Free math worksheets

- 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.

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.

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.

- 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.

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.

\(4\,\text{Hz}\); the part completes \(4\) full oscillations per second.

- 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.

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 \(45-15=30\,^{\circ}\text{F}\).

The lowest temperature occurs at \((0,15)\) and \((24,15)\). The difference between the highest and lowest temperatures is \(30\,^{\circ}\text{F}\).

- The amplitude is the greatest distance from the midline. - Read the value of each graph at \(x=\pi\). - Common zeros are x-intercepts shared by all three graphs.

1. Graph \(f\) has amplitude \(2\), and graph \(g\) has amplitude \(1\). 2. At \(x=\pi\), \(f(\pi)=0\), \(g(\pi)=0\), and \(h(\pi)=0\). Therefore, \(0=0+0\), so the sum relationship is satisfied. 3. All three functions are zero at integer multiples of \(\pi\). In the displayed interval, the common zeros are \(x=0\), \(x=\pi\), and \(x=2\pi\).

a) Amplitude of \(f\): \(2\); amplitude of \(g\): \(1\) b) \(h(\pi)=0=0+0=f(\pi)+g(\pi)\) c) \(x=0\), \(x=\pi\), and \(x=2\pi\)

- Examine only the portion of the graph within the given interval. - Identify the highest and lowest points in that portion. - Include the values at both endpoints when checking for extrema.

1. On \([0, 6]\), the graph increases from \(50\,^{\circ}\text{F}\) at \(t=0\) to \(65\,^{\circ}\text{F}\) at \(t=6\). Therefore, the minimum is \(50\,^{\circ}\text{F}\), and the maximum is \(65\,^{\circ}\text{F}\). 2. On \([6, 18]\), the graph reaches its highest point at \(t=12\), where \(T=80\,^{\circ}\text{F}\). At both endpoints, \(t=6\) and \(t=18\), the temperature is \(65\,^{\circ}\text{F}\). Therefore, the maximum is \(80\,^{\circ}\text{F}\), and the minimum is \(65\,^{\circ}\text{F}\). 3. On \([12, 24]\), the graph decreases from \(80\,^{\circ}\text{F}\) at \(t=12\) to \(50\,^{\circ}\text{F}\) at \(t=24\). Therefore, the maximum is \(80\,^{\circ}\text{F}\), and the minimum is \(50\,^{\circ}\text{F}\).

a) Maximum: \(65\,^{\circ}\text{F}\); minimum: \(50\,^{\circ}\text{F}\) b) Maximum: \(80\,^{\circ}\text{F}\); minimum: \(65\,^{\circ}\text{F}\) c) Maximum: \(80\,^{\circ}\text{F}\); minimum: \(50\,^{\circ}\text{F}\)

- 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 \(k\).

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 \(k=-4\).

a) \(A(k)=\sqrt{2k^2+4k+4}\) b) \(A=\sqrt{10}\), \(\phi\approx0.32\) radians c) \(k=2\) or \(k=-4\)

- 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 \(c\).

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 \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\).

a) Maximum \(2.5\,\text{m}\), minimum \(-0.5\,\text{m}\), amplitude \(1.5\,\text{m}\), midline \(1\,\text{m}\), period \(4\) hours b) \(b=\frac{\pi}{2}\), \(c=1\); \(f(x)=1.5\sin\left(\frac{\pi}{2}(x-1)\right)+1\)

- 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.

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 \(c=8\).

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) \(c=8\)

- 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.

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.

a) \(35+5\sqrt{3}\approx43.66\,\text{ft}\) b) From 3:00 a.m. through 7:00 a.m., inclusive

- 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.

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 \(x=1\), \(f(1)=2\sin\left(\frac{\pi}{4}\right)+3=\sqrt{2}+3\approx4.41\).

a) \(f(x)=2\sin\left(\frac{\pi}{4}x\right)+3\) b) \(3+\sqrt{2}\approx4.41\,\text{in.}\)

- 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.

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 \(g\).

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 \(g\).

- Compare the times of two consecutive lowest points on the graph. - Use the diameter to find the radius and amplitude. - Find the lowest point, then average the highest and lowest heights to locate the midline. - Decide whether the cosine coefficient should be positive or negative from the starting position.

1. Consecutive lowest points occur at \(t=0\) and \(t=4\), so the period is \(4\) minutes. 2. The Ferris wheel has radius \(\frac{100}{2}=50\,\text{ft}\), so the magnitude of the amplitude is \(50\). 3. The lowest point 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, use a negative cosine coefficient. Therefore, \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\).

a) Period: \(4\) minutes b) \(h(t)=-50\cos\left(\frac{\pi}{2}t\right)+60\)

- 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.

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 \(t=\frac{2}{3}\,\text{s}\).

a) Amplitude: \(6\,\text{cm}\); period: \(2\,\text{s}\); frequency: \(\frac{1}{2}\,\text{Hz}\) b) \(d(t)=6\cos(\pi t)\) c) \(t=\frac{2}{3}\,\text{s}\)

- 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.

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 \(0 \leq c < 6\).

\(a=2\), \(b=\frac{\pi}{3}\), \(c=\frac{5}{2}\), \(d=3\)

- 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.

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, \(t=\sqrt{33}\).

a) Maximum: \(\sqrt{t^2+11}\); minimum: \(-\sqrt{t^2+11}\) b) \(A=4\), \(\alpha\approx0.98\) radians c) \(t=\sqrt{33}\)

- Locate points where the graph crosses its varying midline. - Use those points to determine the slope and y-intercept of the trend line. - Use the greatest vertical deviation from the trend line to find the amplitude. - Find the length of one complete oscillation, and use it to determine \(b\).

1. The midline crossings include \((-2, 0)\), \((0, 1)\), \((2, 2)\), \((4, 3)\), and \((6, 4)\). These points lie on a line with slope \(m=\frac{2-1}{2-0}=0.5\) and y-intercept \(1\). Therefore, \(g(x)=0.5x+1\). 2. At \(x=1\), the graph has \(f(1)=3.5\), while the trend line has \(g(1)=1.5\). The maximum deviation is \(3.5-1.5=2\), so the amplitude is \(2\). 3. The sinusoidal pattern repeats every \(4\) units, so the period is \(4\), and \(b=\frac{2\pi}{4}=\frac{\pi}{2}\). 4. Therefore, \(f(x)=0.5x+1+2\sin\left(\frac{\pi}{2}x\right)\).

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