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Sinusoidal and harmonic motion models

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55218712
A simplified tide model is \(H(t)=3\sin\left(\frac{\pi}{6}t\right)+12\), where \(H(t)\) is the water depth in feet and \(t\) is time in hours. State the amplitude and the midline, and explain what each means for the water depth.

Hints

- In a sinusoidal model, amplitude measures the vertical distance from the midline to an extreme. - Look at the coefficient multiplying the sine function. - Look at the constant added outside the sine function. - Interpret both quantities using the depth units given in the model.

Solution

1. The amplitude is the absolute value of the sine coefficient, so the amplitude is \(3\,\text{ft}\). This means the modeled depth varies up to \(3\,\text{ft}\) above or below its central level. 2. The vertical shift is \(12\), so the midline is \(H=12\,\text{ft}\). This is the model's central water depth.

Answer

Amplitude: \(3\,\text{ft}\), meaning the depth varies \(3\,\text{ft}\) above or below the central level. Midline: \(H=12\,\text{ft}\), the model's central water depth.
55218812
A machine part completes one repeating sinusoidal oscillation every \(0.25\,\text{s}\). What is the frequency in hertz, and what does that frequency mean in this context?

Hints

- The period tells how long one full oscillation takes. - One hertz means one complete cycle per second. - Ask how many cycles of the stated duration fit into one second.

Solution

1. Frequency is the reciprocal of period, so \(f=\frac{1}{0.25}=4\,\text{Hz}\). 2. A frequency of \(4\,\text{Hz}\) means the part completes \(4\) full oscillations each second.

Answer

\(4\,\text{Hz}\); the part completes \(4\) full oscillations per second.
55553912
A sensor reading is modeled by \(S(t)=5\cos\left(\frac{\pi}{4}t\right)+9\), where \(t\) is measured in seconds. State the period of the model and explain what that period means for the sensor reading.

Hints

- Identify the coefficient multiplying \(t\) inside the cosine function. - Use the standard relationship between that coefficient and the length of one complete cosine cycle.

Solution

1. For \(\cos(bt)\), the period is \(\frac{2\pi}{|b|}\). 2. Here \(b=\frac{\pi}{4}\), so the period is \(\frac{2\pi}{\pi/4}=8\,\text{s}\). 3. The model repeats the same pattern every \(8\,\text{s}\).

Answer

Period: \(8\,\text{s}\). The modeled sensor-reading pattern repeats every \(8\,\text{s}\).
52659512
The temperature during a clear winter day in a mountain town is modeled by \(T(t)=-15\cos\left(\frac{\pi}{12}t\right)+30\), where \(0\le t\le 24\), \(t\) is the number of hours after midnight, and \(T(t)\) is measured in degrees Fahrenheit. Find the points on the graph where the lowest temperature occurs. Then find the difference between the highest and lowest temperatures that day.

Hints

- What are the maximum and minimum values of cosine? - How does the negative coefficient affect where the temperature is highest and lowest? - Include the endpoints of the \(24\)-hour interval. - Subtract the minimum temperature from the maximum temperature.

Solution

1. The cosine function ranges from \(-1\) to \(1\). Because its coefficient is \(-15\), \(T(t)\) is smallest when \(\cos\left(\frac{\pi}{12}t\right)=1\). 2. On \([0,24]\), this occurs at \(t=0\) and \(t=24\). Therefore, \(T(0)=T(24)=-15+30=15\). 3. The lowest-temperature points are \((0,15)\) and \((24,15)\). 4. The highest temperature occurs when the cosine is \(-1\), at \(t=12\): \(T(12)=15+30=45\). 5. The temperature difference is \(45-15=30\,^{\circ}\text{F}\).

Answer

The lowest temperature occurs at \((0,15)\) and \((24,15)\). The difference between the highest and lowest temperatures is \(30\,^{\circ}\text{F}\).
53358612
In acoustics, sounds can be superimposed. The graph shows two pure sine waves, \(f\) and \(g\), and their sum \(h=f+g\). a) Find the amplitudes of \(f\) and \(g\) from the graph. b) At \(x=\frac{\pi}{2}\), verify from the graph that \(h(x)=f(x)+g(x)\). c) At which x-values in the displayed interval do all three functions have a zero at the same time?
Figure for problem 533586

Hints

- The amplitude is the greatest vertical distance from the x-axis for each pure sine wave. - At \(x=\frac{\pi}{2}\), read the y-value of each of the three curves and compare the values numerically. - Common zeros are x-intercepts shared by all three graphs.

Solution

1. Graph \(f\) has amplitude \(2\), and graph \(g\) has amplitude \(1\). 2. At \(x=\frac{\pi}{2}\), the graph gives \(f\left(\frac{\pi}{2}\right)=2\), \(g\left(\frac{\pi}{2}\right)=-1\), and \(h\left(\frac{\pi}{2}\right)=1\). Therefore, \(1=2+(-1)\), so the displayed values satisfy \(h=f+g\) with both component waves contributing nonzero values. 3. All three functions are zero at integer multiples of \(\pi\). In the displayed interval, the common zeros are \(x=0\), \(x=\pi\), and \(x=2\pi\).

Answer

a) Amplitude of \(f\): \(2\); amplitude of \(g\): \(1\) b) \(h\left(\frac{\pi}{2}\right)=1=2+(-1)=f\left(\frac{\pi}{2}\right)+g\left(\frac{\pi}{2}\right)\) c) \(x=0\), \(x=\pi\), and \(x=2\pi\)
53015112
Consider the family of functions \(f_k(x)=(k+2)\sin(x)+k\cos(x)\), where \(k\in\mathbb{R}\). a) Show that each function can be written as \(f_k(x)=A(k)\sin(x+\phi)\), and find the amplitude \(A(k)\). b) For \(k=1\), find the amplitude and a phase angle \(\phi\in[0, 2\pi)\). Round the angle to the nearest hundredth. c) Find all values of \(k\) for which the amplitude is \(\sqrt{20}\).

Hints

- Expand \(\sin(x+\phi)\) with the angle-addition identity. - Compare the sine and cosine coefficients. - Use the signs of both coefficients to choose the correct quadrant. - Square the amplitude equation to solve for \(k\).

Solution

1. Expanding \(A\sin(x+\phi)\) gives \(A\cos(\phi)\sin(x)+A\sin(\phi)\cos(x)\). Therefore, \(A\cos(\phi)=k+2\) and \(A\sin(\phi)=k\). 2. Squaring and adding gives \(A(k)=\sqrt{(k+2)^2+k^2}=\sqrt{2k^2+4k+4}\). 3. For \(k=1\), \(A=\sqrt{10}\). Also, \(\cos(\phi)=\frac{3}{\sqrt{10}}\) and \(\sin(\phi)=\frac{1}{\sqrt{10}}\), so \(\phi\approx0.32\) radians. 4. Set \(\sqrt{2k^2+4k+4}=\sqrt{20}\). Squaring and simplifying gives \(k^2+2k-8=0\), so \(k=2\) or \(k=-4\).

Answer

a) \(A(k)=\sqrt{2k^2+4k+4}\) b) \(A=\sqrt{10}\), \(\phi\approx0.32\) radians c) \(k=2\) or \(k=-4\)
53240812
The water level in a reservoir is controlled periodically. The graph shows the level during a \(6\)-hour interval. The horizontal axis gives the time in hours since measurements began, and the vertical axis gives the water level in meters relative to a reference elevation. a) Read the maximum level, minimum level, amplitude, midline, and period from the graph. b) Find \(b\) and \(c\) for a model of the form \(f(x)=a\sin(b(x-c))+d\), where \(0 \leq c < 4\). Write the complete model.
Figure for problem 532408

Hints

- Use the maximum and minimum levels to find the amplitude and midline. - Find the period from the time between consecutive maxima. - Use \(b=\frac{2\pi}{p}\). - Locate an increasing midline crossing to determine \(c\).

Solution

1. The maximum level is \(2.5\,\text{m}\), and the minimum level is \(-0.5\,\text{m}\). 2. The amplitude is \(a=\frac{2.5-(-0.5)}{2}=1.5\,\text{m}\). 3. The midline is \(d=\frac{2.5+(-0.5)}{2}=1\,\text{m}\). 4. Consecutive maxima occur at \(x=2\) and \(x=6\), so the period is \(4\) hours. 5. Therefore, \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 6. The graph crosses the midline while increasing at \(x=1\), so \(c=1\). 7. The model is \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\).

Answer

a) Maximum \(2.5\,\text{m}\), minimum \(-0.5\,\text{m}\), amplitude \(1.5\,\text{m}\), midline \(1\,\text{m}\), period \(4\) hours b) \(b=\frac{\pi}{2}\), \(c=1\); \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\)
53353412
A meteorologist models the temperature during a summer day with \(T(t)=a\sin(b(t-c))+d\), where \(t\) is the number of hours after midnight and \(T\) is the temperature in degrees Fahrenheit. a) Find the maximum and minimum temperatures and the times when they occur. b) Find the amplitude \(a\) and vertical shift \(d\). c) Find \(b\), assuming a period of \(24\) hours. d) Find the horizontal shift \(c\) from an appropriate midline crossing.
Figure for problem 533534

Hints

- Average the maximum and minimum temperatures to find \(d\). - The amplitude is the distance from the midline to an extreme temperature. - Use \(b=\frac{2\pi}{p}\). - Find where the graph crosses the midline while increasing.

Solution

1. The maximum temperature is \(80\,^{\circ}\text{F}\) at \(t=14\), or 2:00 p.m. The minimum is \(60\,^{\circ}\text{F}\) at \(t=2\), or 2:00 a.m. 2. The midline is \(d=\frac{80+60}{2}=70\), and the amplitude is \(a=80-70=10\). 3. With a period of \(24\) hours, \(b=\frac{2\pi}{24}=\frac{\pi}{12}\). 4. The graph crosses the midline while increasing at \(t=8\), so \(c=8\).

Answer

a) Maximum \(80\,^{\circ}\text{F}\) at 2:00 p.m.; minimum \(60\,^{\circ}\text{F}\) at 2:00 a.m. b) \(a=10\), \(d=70\) c) \(b=\frac{\pi}{12}\) d) \(c=8\)
53379612
A harbor records the water depth over a \(24\)-hour period. The depth is modeled by \(h(t)=10\sin\left(\frac{\pi}{6}(t-2)\right)+35\), where \(h\) is measured in feet and \(t\) is the number of hours after midnight. The graph shows this model. a) What is the water depth at 4:00 a.m., when \(t=4\)? b) A cargo ship needs a water depth of at least \(40\,\text{ft}\) to enter safely. During which time interval in the first \(12\) hours can the ship enter?
Figure for problem 533796

Hints

- At \(t=4\), read the graph or substitute into the model. - Compare the graph with the horizontal line \(h=40\). - Find where the graph is on or above that line during the first \(12\) hours.

Solution

1. Evaluate the model at \(t=4\): \(h(4)=10\sin\left(\frac{\pi}{6}(4-2)\right)+35=10\sin\left(\frac{\pi}{3}\right)+35=5\sqrt{3}+35\approx43.66\,\text{ft}\). 2. To find when \(h(t) \geq 40\), identify where the graph meets the horizontal line \(h=40\). In the first \(12\) hours, the intersections occur at \(t=3\) and \(t=7\). 3. The graph is at or above \(40\,\text{ft}\) for \(3 \leq t \leq 7\). Therefore, the safe interval is from 3:00 a.m. through 7:00 a.m.

Answer

a) \(35+5\sqrt{3}\approx43.66\,\text{ft}\) b) From 3:00 a.m. through 7:00 a.m., inclusive
53381912
A decorative metal panel has a wavy top edge modeled by the graph. Both coordinates are measured in inches. a) Find an equation in the form \(f(x)=a\sin(bx)+d\). b) Find the height of the panel at \(x=1\).
Figure for problem 533819

Hints

- Find the horizontal midline. - Use the distance from the midline to a maximum. - Find the horizontal length of one complete cycle. - Use the period to calculate the inside coefficient. - Substitute \(x=1\) into your equation.

Solution

1. The graph oscillates about the midline \(y=3\), so \(d=3\). 2. The maximum is \(5\), and the minimum is \(1\), so the amplitude is \(a=2\). 3. One complete cycle has length \(8\), so the period is \(8\), and \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 4. Therefore, \(f(x)=2\sin\left(\frac{\pi}{4}x\right)+3\). 5. At \(x=1\), \(f(1)=2\sin\left(\frac{\pi}{4}\right)+3=\sqrt{2}+3\approx4.41\).

Answer

a) \(f(x)=2\sin\left(\frac{\pi}{4}x\right)+3\) b) \(3+\sqrt{2}\approx4.41\,\text{in.}\)
53382012
A physics lab compares two oscillations on an oscilloscope. The graphs of functions \(g\) and \(h\) show displacement in millimeters over time in seconds. a) Find a suitable equation of the form \(f(t)=a\sin(bt)+d\) for each graph. b) Compare the amplitudes and frequencies of the two oscillations.
Figure for problem 533820

Hints

- For each curve, identify the greatest displacement from its midline. - Use the time required for one complete cycle of each graph. - Use \(b=\frac{2\pi}{P}\), where \(P\) is the period. - Relate frequency to the length of the period.

Solution

1. Both curves oscillate about \(y=2.5\), so \(d=2.5\). 2. For \(g\), the amplitude is \(1.5\,\text{mm}\) and the period is \(4\,\text{s}\). Therefore, \(b_g=\frac{2\pi}{4}=\frac{\pi}{2}\), and \(g(t)=1.5\sin\left(\frac{\pi}{2}t\right)+2.5\). 3. For \(h\), the amplitude is \(0.5\,\text{mm}\) and the period is \(2\,\text{s}\). Therefore, \(b_h=\frac{2\pi}{2}=\pi\), and \(h(t)=0.5\sin(\pi t)+2.5\). 4. The amplitude of \(h\) is one-third the amplitude of \(g\). The period of \(h\) is half the period of \(g\), so the frequency of \(h\) is twice the frequency of \(g\).

Answer

a) \(g(t)=1.5\sin\left(\frac{\pi}{2}t\right)+2.5\) and \(h(t)=0.5\sin(\pi t)+2.5\) b) The amplitude of \(h\) is one-third the amplitude of \(g\), and the frequency of \(h\) is twice the frequency of \(g\).
53404212
A Ferris wheel has a diameter of \(100\,\text{ft}\). Its highest point is \(110\,\text{ft}\) above the ground. The graph shows the rider's normalized vertical position over time; its vertical scale is not a height scale. a) Use the graph to determine the period of the motion and whether the rider starts at the highest point, the lowest point, or the midline. b) Use the Ferris wheel dimensions together with the graph to write an equation of the form \(h(t)=a\cos(bt)+d\), where \(h\) is the rider's height in feet and \(t\) is time in minutes.
Figure for problem 534042

Hints

- Use repeated corresponding positions on the normalized graph to determine one full cycle. - The graph also shows which point in the cycle occurs at \(t=0\). - Use the Ferris wheel diameter and highest height, not the graph's vertical scale, to determine the amplitude and midline of the height model. - Relate the period to the coefficient inside cosine, then choose the sign of the cosine coefficient from the starting position.

Solution

1. Consecutive minima on the normalized graph occur at \(t=0\) and \(t=4\), so the period is \(4\) minutes. At \(t=0\), the graph is at a minimum, so the rider starts at the lowest point. 2. The Ferris wheel radius is \(\frac{100}{2}=50\,\text{ft}\), so the amplitude of the height model is \(50\,\text{ft}\). 3. The lowest height is \(110-100=10\,\text{ft}\). The midline is \(d=\frac{110+10}{2}=60\,\text{ft}\). 4. With period \(4\), \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 5. Because the rider starts at the lowest point, the cosine coefficient is negative. Therefore, \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\).

Answer

a) Period: \(4\) minutes; the rider starts at the lowest point. b) \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\)
53497512
A thermometer records the outdoor temperature \(T\), in degrees Fahrenheit, over a \(24\)-hour period. The graph shows a sinusoidal daily cycle as a function of time \(t\), in hours. a) Determine the amplitude, midline, and period of the cycle. b) Write a model in the form \(T(t)=a\cos(bt)+d\), with \(b>0\) and no horizontal shift. c) Use the period of your model to predict the first two times after \(t=24\) when the temperature reaches its maximum value of \(80\,^{\circ}\text{F}\).
Figure for problem 534975

Hints

- Use the highest and lowest temperatures to determine the vertical features of the cycle. - Compare matching positions in consecutive cycles to determine the repeat length. - With no horizontal shift, inspect whether the graph begins at a maximum or a minimum to determine the sign of the cosine coefficient. - Extend a known maximum time by whole periods rather than extending the graph visually.

Solution

1. The graph has maximum \(80\,^{\circ}\text{F}\) and minimum \(50\,^{\circ}\text{F}\). The amplitude is \(\frac{80-50}{2}=15\,^{\circ}\text{F}\), and the midline is \(\frac{80+50}{2}=65\,^{\circ}\text{F}\). 2. The cycle runs from a minimum at \(t=0\) to the next minimum at \(t=24\), so the period is \(24\,\text{h}\). Therefore, \(b=\frac{2\pi}{24}=\frac{\pi}{12}\). 3. With no horizontal shift, \(\cos(0)=1\). Because the graph starts at the minimum, the cosine coefficient must be negative: \(a=-15\). Thus, \(T(t)=-15\cos\left(\frac{\pi}{12}t\right)+65\). 4. A maximum occurs at \(t=12\). Repeating every \(24\) hours gives later maxima at \(t=12+24k\). The first two after \(t=24\) are \(t=36\) and \(t=60\).

Answer

a) Amplitude: \(15\,^{\circ}\text{F}\); midline: \(T=65\,^{\circ}\text{F}\); period: \(24\,\text{h}\) b) \(T(t)=-15\cos\left(\frac{\pi}{12}t\right)+65\) c) \(t=36\,\text{h}\) and \(t=60\,\text{h}\)
55097312
A mass attached to a spring oscillates about its equilibrium position. The graph shows its signed displacement \(d\), in centimeters, as a function of time \(t\), in seconds. Positive displacement is one direction from equilibrium and negative displacement is the opposite direction. a) Read the amplitude and period from the graph. Then find the frequency in hertz. b) Write a model of the form \(d(t)=A\cos(\omega t)\) that matches the graph. c) Find the first time \(t>0\) when the displacement is \(-3\,\text{cm}\).
Figure for problem 550973

Hints

- Measure the greatest distance from the equilibrium line for the amplitude, and compare consecutive matching peaks for the period. - Frequency tells how many cycles occur per second, while angular frequency is the coefficient of time inside the cosine function. - The graph starts at an extreme displacement rather than at equilibrium; use that to choose between a basic sine and cosine model. - For the requested displacement, set the model equal to the target value and choose the earliest positive solution.

Solution

1. The graph reaches \(6\,\text{cm}\) above and below equilibrium, so the amplitude is \(6\,\text{cm}\). Consecutive maxima occur \(2\,\text{s}\) apart, so the period is \(2\,\text{s}\). The frequency is the reciprocal of the period: \(\frac{1}{2}\,\text{Hz}\). 2. The graph begins at a positive maximum, so a cosine model with no phase shift is appropriate. With period \(2\), \(\omega=\frac{2\pi}{2}=\pi\). Thus \(d(t)=6\cos(\pi t)\). 3. Set \(6\cos(\pi t)=-3\), so \(\cos(\pi t)=-\frac{1}{2}\). The first positive angle with cosine \(-\frac{1}{2}\) is \(\frac{2\pi}{3}\). Therefore, \(\pi t=\frac{2\pi}{3}\), giving \(t=\frac{2}{3}\,\text{s}\).

Answer

a) Amplitude: \(6\,\text{cm}\); period: \(2\,\text{s}\); frequency: \(\frac{1}{2}\,\text{Hz}\) b) \(d(t)=6\cos(\pi t)\) c) \(t=\frac{2}{3}\,\text{s}\)
55553012
A sensor records a quantity at equal time intervals. The plotted measurements are noisy but appear periodic. Estimate the midline and amplitude to the nearest whole unit and the period to the nearest whole time unit. Then write a sine model using those estimates and a rising midline crossing at \(t=0\).
Figure for problem 555530

Hints

- Use the overall high and low clusters rather than treating one noisy measurement as an exact extreme. - The midline is halfway between the typical high and low levels. - Compare similar peaks to estimate the repeat length. - Use the direction of the data near \(t=0\) to choose the phase of a sine model.

Solution

1. The measurements have highs near \(25\) and lows near \(15\), so the estimated midline is \(\frac{25+15}{2}=20\) and the estimated amplitude is \(\frac{25-15}{2}=5\). 2. High points occur near \(t=2\) and \(t=10\), so the estimated period is \(8\) time units. 3. For period \(8\), the angular coefficient is \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 4. The data are near the midline and increasing at \(t=0\), so a reasonable estimated model is \(S(t)=5\sin\left(\frac{\pi}{4}t\right)+20\).

Answer

Estimated midline: \(20\); estimated amplitude: \(5\); estimated period: \(8\) One reasonable model is \(S(t)=5\sin\left(\frac{\pi}{4}t\right)+20\).
55553112
Two data sets are shown as unconnected measurements. a) Which data set provides stronger evidence for a sinusoidal model? Give two features of the data that support your choice. b) Estimate the period and midline of the data set you chose. c) Explain why the other data set does not yet provide strong evidence of sinusoidal periodic behavior.
Figure for problem 555531

Hints

- A sinusoidal pattern should repeatedly move above and below a central level. - Look for recurring peaks or troughs at approximately equal horizontal spacing. - Distinguish “not enough evidence yet” from proving that a sinusoidal model is impossible.

Solution

1. Data set a provides stronger evidence for a sinusoidal model. It repeatedly rises above and falls below a stable central level, and its peaks recur at approximately regular horizontal intervals. 2. The values are centered near \(5\), so the estimated midline is \(y=5\). Peaks occur near \(t=2\) and \(t=10\), giving an estimated period of \(8\). 3. Data set b rises throughout the displayed measurements and does not show a repeated peak-trough-midline pattern. A portion of a sinusoid can rise over a limited interval, so these observations alone are insufficient to establish repeating sinusoidal behavior.

Answer

a) Data set a; it oscillates around a stable center and repeats a peak-trough pattern at roughly regular intervals. b) Estimated period: \(8\); estimated midline: \(y=5\) c) Data set b shows only sustained increase over the observed interval, with no repeated cycle, so the data do not yet establish sinusoidal periodic behavior.
55553212
A proposed model is \(M(t)=10+3\sin\left(\frac{\pi}{2}t\right)\). For this problem, define the residual as observed value minus model value. <table> <tr><th>\(t\)</th><th>Observed value</th></tr> <tr><td>\(0\)</td><td>\(10.4\)</td></tr> <tr><td>\(1\)</td><td>\(12.6\)</td></tr> <tr><td>\(2\)</td><td>\(9.7\)</td></tr> <tr><td>\(3\)</td><td>\(7.4\)</td></tr> <tr><td>\(4\)</td><td>\(10.2\)</td></tr> </table> a) Find the residual at each listed time. b) Find the maximum absolute residual. c) A model is considered acceptable for this application if every listed observation is within \(0.5\) unit of the model. Does \(M\) meet that criterion?

Hints

- Evaluate the model at the listed integer inputs before comparing it with the observations. - Keep the specified subtraction order for every residual. - For the tolerance test, compare absolute residuals rather than signed residuals.

Solution

1. The model values at \(t=0,1,2,3,4\) are \(10\), \(13\), \(10\), \(7\), and \(10\), respectively. 2. Subtracting model value from observed value gives residuals \(0.4\), \(-0.4\), \(-0.3\), \(0.4\), and \(0.2\). 3. The absolute residuals are \(0.4\), \(0.4\), \(0.3\), \(0.4\), and \(0.2\), so the maximum absolute residual is \(0.4\). 4. Since \(0.4\le0.5\), every listed observation is within the required tolerance. The model meets the criterion.

Answer

a) Residuals for \(t=0,1,2,3,4\): \(0.4,-0.4,-0.3,0.4,0.2\) b) \(0.4\) c) Yes. The maximum absolute residual is \(0.4\), which is at most \(0.5\).
55553312
A laboratory controller follows an exact sinusoidal cycle. The table gives exact measurements from the cycle. <table> <tr><th>Time \(t\)</th><th>Output \(R(t)\)</th></tr> <tr><td>\(1\)</td><td>\(16\)</td></tr> <tr><td>\(3\)</td><td>\(20\)</td></tr> <tr><td>\(5\)</td><td>\(16\)</td></tr> <tr><td>\(7\)</td><td>\(12\)</td></tr> <tr><td>\(9\)</td><td>\(16\)</td></tr> </table> a) Determine the amplitude, midline, and period. b) Write a model in the form \(R(t)=a\sin(b(t-c))+d\) with \(a>0\) and \(0\le c<P\), where \(P\) is the period. c) Write an equivalent cosine model with a positive coefficient whose phase shift is the time of a maximum. d) Use periodicity to find \(R(11)\) without substituting into either formula.

Hints

- Use the maximum and minimum table values to determine the vertical features. - Look for matching midline positions with the same direction of motion to identify a full period. - A positive sine model naturally starts at a rising midline crossing, while a positive cosine model naturally starts at a maximum. - For the final part, compare \(11\) with a time in the table that differs by exactly one period.

Solution

1. The maximum is \(20\) and the minimum is \(12\). Thus, the amplitude is \(\frac{20-12}{2}=4\), and the midline is \(\frac{20+12}{2}=16\). 2. The table goes from a rising midline value at \(t=1\) to the next rising midline value at \(t=9\), so the period is \(P=8\). Therefore, \(b=\frac{2\pi}{8}=\frac{\pi}{4}\). 3. With positive sine amplitude, the rising midline crossing gives \(c=1\). Thus, \(R(t)=4\sin\left(\frac{\pi}{4}(t-1)\right)+16\). 4. The maximum occurs at \(t=3\), so an equivalent positive-cosine model is \(R(t)=4\cos\left(\frac{\pi}{4}(t-3)\right)+16\). 5. Since the period is \(8\), \(R(11)=R(3)=20\).

Answer

a) Amplitude: \(4\); midline: \(R=16\); period: \(8\) b) \(R(t)=4\sin\left(\frac{\pi}{4}(t-1)\right)+16\) c) \(R(t)=4\cos\left(\frac{\pi}{4}(t-3)\right)+16\) d) \(R(11)=20\)
52375112
A periodic process is modeled by \(f(x)=a\sin(b(x-c))+d\), where \(a>0\) and \(0 \leq c < p\), and \(p\) is the period. A minimum occurs at \((1, 1)\), and the next maximum occurs at \((4, 5)\). There are no other extrema between these points. Find \(a\), \(b\), \(c\), and \(d\).

Hints

- Use the two y-values to find the midline and amplitude. - The distance from a minimum to the next maximum is half a period. - Use \(b=\frac{2\pi}{p}\). - For \(a>0\), locate the upward midline crossing one-fourth of a period before the maximum.

Solution

1. The midline is halfway between the minimum and maximum values: \(d=\frac{1+5}{2}=3\). 2. The amplitude is the distance from the midline to either extreme: \(a=5-3=2\). 3. The horizontal distance from a minimum to the next maximum is half a period. Thus, \(\frac{p}{2}=4-1=3\), so \(p=6\). 4. Therefore, \(b=\frac{2\pi}{p}=\frac{\pi}{3}\). 5. For a positive sine coefficient, the maximum occurs one-fourth of a period after the upward midline crossing at \(x=c\). Hence, \(4=c+\frac{6}{4}\), so \(c=\frac{5}{2}\), which satisfies \(0 \leq c < 6\).

Answer

\(a=2\), \(b=\frac{\pi}{3}\), \(c=\frac{5}{2}\), \(d=3\)
53015212
Consider the family of functions \(g_t(x)=t\sin(x)+\sqrt{11}\cos(x)\), where \(t\in\mathbb{R}\). a) Find the maximum and minimum values in terms of \(t\). b) For \(t=\sqrt{5}\), write \(g_{\sqrt{5}}(x)=A\sin(x+\alpha)\), where \(A>0\) and \(\alpha\in[0, 2\pi)\). Find \(A\) and round \(\alpha\) to the nearest hundredth. c) Find \(t>0\) so that the first maximum in \([0, 2\pi)\) occurs at \(x=\frac{\pi}{3}\).

Hints

- Use the amplitude formula for a sine-cosine combination. - Match coefficients after expanding \(A\sin(x+\alpha)\). - The sine function reaches a maximum when its argument is \(\frac{\pi}{2}\). - Relate the phase angle to the coefficient ratio with tangent.

Solution

1. The amplitude of \(p\sin(x)+q\cos(x)\) is \(\sqrt{p^2+q^2}\). Thus the maximum is \(\sqrt{t^2+11}\), and the minimum is \(-\sqrt{t^2+11}\). 2. For \(t=\sqrt{5}\), \(A=\sqrt{5+11}=4\). Matching coefficients gives \(\cos(\alpha)=\frac{\sqrt{5}}{4}\) and \(\sin(\alpha)=\frac{\sqrt{11}}{4}\). Both are positive, so \(\alpha\approx0.98\) radians. 3. A maximum of \(A\sin(x+\alpha)\) occurs when \(x+\alpha=\frac{\pi}{2}\). Requiring the first maximum at \(x=\frac{\pi}{3}\) gives \(\alpha=\frac{\pi}{6}\). 4. Since \(\tan(\alpha)=\frac{\sqrt{11}}{t}\), \(\frac{1}{\sqrt{3}}=\frac{\sqrt{11}}{t}\). Therefore, \(t=\sqrt{33}\).

Answer

a) Maximum: \(\sqrt{t^2+11}\); minimum: \(-\sqrt{t^2+11}\) b) \(A=4\), \(\alpha\approx0.98\) radians c) \(t=\sqrt{33}\)
53265112
The solid graph of \(f\) is the sum of a linear trend \(g(x)=mx+c\) and a sinusoidal component \(h(x)=a\sin(bx)\), so \(f(x)=g(x)+h(x)\). The dashed line shows the trend \(g\). a) Use two points where the solid curve and dashed trend line intersect to find the equation of \(g\). b) Find the amplitude and period of \(h\), and use the period to calculate \(b\). c) Write the complete equation for \(f(x)\).
Figure for problem 532651

Hints

- The sinusoidal component is zero wherever the solid curve meets the dashed trend line. - Use two such intersections to determine the slope and y-intercept of the dashed line. - Measure the greatest vertical separation between the solid curve and the trend line to determine the amplitude of the oscillation. - Compare matching positions of the oscillation relative to the dashed line to find the period.

Solution

1. The solid curve and dashed trend line intersect at points including \((0,1)\) and \((2,2)\). The slope is \(m=\frac{2-1}{2-0}=0.5\), and the y-intercept is \(1\). Therefore, \(g(x)=0.5x+1\). 2. At \(x=1\), the solid graph has \(f(1)=3.5\), while the trend line has \(g(1)=1.5\). Their vertical difference is \(2\), so the amplitude of \(h\) is \(2\). 3. The oscillation relative to the trend repeats every \(4\) units, so the period of \(h\) is \(4\). Thus, \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 4. Therefore, \(f(x)=0.5x+1+2\sin\left(\frac{\pi}{2}x\right)\).

Answer

a) \(g(x)=0.5x+1\) b) Amplitude \(2\), period \(4\), \(b=\frac{\pi}{2}\) c) \(f(x)=0.5x+1+2\sin\left(\frac{\pi}{2}x\right)\)

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