- An average rate uses a total change over a time interval. - An instantaneous rate describes what is happening at one moment. - Keep the direction as part of each position rate.

1. The average rate of change of position is \(\frac{120\,\text{mi}}{2\,\text{h}}=60\,\text{mph}\) east. 2. The stated velocity \(55\,\text{mph}\) east describes the rate of change of position at one instant. 3. Therefore, \(60\,\text{mph}\) east is the average rate and \(55\,\text{mph}\) east is the instantaneous rate.

Average rate of change of position: \(60\,\text{mph}\) east. Instantaneous rate of change of position: \(55\,\text{mph}\) east.

- Average rate of change is change in output divided by change in input. - Use the later height minus the earlier height. - Use hours for the elapsed time so the rate has units of centimeters per hour.

1. The change in height is \(30-18=12\,\text{cm}\). 2. The elapsed time is \(2\,\text{h}\). 3. The average rate of change is \(\frac{12\,\text{cm}}{2\,\text{h}}=6\,\text{cm/h}\).

\(6\,\text{cm/h}\)

- A smooth connection requires the two graphs to share a point and a slope. - Find the point on the hill at \(x=2\). - Use the derivative to find the hill's slope at that point. - Write the line using the point and slope.

1. Find the connection point: \(f(2)=-0.2(2)^2+2(2)+1=4.2\), so the point is \((2,4.2)\). 2. Differentiate: \(f'(x)=-0.4x+2\). 3. Find the slope at the connection point: \(f'(2)=-0.4(2)+2=1.2\). 4. A smooth connection requires the line to have the same point and slope. Using point-slope form, \(y-4.2=1.2(x-2)\). 5. Simplify: \(y=1.2x+1.8\).

\(y=1.2x+1.8\)

- Differentiate using the power rule. - Compare each derivative directly with the corresponding perimeter or surface-area formula. - Find the constant factor that multiplies the boundary measure to produce the derivative. - Check whether the factor is the same in both cases.

1. For the square, \(A'(x)=2x\). Since \(P(x)=4x\), \(A'(x)=\frac{1}{2}P(x)\). 2. For the cube, \(V'(x)=3x^2\). Since \(S(x)=6x^2\), \(V'(x)=\frac{1}{2}S(x)\). 3. In both cases, the ratio of the derivative to the boundary measure is \(\frac{1}{2}\).

a) \(A'(x)=2x=\frac{1}{2}P(x)\) b) \(V'(x)=3x^2=\frac{1}{2}S(x)\) c) The ratio is \(\frac{1}{2}\), or \(1:2\).

- Rearrange the average-rate relationship to recover total change. - The interval length is the difference between the two times. - Interpret the sign using the direction of the volume change.

1. The time interval has length \(7-4=3\,\text{min}\). 2. Total change equals average rate times elapsed time: \((-2.5\,\text{gal/min})(3\,\text{min})=-7.5\,\text{gal}\). 3. The negative sign means the tank contains less water at \(t=7\) than at \(t=4\); the volume decreased by \(7.5\,\text{gal}\).

a) \(-7.5\,\text{gal}\) b) The tank's volume decreased by \(7.5\,\text{gal}\) over the interval.

- Compute the hour-long average from total distance and total time. - Ask which information an interval average ignores about what happens inside the interval. - Distinguish a whole-interval quantity from a one-moment quantity.

1. Each cyclist's average speed over the hour is \(\frac{12\,\text{mi}}{1\,\text{h}}=12\,\text{mph}\). 2. An average speed depends only on the total distance and total elapsed time over the interval. 3. An instantaneous speed describes one moment, so the cyclists can have different values at \(1{:}30\) p.m. even though their one-hour averages are equal.

Both average \(12\,\text{mph}\) over the hour, but average speed does not determine the speed at a particular instant; the instantaneous speeds can therefore be \(8\,\text{mph}\) and \(16\,\text{mph}\).

- Compare how each line meets the function graph rather than using its label as a clue. - The average rate over an interval comes from a line through the two endpoint values. - The instantaneous rate at a point comes from the tangent slope there.

1. Line \(p\) passes through the two function points \((0,0)\) and \((2,4)\), so it is the secant for the interval and represents the average rate of change. 2. Its slope is \(\frac{4-0}{2-0}=2\). 3. Line \(q\) touches the graph at \((2,4)\) with the same local direction as the curve, so it is the tangent and represents the instantaneous rate at \(x=2\). 4. Using \((1,0)\) and \((2,4)\) on \(q\), its slope is \(\frac{4-0}{2-1}=4\).

a) \(p\), because it is the secant through the two interval endpoints. b) \(q\), because it is tangent to the graph at \(x=2\). c) Average rate: \(2\); instantaneous rate at \(x=2\): \(4\)

- Write the average-rate formula with the unknown endpoint value still in place. - Use the full input change from \(2\) to \(5\). - Solve the resulting linear equation for the unknown output.

1. Use the average-rate equation \(\frac{R(5)-R(2)}{5-2}=-3\). 2. Substitute \(R(2)=14\): \(\frac{R(5)-14}{3}=-3\). 3. Thus \(R(5)-14=-9\), so \(R(5)=5\).

\(R(5)=5\)

- What equation identifies the point where the object reaches the ground? - The derivative at the impact point gives the tangent slope. - Use the magnitude of the slope when finding the acute angle with the horizontal. - Round the angle as requested.

1. Find where the shot put reaches the ground by solving \(h(x)=0\). The solutions are \(x=-2\) and \(x=10\), so the physically relevant impact point is \(x=10\). 2. Differentiate: \(h'(x)=-0.2x+0.8\). 3. Find the slope at impact: \(h'(10)=-0.2(10)+0.8=-1.2\). 4. The acute angle \(\alpha\) with the horizontal satisfies \(\tan(\alpha)=|-1.2|=1.2\). Thus, \(\alpha=\arctan(1.2)\approx 50.2^\circ\).

\(50.2^\circ\)

- What does the derivative tell you about the graph at a point? - For a smooth transition to the ground, which two quantities must match? - Use the relationship between slope and angle.

1. Differentiate: \(f'(x)=\frac{1}{4}x-1\). 2. At \(x=0\), the slope is \(f'(0)=-1\). The acute angle \(\alpha\) satisfies \(\tan(\alpha)=|-1|=1\), so \(\alpha=45^\circ\). 3. At \(x=4\), \(f(4)=\frac{1}{8}\cdot 4^2-4+2=0\), so the ramp reaches the ground. 4. Also, \(f'(4)=\frac{1}{4}\cdot 4-1=0\), which matches the slope of the horizontal line \(y=0\). Since both the height and slope agree, the ramp joins the ground smoothly at \((4,0)\).

a) \(45^\circ\) b) \(f(4)=0\) and \(f'(4)=0\), so the ramp and the ground line \(y=0\) have the same point and slope at \((4,0)\).

- A tangent continuation has the same value and slope as the curve at the transition point. - Differentiate the given function. - Use point-slope form for the tangent line. - Verify that the line passes through the point on the curve.

1. Find the transition point: \(g(2)=2^3-4\cdot 2^2+5\cdot 2=2\), so the point is \((2,2)\). 2. Differentiate: \(g'(x)=3x^2-8x+5\). 3. Find the slope at \(x=2\): \(g'(2)=3\cdot 2^2-8\cdot 2+5=1\). 4. Use point-slope form: \(y-2=1(x-2)\). 5. Simplify: \(y=x\).

\(y=x\)

- Find the point where the straight continuation begins. - A smooth connection requires the line to have the same slope as the curve. - Write the tangent line at \(x=10\). - Set the line's output equal to the target height.

1. Find the transition point: \(f(10)=0.01(10)^3-0.2(10)^2+1.5(10)+2=7\), so the point is \((10,7)\). 2. Differentiate: \(f'(x)=0.03x^2-0.4x+1.5\). 3. Find the slope at the transition: \(f'(10)=0.03(100)-0.4(10)+1.5=0.5\). 4. A smooth straight continuation is the tangent line: \(y-7=0.5(x-10)\), so \(y=0.5x+2\). 5. Set the height equal to \(12\): \(12=0.5x+2\). Solving gives \(x=20\).

\(x=20\,\text{m}\)

- A smooth transition means the linear model has the same value and slope at the joining time. - Find the temperature and rate of change at \(t=10\). - Write the linear model for times after \(10\) minutes. - Convert the final time to the requested unit.

1. Find the temperature at the transition: \(T(10)=0.1(10)^2-4(10)+80=50\). 2. Differentiate: \(T'(t)=0.2t-4\). The rate at \(t=10\) is \(T'(10)=-2\) degrees Celsius per minute. 3. A smooth linear continuation has the same value and rate at \(t=10\): \(L(t)-50=-2(t-10)\), so \(L(t)=-2t+70\). 4. Solve \(25=-2t+70\). This gives \(t=22.5\) minutes. 5. Convert to seconds: \(22.5\cdot 60=1350\) seconds.

\(1350\,\text{s}\)

- The derivative gives the local slope of the trail. - How are slope, percent grade, and angle of inclination related? - A tangent line requires both the point and the slope. - Use point-slope form for the tangent line.

1. Differentiate: \(h'(x)=-0.006x^2+0.12x\). 2. At \(x=5\), \(h'(5)=-0.006(25)+0.12(5)=0.45\). The percent grade is \(0.45\cdot 100\%=45\%\). 3. At \(x=15\), \(h'(15)=0.45\). Therefore, \(\alpha=\arctan(0.45)\approx 24.23^\circ\). 4. At \(x=25\), the slope is \(h'(25)=-0.75\), and \(h(25)=16.25\). The tangent line is \(y-16.25=-0.75(x-25)\), so \(y=-0.75x+35\).

a) \(45\%\) b) \(24.23^\circ\), approximately c) \(y=-0.75x+35\)

- A line perpendicular to the curve at a point is the normal line. - Use the derivative to find the curve's tangent slope. - Tangent and normal slopes are negative reciprocals. - Intersect the normal line with \(y=7\).

1. Differentiate: \(f'(x)=0.4x\). At \(x=5\), the tangent slope is \(f'(5)=2\). 2. The support follows the normal line, whose slope is \(-\frac{1}{2}\). 3. Through \(P(5,3)\), the normal line is \(y-3=-0.5(x-5)\), or \(y=-0.5x+5.5\). 4. Set \(y=7\): \(7=-0.5x+5.5\), so \(x=-3\). 5. Therefore, \(Q=(-3,7)\).

\(Q=(-3,7)\)

- Rewrite the square root as a rational power. - Interpret a derivative of speed with respect to distance using quotient units. - To show a rate decreases, examine the sign of its derivative. - Simplify the units after dividing speed units by distance units.

1. a) Rewrite \(v(s)=\sqrt{2g}\,s^{\frac{1}{2}}\). Then \(v'(s)=\frac{\sqrt{2g}}{2\sqrt{s}}=\sqrt{\frac{g}{2s}}\), for \(s>0\). 2. b) \(v'(2)=\sqrt{\frac{9.81}{4}}\approx1.566\,\text{s}^{-1}\), and \(v'(8)=\sqrt{\frac{9.81}{16}}\approx0.783\,\text{s}^{-1}\). 3. The derivative measures the increase in speed, in meters per second, per additional meter fallen. Its units simplify to \(\text{s}^{-1}\). 4. c) Differentiate again: \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}\). 5. Since \(v''(s)<0\) for every \(s>0\), \(v'(s)\) is strictly decreasing. Therefore, the speed gained per additional meter becomes smaller as the object falls farther.

a) \(v'(s)=\sqrt{\frac{g}{2s}}\) b) \(v'(2)\approx1.566\,\text{s}^{-1}\); \(v'(8)\approx0.783\,\text{s}^{-1}\). It is the increase in speed per additional meter fallen. c) \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}<0\), so \(v'(s)\) decreases as \(s\) increases.

- Use symmetry to locate the right edge from the total width. - Convert a percent grade to a decimal slope. - The derivative at the edge gives a condition on \(a\). - Evaluate the model at the edge to find the depth.

1. Because the bowl is \(12\,\text{m}\) wide and symmetric, the right edge is at \(x=6\). 2. A grade of \(120\%\) corresponds to slope \(1.2\), so \(f'(6)=1.2\). 3. Since \(f'(x)=2ax\), the condition gives \(12a=1.2\), so \(a=0.1\). Thus \(f(x)=0.1x^2\). 4. The depth is the height of the edge above the origin: \(f(6)=0.1(6)^2=3.6\,\text{m}\).

a) \(f(x)=0.1x^2\) b) \(3.6\,\text{m}\)

- Write the tangent-line equation at a general input \(x_0\). - Use the derivative for the tangent slope. - Substitute the sensor coordinates into the tangent-line equation. - Solve the resulting quadratic equation. - Check each solution against the stated domain.

1. Differentiate: \(f'(x)=-0.5x\). 2. The tangent line at \(x=x_0\) is \(t(x)=f'(x_0)(x-x_0)+f(x_0)\). Substitution and simplification give \(t(x)=-0.5x_0x+0.25x_0^2+4\). 3. Since the line must pass through \(S(4,1)\), set \(t(4)=1\): \(1=-2x_0+0.25x_0^2+4\). 4. Rearranging gives \(x_0^2-8x_0+12=0\), so \((x_0-2)(x_0-6)=0\). Thus, \(x_0=2\) or \(x_0=6\). 5. Only \(x_0=2\) lies in the rail's domain \([0,4]\).

The laser must be at \(x_0=2\).

- Find the vertex of the parabola. - Translate “stay at or above the roadway” into an inequality for the minimum value. - Differentiate to find the slope at the tower. - Use the tangent relationship between slope and angle.

1. The parabola's lowest point occurs halfway between the towers, at \(x=50\). Its height is \(f_a(50)=30-2500a\). 2. To stay at or above the roadway, the minimum height must satisfy \(30-2500a\ge0\). Therefore, \(a\le0.012\). Together with \(a>0\), this gives \(0<a\le0.012\). 3. If the minimum height is \(5\,\text{m}\), then \(30-2500a=5\), so \(a=0.01\). 4. For \(a=0.01\), \(f_a'(x)=0.02x-1\). At \(x=0\), the slope is \(-1\). The acute angle relative to the horizontal is \(\arctan(1)=45^\circ\).

a) \(0<a\le0.012\) b) \(a=0.01\) c) \(45^\circ\)

- Perpendicular lines have slopes that are negative reciprocals. - Use the derivative to find the walking path's tangent slope at \(A\). - Write the connector line through \(A\). - Set the connector and creek equations equal to find their intersection.

1. Differentiate: \(f'(x)=-x+2\). At \(x=3\), the tangent slope is \(f'(3)=-1\). 2. The perpendicular connector has slope \(1\). 3. Through \(A(3,3)\), its equation is \(y-3=x-3\), or \(y=x\). 4. Intersect this line with the creek: \(x=-2x+6\). Thus, \(3x=6\), so \(x=2\) and \(y=2\). 5. Therefore, the bridge location is \(B=(2,2)\).

The connector path is \(y=x\), and the bridge is at \(B=(2,2)\).

- First find the exact takeoff point on the ramp. - Use the derivative to find the ramp's slope at takeoff. - Model the initial path with the tangent line. - Substitute the wall's x-coordinate into the tangent-line equation.

1. Find the takeoff point: \(f(2)=-0.1\cdot2^3+0.6\cdot2^2=1.6\), so the point is \((2,1.6)\). 2. Differentiate: \(f'(x)=-0.3x^2+1.2x\). 3. Find the slope at takeoff: \(f'(2)=-0.3\cdot2^2+1.2\cdot2=1.2\). 4. The tangent line is \(y-1.6=1.2(x-2)\), or \(y=1.2x-0.8\). 5. At the wall, \(y=1.2\cdot6-0.8=6.4\). Therefore, \(h=6.4\,\text{m}\).

\(h=6.4\,\text{m}\)

- Use the midpoint coordinates in the cable model. - The derivative gives the slope of the cable at the tower. - Relate slope to angle using the tangent function. - For an upward-opening parabola, the minimum occurs at the vertex.

1. At the midpoint, \(x=0\). Set the height equal to \(30\): \(50-2500k=30\). Thus, \(k=\frac{20}{2500}=\frac1{125}\). 2. The right tower is at \(x=50\). Differentiate: \(f_k'(x)=2kx\). For \(k=\frac1{125}\), the slope at the tower is \(f_k'(50)=\frac45\). 3. If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 4. Because \(k>0\), the lowest point is at \(x=0\). The cable touches or passes below the roadway when \(50-2500k\le0\). Therefore, \(k\ge\frac1{50}\). It first touches the roadway when \(k=\frac1{50}\).

a) \(k=\frac1{125}\) b) \(\theta\approx38.66^\circ\) c) \(k\ge\frac1{50}\); it first touches the roadway at \(k=\frac1{50}\).

- Landing at the same height means the function value is zero at the basin. - Use the vertex of the downward-opening parabola. - Convert slope to angle with the inverse tangent.

1. The basin condition is \(h_c(4)=0\): \(-0.5(4)^2+4c=0\). Thus \(-8+4c=0\), so \(c=2\). 2. For \(c=2.5\), the vertex occurs at \(x=2.5\). The maximum height is \(h_{2.5}(2.5)=3.125\,\text{m}\). 3. The initial slope is \(h_c'(0)=c\). For \(c=1.5\), \(\tan(\alpha)=1.5\), so \(\alpha=\arctan(1.5)\approx56.31^\circ\).

a) \(c=2\) b) \(3.125\,\text{m}\) c) \(56.31^\circ\)

- Interpret a function value as height and a derivative value as slope. - Use vertex form because the highest point is given. - The water reaches the surface at a zero of the function. - The tangent of the launch angle equals the slope at the nozzle.

1. The condition \(f(0)=2\) means the water leaves the nozzle \(2\,\text{ft}\) above the surface. The condition \(f'(4)=0\) means the path has a horizontal tangent at a horizontal distance of \(4\,\text{ft}\), where it reaches its maximum height. 2. Using vertex form with vertex \((4, 10)\), write \(f(x)=a(x-4)^2+10\). Since \(f(0)=2\), \(16a+10=2\), so \(a=-\frac{1}{2}\). Thus, \(f(x)=-\frac{1}{2}(x-4)^2+10=-\frac{1}{2}x^2+4x+2\). 3. Set \(f(x)=0\): \(-\frac{1}{2}x^2+4x+2=0\), or \(x^2-8x-4=0\). The positive solution is \(x=4+2\sqrt{5}\approx8.47\). The stream hits the water about \(8.47\,\text{ft}\) from the nozzle. 4. The initial slope is \(f'(0)\). Since \(f'(x)=-x+4\), \(f'(0)=4\). Therefore, \(\tan(\theta)=4\), so \(\theta=\arctan(4)\approx76.0^\circ\).

a) \(f(0)=2\) gives the nozzle's height, and \(f'(4)=0\) identifies the horizontal tangent at the maximum height. b) \(f(x)=-\frac{1}{2}x^2+4x+2\) c) \(4+2\sqrt{5}\,\text{ft}\approx8.47\,\text{ft}\) d) \(\theta\approx76.0^\circ\)

- Focus on the rates from right-side intervals that shrink toward \(t=3\). - Compare the limiting patterns for the two functions separately. - A whole-interval average summarizes net change over that interval, not the detailed behavior inside it.

1. The shrinking right-side interval rates for \(f\) approach \(6\). 2. The shrinking right-side interval rates for \(g\) approach \(4\). 3. Thus the right-hand instantaneous rates are \(6\) for \(f\) and \(4\) for \(g\). 4. Equal average rates on the larger interval \([2, 4]\) do not force equal local behavior at its midpoint.

At \(t=3\), \(f\) has right-hand instantaneous rate \(6\) and \(g\) has right-hand instantaneous rate \(4\). Their equal average rate on \([2, 4]\) does not determine these local rates.

- Use endpoint position change over endpoint time change for part a). - For part b), use the two times that bracket \(t=2\) most closely. - Keep the time interval used for each rate clear when comparing them.

1. The average velocity from \(t=1\) to \(t=3\) is \(\frac{27-1}{3-1}=13\,\text{m/s}\). 2. A local secant estimate around \(t=2\) is \(\frac{9.261-6.859}{2.1-1.9}=12.01\,\text{m/s}\). 3. The local estimate is slightly smaller than the longer-interval average: \(12.01\,\text{m/s}\) versus \(13\,\text{m/s}\).

a) \(13\,\text{m/s}\) b) Approximately \(12.01\,\text{m/s}\) c) The estimated instantaneous velocity is slightly smaller than the average velocity over \([1,3]\).

- Treat radius as the input and volume as the output. - Compute output change before dividing by input change. - The units should be volume units per radius unit.

1. The change in volume is \(\frac{32\pi}{3}-\frac{4\pi}{3}=\frac{28\pi}{3}\,\text{in}^3\). 2. The change in radius is \(2-1=1\,\text{in}\). 3. The average rate is \(\frac{28\pi}{3}\,\text{in}^3/\text{in}\).

\(\frac{28\pi}{3}\,\text{in}^3/\text{in}\)

- Represent each average rate as a secant slope. - Equal average rates correspond to equal secant slopes. - Keep the input differences in the same order as the output differences.

1. The slope of \(\overline{AB}\) is \(\frac{10-2}{5-1}=2\). 2. The slope of \(\overline{CD}\) is \(\frac{y-3}{2-(-2)}=\frac{y-3}{4}\). 3. Set the average rates equal: \(\frac{y-3}{4}=2\). 4. Solving gives \(y=11\).

\(y=11\)

- The upper edges lie on the x-axis, so use their coordinates in the model. - A \(50\%\) grade corresponds to derivative value \(0.5\). - The parabola's lowest point occurs at its vertex. - Report depth as a positive distance.

1. The top edges are at \(x=\pm10\), and \(f(10)=0\). 2. Since \(f'(x)=2ax\) and \(f'(10)=0.8\), \(20a=0.8\), so \(a=0.04\). 3. Use \(f(10)=0\): \(0.04(10)^2+c=0\), giving \(c=-4\). Thus \(f(x)=0.04x^2-4\). 4. The lowest point is \(f(0)=-4\), so the maximum depth is \(4\,\text{m}\). 5. A \(50\%\) slope is \(0.5\). Solve \(f'(x)=0.08x=0.5\), giving \(x=6.25\).

a) \(f(x)=0.04x^2-4\) b) \(4\,\text{m}\) c) \(6.25\,\text{m}\) from the center

- A tangential pulse follows the tangent line at the release point. - Use the derivative to find the tangent slope at each candidate input. - Write each tangent line using its release point and slope. - Evaluate each tangent line at the sensor's x-coordinate.

1. Differentiate: \(g'(x)=0.5\cos(0.5x)\). 2. At \(x_1=\frac{3\pi}{2}\approx4.712\), \(g(x_1)\approx2.707\) and \(g'(x_1)\approx-0.354\). The tangent gives \(y\approx-0.354(10-4.712)+2.707\approx0.835\), which does not match the sensor height. 3. At \(x_2=2\pi\approx6.283\), \(g(x_2)\approx2.000\) and \(g'(x_2)\approx-0.500\). The tangent gives \(y\approx-0.500(10-6.283)+2.000\approx0.142\), which matches the sensor height. 4. At \(x_3=\frac{5\pi}{2}\approx7.854\), \(g(x_3)\approx1.293\) and \(g'(x_3)\approx-0.354\). The tangent gives \(y\approx-0.354(10-7.854)+1.293\approx0.533\), which does not match the sensor height. 5. Therefore, the pulse should be released at \(x_2=2\pi\).

\(x_2=2\pi\)

- A point on the path gives a condition on the function value. - At the highest point, the tangent is horizontal. - Start with a general quadratic and solve for its coefficients. - The tangent slope and the angle with the horizontal are related by tangent.

1. The takeoff condition is \(g(0)=0\). A horizontal tangent at the highest point gives \(g'(6)=0\). The remaining given point gives \(g(10)=5\). 2. Let \(g(x)=ax^2+bx+c\). From \(g(0)=0\), \(c=0\). Since \(g'(x)=2ax+b\), the condition \(g'(6)=0\) gives \(12a+b=0\), so \(b=-12a\). Using \(g(10)=5\), \(100a+10b=5\). Substitution gives \(-20a=5\), so \(a=-\frac{1}{4}\) and \(b=3\). Therefore, \(g(x)=-\frac{1}{4}x^2+3x\). 3. The maximum occurs at \(x=6\): \(g(6)=-\frac{1}{4}(36)+18=9\). The maximum height is \(9\,\text{m}\). 4. The slope at landing is \(g'(12)=-\frac{1}{2}(12)+3=-3\). If \(\theta\) is the magnitude of the angle with the horizontal ground, then \(\tan(\theta)=3\). Thus \(\theta=\arctan(3)\approx71.6^\circ\).

a) \(g(0)=0\) and \(g'(6)=0\). b) \(g(x)=-\frac{1}{4}x^2+3x\). c) \(9\,\text{m}\). d) \(\arctan(3)\approx71.6^\circ\).