distance.

1. For a horizontal ellipse measured from the right focus, the nearest and farthest focal distances are \(a-c\) and \(a+c\). 2. Thus \(a-c=7000\) and \(a+c=13{,}000\). Adding gives \(a=10{,}000\), and subtracting gives \(c=3000\). 3. Then \(b^2=a^2-c^2=100{,}000{,}000-9{,}000{,}000=91{,}000{,}000\). 4. The equation is \(\frac{x^2}{100000000}+\frac{y^2}{91000000}=1\). 5. The foci are \((-3000, 0)\) and \((3000, 0)\), and the eccentricity is \(e=\frac{c}{a}=0.3\).

\(\frac{x^2}{100000000}+\frac{y^2}{91000000}=1\) Foci: \((-3000, 0)\) and \((3000, 0)\) Eccentricity: \(0.3\)

- Use vertex form with the cable’s lowest point as the vertex. - Use one tower point to determine the remaining coefficient. - Evaluate the model at the listed symmetric positions before adding hanger lengths.

1. With the vertex at \((0, 20)\), write the cable as \(y=ax^2+20\). 2. A tower point is \((200, 100)\), so \(100=40{,}000a+20\), giving \(a=\frac{1}{500}\). 3. The cable equation is \(y=\frac{x^2}{500}+20\). 4. The hanger lengths are \(20\) at \(x=0\), \(25\) at \(x=\pm50\), \(40\) at \(x=\pm100\), and \(65\) at \(x=\pm150\). 5. Their total length is \(20+2(25+40+65)=280\,\text{ft}\).

Cable equation: \(y=\frac{x^2}{500}+20\) Total hanger length: \(280\,\text{ft}\)

- Use half-widths as x-coordinates in the cross-sectional model. - The narrowest width determines the transverse semi-axis. - After finding the remaining scale, evaluate the model at the requested height and double the positive x-value.

1. Use \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), where \(y\) measures height from the narrowest level. 2. The minimum half-width is \(a=30\), so \(a^2=900\). 3. The point \((50, 40)\) lies on the profile. Substitution gives \(\frac{2500}{900}-\frac{1600}{b^2}=1\), so \(b^2=900\). 4. The model is \(\frac{x^2}{900}-\frac{y^2}{900}=1\). 5. At \(y=60\), \(\frac{x^2}{900}-4=1\), so \(x=30\sqrt{5}\) on the right side. 6. The full width is \(2x=60\sqrt{5}\,\text{ft}\approx134.16\,\text{ft}\).

Model: \(\frac{x^2}{900}-\frac{y^2}{900}=1\) Predicted width \(60\,\text{ft}\) above the narrowest level: \(60\sqrt{5}\,\text{ft}\approx134.16\,\text{ft}\)

- Use half the entrance width and the center height as ellipse semi-axes. - Check the roof height directly above a top corner of the centered vehicle. - Compare that height with the vehicle height.

1. With the ellipse centered at ground level, its equation is \(\frac{x^2}{400}+\frac{y^2}{144}=1\), with \(y\ge0\). 2. The vehicle’s top corners are at \(x=\pm7\). 3. At \(x=7\), the tunnel height is \(y=12\sqrt{1-\frac{49}{400}}=\frac{3\sqrt{351}}{5}\approx11.241\,\text{ft}\). 4. This exceeds the vehicle height of \(11\,\text{ft}\), so the vehicle clears. 5. The smallest clearance occurs above the top corners and is \(\frac{3\sqrt{351}}{5}-11\approx0.241\,\text{ft}\).

Yes. The minimum vertical clearance is \(\frac{3\sqrt{351}}{5}-11\,\text{ft}\approx0.241\,\text{ft}\).

- Fit each model from the shared center and endpoint data. - Evaluate both equations at the same horizontal location. - Compare the two resulting heights rather than only comparing their equations.

1. Model A has form \(y=8-ax^2\). Using \((10, 0)\) gives \(a=\frac{2}{25}\), so \(y=8-\frac{2}{25}x^2\). 2. Model B has semi-axes \(a=10\) and \(b=8\), so \(\frac{x^2}{100}+\frac{y^2}{64}=1\), with \(y\ge0\). 3. At \(x=6\), Model A has height \(8-\frac{72}{25}=\frac{128}{25}=5.12\). 4. Model B has height \(8\sqrt{1-\frac{36}{100}}=\frac{32}{5}=6.4\). 5. The semielliptical model is higher by \(\frac{32}{5}-\frac{128}{25}=\frac{32}{25}=1.28\).

Model A: \(y=8-\frac{2}{25}x^2\) Model B: \(\frac{x^2}{100}+\frac{y^2}{64}=1\), \(y\ge0\) At \(x=6\), Model B gives \(1.28\) more units of clearance.

- Identify the focus and directrix from the parabola equation. - Express the incoming horizontal distance in terms of the reflection point’s x-coordinate. - Replace the focus distance using the defining property of a parabola.

1. The parabola has \(4p=16\), so its focus is \(F=(4, 0)\) and its directrix is \(x=-4\). 2. The incoming horizontal segment from the wavefront point \((12, y)\) to \(P=(x, y)\) has length \(12-x\). 3. By the focus–directrix definition of the parabola, \(PF\) equals the perpendicular distance from \(P\) to \(x=-4\), which is \(x+4\). 4. The total distance is \((12-x)+(x+4)=16\,\text{m}\), independent of \(P\). 5. The travel time is \(\frac{16}{343}\,\text{s}\approx0.0466\,\text{s}\), or approximately \(46.6\,\text{ms}\).

Every reflected path has total length \(16\,\text{m}\). Travel time: \(\frac{16}{343}\,\text{s}\approx0.0466\,\text{s}\), or approximately \(46.6\,\text{ms}\)

- Represent a boundary point on the follower ray using polar-style coordinates centered at the ellipse's center. - Substitute the ray coordinates into the ellipse equation. - After the cam rotates, update the angle between the fixed follower direction and the major axis.

1. Along a ray making angle \(\theta\) with the major axis, write \((x, y)=(r\cos\theta,r\sin\theta)\). 2. Substitution gives \(r=\frac{1}{\sqrt{\frac{\cos^2\theta}{100}+\frac{\sin^2\theta}{36}}}\). 3. At \(\theta=30^\circ\), \(r=\frac{1}{\sqrt{\frac{3}{400}+\frac{1}{144}}}=\frac{30}{\sqrt{13}}\approx8.32\). 4. After a \(90^\circ\) rotation, the fixed follower direction makes an angle of \(60^\circ\) in magnitude with the major axis. 5. Thus, \(r=\frac{1}{\sqrt{\frac{1}{400}+\frac{3}{144}}}=\sqrt{\frac{300}{7}}=\frac{10\sqrt{21}}{7}\approx6.55\). 6. The follower moves inward by \(\frac{30}{\sqrt{13}}-\frac{10\sqrt{21}}{7}\approx1.77\) units.

Initial distance: \(\frac{30}{\sqrt{13}}\approx8.32\) units Distance after rotation: \(\frac{10\sqrt{21}}{7}\approx6.55\) units Inward change: approximately \(1.77\) units

- Write the translated ellipse equation from the center and two semi-axis lengths. - Substitute the line equation into the ellipse and solve the resulting quadratic. - Use the distance formula between the two boundary crossings, keeping the exact fractional coordinates until the end.

1. The boundary of the no-fly zone is \(\frac{(x-4)^2}{100}+\frac{(y+2)^2}{36}=1\). 2. Substitute the flight path \(y=x\): \(\frac{(x-4)^2}{100}+\frac{(x+2)^2}{36}=1\). 3. Multiplying by \(900\) and simplifying gives \(17x^2+14x-328=0\). 4. The roots are \(x=4\) and \(x=-\frac{82}{17}\). Since \(y=x\), the boundary points are \((4,4)\) and \(\left(-\frac{82}{17},-\frac{82}{17}\right)\). 5. The change in each coordinate is \(4+\frac{82}{17}=\frac{150}{17}\). 6. The distance traveled inside the ellipse is \(\sqrt{\left(\frac{150}{17}\right)^2+\left(\frac{150}{17}\right)^2}=\frac{150\sqrt{2}}{17}\,\text{km}\approx12.5\,\text{km}\).

Entry and exit points: \(\left(-\frac{82}{17}, -\frac{82}{17}\right)\) and \((4, 4)\) Distance inside the zone: \(\frac{150\sqrt{2}}{17}\,\text{km}\approx12.5\,\text{km}\)

- Half the constant focal-distance difference is \(a\), while half the station separation is \(c\). - Use the hyperbola relation \(c^2=a^2+b^2\) to determine the other denominator. - Substitute the reported y-coordinate and select the right-branch solution before checking both focal distances.

1. The foci are \((-5,0)\) and \((5,0)\), so \(c=5\). 2. The constant distance difference is \(6\), so \(2a=6\) and \(a=3\). 3. For a hyperbola, \(c^2=a^2+b^2\), so \(b^2=25-9=16\). 4. The locus is \(\frac{x^2}{9}-\frac{y^2}{16}=1\). 5. Substitute \(y=12\): \(\frac{x^2}{9}-9=1\), so \(x^2=90\). 6. The vessel is on the right branch, so \(x=3\sqrt{10}\). Its coordinates are \((3\sqrt{10},12)\). 7. Its distances to the left and right stations are \(5\sqrt{10}+3\) and \(5\sqrt{10}-3\), respectively, whose difference is \(6\,\text{km}\).

a) \(\frac{x^2}{9}-\frac{y^2}{16}=1\) b) \((3\sqrt{10}, 12)\); the focal distances are \(5\sqrt{10}+3\,\text{km}\) and \(5\sqrt{10}-3\,\text{km}\), differing by \(6\,\text{km}\)

- Use the vertex and focus distances to determine \(a\), \(c\), and then \(b\). - Substitute the given x-coordinate and choose the positive y-value for the upper point. - Compute both focal distances and compare their difference with \(2a\).

1. The center-to-vertex distance is \(a=5\), and the focal distance is \(c=13\). 2. For a hyperbola, \(c^2=a^2+b^2\), so \(b^2=169-25=144\). 3. The standard equation is \(\frac{x^2}{25}-\frac{y^2}{144}=1\), and the asymptotes are \(y=\pm\frac{12}{5}x\). 4. At \(x=13\), \(\frac{169}{25}-\frac{y^2}{144}=1\). 5. Thus \(\frac{y^2}{144}=\frac{144}{25}\), so the upper point is \(P=\left(13,\frac{144}{5}\right)\). 6. The distance to the right focus \((13,0)\) is \(\frac{144}{5}\,\text{cm}\). 7. The distance to the left focus \((-13,0)\) is \(\sqrt{26^2+\left(\frac{144}{5}\right)^2}=\frac{194}{5}\,\text{cm}\). 8. Their difference is \(\frac{194}{5}-\frac{144}{5}=10\,\text{cm}=2a\), which verifies the constant-difference property.

a) \(\frac{x^2}{25}-\frac{y^2}{144}=1\); asymptotes \(y=\pm\frac{12}{5}x\) b) \(P=\left(13,\frac{144}{5}\right)\) c) Distances \(\frac{194}{5}\,\text{cm}\) and \(\frac{144}{5}\,\text{cm}\); difference \(10\,\text{cm}\)

- The midpoint of the nearest and farthest axial points is the ellipse center. - Use the star's location to find the center-to-focus distance \(c\). - Apply \(b^2=a^2-c^2\) and then compute \(e=\frac{c}{a}\).

1. The endpoints of the major axis are \((2,0)\) and \((-18,0)\), so the center is their midpoint, \((-8,0)\). 2. The semi-major axis is \(a=10\). 3. The star at the origin is \(8\) units from the center, so \(c=8\). 4. For an ellipse, \(b^2=a^2-c^2=100-64=36\). 5. The orbit equation is \(\frac{(x+8)^2}{100}+\frac{y^2}{36}=1\). 6. The eccentricity is \(e=\frac{c}{a}=\frac{4}{5}\). 7. The other focus is \(8\) units left of the center, at \((-16,0)\).

a) \(\frac{(x+8)^2}{100}+\frac{y^2}{36}=1\) b) Eccentricity: \(\frac{4}{5}\); other focus: \((-16, 0)\)

- Translate each candidate position so the reported estimate becomes the origin. - Resolve each translated vector along the major-axis and minor-axis directions. - Compare the resulting coordinates with the standard ellipse inequality.

1. Relative to the estimate, use principal coordinates \(u=(x-20)\cos30^\circ+(y-10)\sin30^\circ\) and \(v=-(x-20)\sin30^\circ+(y-10)\cos30^\circ\). 2. The uncertainty region is \(\frac{u^2}{36}+\frac{v^2}{4}\leq1\). 3. For \(A\), the relative vector is \((3\sqrt3, 3)\), giving \((u, v)=(6, 0)\). Thus \(\frac{u^2}{36}+\frac{v^2}{4}=1\), so \(A\) is on the ellipse. 4. For \(B\), the relative vector is \((2\sqrt3, 2)\), giving \((u, v)=(4, 0)\). Thus the value is \(\frac{16}{36}=\frac49<1\), so \(B\) is inside. 5. For \(C\), the relative vector is \(\left(-\frac32, \frac{3\sqrt3}{2}\right)\), giving \((u, v)=(0, 3)\). Thus the value is \(\frac94>1\), so \(C\) is outside.

a) \(A\) is on the uncertainty ellipse. b) \(B\) is inside the uncertainty ellipse. c) \(C\) is outside the uncertainty ellipse.

- Convert each full lake dimension into a semi-axis length before writing the ellipse equation. - Substitute the ferry route's constant y-coordinate into the shoreline equation. - Compare the distance between the two intersection x-coordinates with the full major-axis length.

1. The semi-major axis is \(a=13\), and the semi-minor axis is \(b=5\). 2. The shoreline is modeled by \(\frac{x^2}{169}+\frac{y^2}{25}=1\). 3. On the ferry route, \(y=3\). Substitute: \(\frac{x^2}{169}+\frac{9}{25}=1\). 4. Then \(\frac{x^2}{169}=\frac{16}{25}\), so \(x=\pm\frac{52}{5}\). 5. The entry and exit points are \(\left(-\frac{52}{5}, 3\right)\) and \(\left(\frac{52}{5}, 3\right)\). 6. The route length across the lake is \(\frac{52}{5}-\left(-\frac{52}{5}\right)=\frac{104}{5}=20.8\,\text{km}\). 7. The maximum east-west width is \(26\,\text{km}\), so the percentage is \(\frac{20.8}{26}\cdot100\%=80\%\).

a) \(\frac{x^2}{169}+\frac{y^2}{25}=1\) b) \(\left(-\frac{52}{5}, 3\right)\) and \(\left(\frac{52}{5}, 3\right)\) c) \(20.8\,\text{km}\), which is \(80\%\) of the maximum east-west width

- Compute each broken path as the sum of two ordinary distances. - Compare the result with the constant focal-distance sum of the ellipse. - Use the comparison to explain why the exact focus location matters.

1. Through \(A=(10, 0)\), the speaker-to-wall distance is \(3\) and the wall-to-listener distance is \(18\), so the total path length is \(21\,\text{m}\). 2. Through \(B=(0, 6)\), the speaker-to-wall distance is \(\sqrt{(7-0)^2+(0-6)^2}=\sqrt{85}\). 3. The wall-to-listener distance is \(\sqrt{(0+8)^2+(6-0)^2}=10\). 4. The second total path length is \(10+\sqrt{85}\,\text{m}\). 5. The difference is \(21-(10+\sqrt{85})=11-\sqrt{85}\approx1.78\,\text{m}\). 6. If the speaker were at the right focus \((8, 0)\), every one-reflection broken path to the left focus would have length \(2a=20\,\text{m}\). The unequal lengths show that moving the speaker off the focus destroys the constant-path property.

a) Through \(A\): \(21\,\text{m}\); through \(B\): \(10+\sqrt{85}\,\text{m}\approx19.22\,\text{m}\) b) \(11-\sqrt{85}\,\text{m}\approx1.78\,\text{m}\) c) The speaker is not at a focus, so the broken path length to the other focus is no longer constant.

- Use \(c^2=a^2+b^2\) to relate the focus to the nearest vertex. - Read the asymptote slopes from the standard horizontal-hyperbola form. - Substitute the given x-coordinate and use symmetry for the two possible y-coordinates.

1. The hyperbola has \(a=8\), \(b=6\), and \(c=\sqrt{a^2+b^2}=10\), so the stated planet location is a focus. 2. The closest point on the right branch to the right focus is the right vertex \((8, 0)\). 3. The closest distance is \(c-a=10-8=2\) thousand kilometers. 4. The asymptotes are \(y=\pm\frac{b}{a}x=\pm\frac34x\). 5. Their direction angles are \(\theta=\pm\arctan\left(\frac34\right)\approx\pm36.9^\circ\). 6. At \(x=20\), \(\frac{400}{64}-\frac{y^2}{36}=1\). 7. Thus \(\frac{y^2}{36}=\frac{21}{4}\), so \(y=\pm3\sqrt{21}\). 8. The distance between the two positions is \(3\sqrt{21}-(-3\sqrt{21})=6\sqrt{21}\) thousand kilometers, approximately \(27.5\) thousand kilometers.

a) \(2\) thousand kilometers b) \(y=\pm\frac34x\); direction angles \(\pm\arctan\left(\frac34\right)\approx\pm36.9^\circ\) c) \(y=\pm3\sqrt{21}\); separation \(6\sqrt{21}\) thousand kilometers \(\approx27.5\) thousand kilometers

- Use the trimming line to find the two symmetric boundary points. - Locate the focus from the standard parabola coefficient. - Use a dot product between the two focus-to-rim vectors to find the included angle.

1. Since \(y^2=4ax\), the focal parameter is \(a=3\), so the lamp is at \(F=(3, 0)\). 2. At the trimming line \(x=12\), \(y^2=144\), so the rim points are \(R_1=(12, 12)\) and \(R_2=(12, -12)\). 3. Each focus-to-rim distance is \(FR_i=\sqrt{(12-3)^2+12^2}=\sqrt{225}=15\,\text{cm}\). 4. The vectors from the focus to the rim points are \(\mathbf v_1=(9, 12)\) and \(\mathbf v_2=(9, -12)\). 5. Their dot product is \(\mathbf v_1\cdot\mathbf v_2=81-144=-63\), and each vector has length \(15\). 6. If \(\phi\) is the opening angle at the lamp, then \(\cos\phi=\frac{-63}{225}=-\frac7{25}\). Thus \(\phi=\arccos\left(-\frac7{25}\right)\approx106.3^\circ\).

Rim points: \((12, 12)\) and \((12, -12)\) Distance from the lamp to either rim point: \(15\,\text{cm}\) Opening angle: \(\arccos\left(-\frac7{25}\right)\approx106.3^\circ\)

- Eliminate the height variable by combining the cone and plane equations. - Complete the square before classifying the projected curve. - Read the semi-axis lengths from standard form before computing area.

1. Points on the cut satisfy both equations, so substitute \(z=2+\frac{x}{2}\) into \(z^2=x^2+y^2\). 2. This gives \(\left(2+\frac{x}{2}\right)^2=x^2+y^2\), so \(\frac34x^2-2x+y^2-4=0\). 3. Complete the square in \(x\): \(\frac34\left(x-\frac43\right)^2+y^2=\frac{16}{3}\). 4. Dividing by \(\frac{16}{3}\) gives \(\frac{\left(x-\frac43\right)^2}{\frac{64}{9}}+\frac{y^2}{\frac{16}{3}}=1\). 5. This ellipse has \(-\frac43\le x\le4\), so \(z=2+\frac{x}{2}\ge\frac43>0\) everywhere on the curve. Thus every projected point belongs to the upper cone. 6. The top-view curve is an ellipse centered at \(\left(\frac43, 0\right)\), with horizontal semi-axis \(\frac83\) and vertical semi-axis \(\frac{4}{\sqrt3}\). 7. Its area is \(\pi\left(\frac83\right)\left(\frac{4}{\sqrt3}\right)=\frac{32\pi}{3\sqrt3}=\frac{32\pi\sqrt3}{9}\) square centimeters.

Top-view equation: \(\frac{\left(x-\frac43\right)^2}{\frac{64}{9}}+\frac{y^2}{\frac{16}{3}}=1\) Conic: ellipse Center: \(\left(\frac43, 0\right)\) Semi-axis lengths: \(\frac83\) and \(\frac{4}{\sqrt3}\) Area: \(\frac{32\pi\sqrt3}{9}\,\text{cm}^2\)