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KC has a piece of gum stuck to her bike tire, which applies a forward torque on the tire's movement. The torque varies periodically and can be modeled approximately by a trigonometric function.

When the gum is on the front of the tire, its weight is pulling forwards with a maximum torque of 0.01 Nm (Newton metre). When it's on the back of the tire, it's pulling backwards with a minimum torque of -0.01 Nm. The maximum torque is reached once in every rotation, which is every 1.2 meters. The first time it reaches its maximum torque is 0.3 meters into the race.

a. Find the formula of the trigonometric function that models the torque [tex]$\tau$[/tex] the gum applies on the tire [tex]$d$[/tex] meters into the race. Define the function using radians.
[tex]\[
\tau(d) = \square
\][/tex]

b. What is the torque when KC has ridden 4 meters? Round your answer, if necessary, to four decimal places.
[tex]\[
\text{Torque} = \square \quad \text{Nm}
\][/tex]



Answer :

GinnyAnswer
To model the torque [tex]\(\tau\)[/tex], we can represent it as a trigonometric function. Given the nature of the torque mentioned in the problem, a sine function is most suitable because it naturally oscillates between maximum and minimum values.

Here's how we derive the formula step-by-step:

1. Determine the amplitude [tex]\(A\)[/tex]:
- The amplitude of the sine function is half the difference between the maximum and minimum torque values.
- Given the maximum torque [tex]\(\tau_{\text{max}} = 0.01 \, \text{Nm}\)[/tex] and the minimum torque [tex]\(\tau_{\text{min}} = -0.01 \, \text{Nm}\)[/tex], we find:
[tex]\[ A = \frac{\tau_{\text{max}} - \tau_{\text{min}}}{2} = \frac{0.01 - (-0.01)}{2} = \frac{0.02}{2} = 0.01 \, \text{Nm} \][/tex]

2. Determine the angular frequency [tex]\(\omega\)[/tex]:
- The angular frequency is related to the period of the function. The period [tex]\(T\)[/tex] is the distance over which the function completes one full cycle, which is given as 1.2 meters.
- The angular frequency [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{1.2 \, \text{m}} \approx 5.235987755982989 \, \text{radians per meter} \][/tex]

3. Determine the phase shift [tex]\(\phi\)[/tex]:
- The phase shift is the horizontal shift of the function to the right. The function reaches its maximum torque for the first time at 0.3 meters, indicating that the phase shift [tex]\(\phi\)[/tex] is 0.3 meters.
- Therefore, the phase shift [tex]\(\phi \approx 0.3 \, \text{meters}\)[/tex]

4. Form the torque function [tex]\(\tau(d)\)[/tex]:
- The general formula for the sine function considering amplitude [tex]\(A\)[/tex], angular frequency [tex]\(\omega\)[/tex], and phase shift [tex]\(\phi\)[/tex] is:
[tex]\[ \tau(d) = A \sin(\omega d - \phi) \][/tex]
- Substituting [tex]\(A\)[/tex], [tex]\(\omega\)[/tex], and [tex]\(\phi\)[/tex] we get:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]

Now, let's find the torque when KC has ridden [tex]\(d = 4\)[/tex] meters into the race:

[tex]\[ \tau(4) = 0.01 \sin(5.235987755982989 \times 4 - 0.3) \][/tex]

Calculating the sine term:

[tex]\[ \tau(4) \approx 0.01 \sin(5.235987755982989 \times 4 - 0.3) \approx 0.0098 \, \text{Nm} \][/tex]

So, the torque when KC has ridden 4 meters is approximately 0.0098 Nm.

In conclusion, the formula for the torque [tex]\(\tau\)[/tex] applied by the gum on the tire as a function of the distance [tex]\(d\)[/tex] is:
[tex]\[ \tau(d) = 0.01 \sin(5.235987755982989 \, d - 0.3) \][/tex]

The torque when KC has ridden 4 meters into the race is approximately 0.0098 Nm.