Cheryl attaches a heart on the middle of her bike tire. The heart will go from the midline up and then down. Is a sine or cosine better. Use the given information to write a sine or cosine function.

Amplitude: 12
Period: ?
Revolutions per minute: 70
Phase shift: ?
D: 12

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
2 hours ago

To determine whether to use a sine or cosine function for Cheryl's heart on the bike tire, we note the following:

  1. Amplitude: The heart has an amplitude of 12. This means that the heart will reach a maximum height of 12 units above and below the midline.

  2. Midline (D): The midline is given as \(D = 12\). This indicates that the vertical midpoint of the function is at \(y = 12\).

  3. Revolutions per minute: The bike tire makes 70 revolutions per minute.

  4. Period: The period of a function in general can be determined by the formula: \[ \text{Period} = \frac{60}{\text{revolutions per minute}} \times 2\pi \] Substituting the values: \[ \text{Period} = \frac{60}{70} \times 2\pi \approx 5.366 \text{ seconds} \]

  5. Phase shift: There is no phase shift specified, so we will consider it to be 0.

Choosing Sine or Cosine

Since the heart starts at the midline and goes upward (if we visualize it), a sine function is usually the best fit given that sine starts at (0, 0) and then goes up.

Writing the Function

The general form for a sine function with the given parameters is: \[ y(t) = A \sin(B(t - C)) + D \]

Where:

  • \(A\) is the amplitude; \(A = 12\).
  • \(B\) relates to the period; \(B = \frac{2\pi}{P}\).
  • \(C\) is the phase shift; here, \(C = 0\).
  • \(D\) is the midline; \(D = 12\).

Calculating \(B\): \[ B = \frac{2\pi}{\text{Period}} \approx \frac{2\pi}{5.366} \approx 1.169 \]

Putting it all together, the function would be: \[ y(t) = 12 \sin\left(1.169t\right) + 12 \]

This sine function models the position of the heart on the bike tire as it rotates, rising and falling with an amplitude of 12 above and below the midline of 12, completing a full cycle in approximately 5.366 seconds.