A Wave That Starts at a Maximum

A graph has a maximum of $5$ at $x = 0$, a minimum of $1$, and a period of $2\pi$. Write its equation.

Find the midline D: $D = \frac{5 + 1}{2} = \frac{6}{2}$ = $D = 3$

Find the amplitude A: $A = \frac{5 - 1}{2} = \frac{4}{2}$ = $A = 2$

Find B from the period: $B = \frac{2\pi}{2\pi}$ = $B = 1$

Choose the form and C: The wave starts at a maximum at $x = 0$, so use cosine with $C = 0$ = $C = 0$

Answer: $y = 2\cos(x) + 3$

A Cosine Equation from a Description

A wave has a maximum of $4$ at $x = \frac{\pi}{4}$, a minimum of $-2$, and a period of $\pi$. Write a cosine equation.

Find the midline D: $D = \frac{4 + (-2)}{2} = \frac{2}{2}$ = $D = 1$

Find the amplitude A: $A = \frac{4 - (-2)}{2} = \frac{6}{2}$ = $A = 3$

Find B: $B = \frac{2\pi}{\pi}$ = $B = 2$

Find C from the maximum: Cosine starts at a maximum, and the maximum is at $x = \frac{\pi}{4}$ = $C = \frac{\pi}{4}$

Check one point: At $x = \frac{\pi}{4}$: $3\cos(0) + 1 = 3 + 1$ = $4$, the maximum

Answer: $y = 3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$

The Same Wave as a Sine Equation

Write the wave from the previous example (max $4$ at $x = \frac{\pi}{4}$, min $-2$, period $\pi$) as a sine equation.

Keep A, B and D: They do not depend on the form = $A = 3$, $B = 2$, $D = 1$

Find a quarter period: One quarter of the period: $\frac{1}{4} \cdot \pi$ = $\frac{\pi}{4}$

Find where the wave crosses the midline going up: Sine reaches its maximum a quarter period after it starts: $\frac{\pi}{4} - \frac{\pi}{4}$ = $C = 0$

Check one point: At $x = \frac{\pi}{4}$: $3\sin\left(\frac{\pi}{2}\right) + 1 = 3 + 1$ = $4$, the maximum

Answer: $y = 3\sin(2x) + 1$ draws exactly the same wave as $y = 3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$

Ferris Wheel Height

A Ferris wheel has a diameter of $40$ m and its center is $22$ m above the ground. It turns once every $8$ minutes. You board at the bottom at $t = 0$. Write your height $h$ (in m) after $t$ minutes.

Find the highest and lowest height: Radius $= 20$ m: lowest $22 - 20$, highest $22 + 20$ = min $= 2$ m, max $= 42$ m

Find D and the amplitude: $D = \frac{42 + 2}{2}$, amplitude $= \frac{42 - 2}{2}$ = $D = 22$, amplitude $= 20$

Find B: $B = \frac{2\pi}{8}$ = $B = \frac{\pi}{4}$

Choose the form: At $t = 0$ you are at the minimum. A cosine with a negative $A$ starts at a minimum = $A = -20$, $C = 0$

Check: $t = 0$: $-20\cos(0) + 22 = 2$; $t = 4$: $-20\cos(\pi) + 22 = 42$ = Bottom at the start, top after half a turn

Answer: $h(t) = -20\cos\left(\frac{\pi}{4}t\right) + 22$, or the same wave as $h(t) = 20\cos\left(\frac{\pi}{4}(t - 4)\right) + 22$

Tide Model

In a harbor the water is $7$ m deep at high tide at 2:00 and $3$ m deep at the next low tide at 8:00. Write a model for the depth $y$ after $x$ hours (after midnight) and find the depth at 6:00.

Find the period: High to low tide is half a cycle: $8 - 2 = 6$ hours, so the period is $2 \cdot 6$ = period $= 12$ hours

Find A and D: $A = \frac{7 - 3}{2}$, $D = \frac{7 + 3}{2}$ = $A = 2$, $D = 5$

Find B: $B = \frac{2\pi}{12}$ = $B = \frac{\pi}{6}$

Find C: Use cosine: the maximum is at $x = 2$ = $C = 2$

Find the depth at 6:00: $y = 2\cos\left(\frac{\pi}{6}(6 - 2)\right) + 5 = 2\cos\left(\frac{2\pi}{3}\right) + 5 = 2 \cdot \left(-\frac{1}{2}\right) + 5$ = $y = 4$ m

Answer: $y = 2\cos\left(\frac{\pi}{6}(x - 2)\right) + 5$; at 6:00 the water is $4$ m deep

Mistake: Using the maximum as the amplitude (max $7$, min $3$, so $A = 7$)

Why: The amplitude is measured from the midline, not from the x-axis. The wave is only $2$ units above its midline $y = 5$.

Correct: $A = \frac{\text{max} - \text{min}}{2} = \frac{7 - 3}{2} = 2$, and the $5$ goes into $D$.

Mistake: Writing the period as B (period $12$, so $B = 12$)

Why: B is not the period. A bigger B makes the wave repeat faster, so the period gets shorter.

Correct: $B = \frac{2\pi}{\text{period}} = \frac{2\pi}{12} = \frac{\pi}{6}$.

Mistake: Getting the sign of C wrong: a maximum at $x = 2$ written as $\cos(B(x + 2))$

Why: Inside the bracket the shift has the opposite sign. $(x + 2)$ moves the wave to the left.

Correct: A maximum at $x = 2$ means $\cos(B(x - 2))$. Check: at $x = 2$ the bracket is $0$ and $\cos(0) = 1$.

Mistake: Using sine and putting C at the maximum

Why: Sine does not start at a maximum. At $x = C$ the sine wave is on the midline, going up.

Correct: For sine take the midline crossing going up, which is a quarter period before the maximum. Or use cosine and take the maximum.

Hours of Daylight

Astronomers and solar engineers model the length of the day with a cosine wave that repeats every year.

In a northern city the longest day (day $172$, around June 21) has $16$ hours of daylight and the shortest has $8$. A model is $y = 4\cos\left(\frac{2\pi}{365}(d - 172)\right) + 12$.

Ferris Wheels

Ride designers use the height equation to check how fast riders climb and how long they spend near the top.

A wheel $50$ m across with its center $30$ m high turns once every $12$ minutes. Boarding at the bottom gives $h(t) = -25\cos\left(\frac{\pi}{6}t\right) + 30$.

A Weight on a Spring

Engineers model vibrations (car suspensions, bridges, sensors) with the same equations as a weight bouncing on a spring.

A weight bounces between $4$ cm and $16$ cm above a table and needs $2$ seconds for one full bounce.

Q: Should I use sine or cosine?

A: Both are correct. Use the one that makes $C$ easiest: cosine if you can read a maximum clearly, sine if you can read where the wave crosses the midline going up.

Q: My answer is different from the book's answer. Is it wrong?

A: Not necessarily. Check a maximum and a minimum in both equations. If they give the same values at the same $x$, they draw the same wave. For example $y = 3\sin(2x) + 1$ and $y = 3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$ are the same graph.

Q: How do I find the period if I only see part of a wave?

A: Measure the horizontal distance between two neighboring maxima (one full period), or between a maximum and the next minimum (half a period) and double it.

Q: What if the problem uses time in hours or days instead of radians?

A: The recipe does not change: $B = \frac{2\pi}{\text{period}}$ with the period in the same unit as $x$. A tide with period $12$ hours has $B = \frac{\pi}{6}$, a yearly pattern has $B = \frac{2\pi}{365}$.