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At 11:55 p.m., Thomas ties a weight to the minute hand of a clock. The clockwise torque applied by the weight (i.e. the force it applies on the clock's hand to move clockwise) varies in a periodic way that can be modeled by a trigonometric function.
The torque peaks 15 minutes after each whole hour, when the minute hand is pointing directly to the right, at 
3Nm (Newton metre, the SI unit for torque). The minimum torque of 
-3Nm occurs 15 minutes before each whole hour, when the minute hand is pointing directly to the left.
Find the formula of the trigonometric function that models the torque 
tau applied by the weight 
t minutes after Thomas attached the weight. Define the function using radians.

tau(t)=◻

At 1111:5555 p.m., Thomas ties a weight to the minute hand of a clock. The clockwise torque applied by the weight (i.e. the force it applies on the clock's hand to move clockwise) varies in a periodic way that can be modeled by a trigonometric function.\newlineThe torque peaks 1515 minutes after each whole hour, when the minute hand is pointing directly to the right, at 3 3 Nm\mathrm{Nm} (Newton metre, the SI unit for torque). The minimum torque of 3 -3 Nm\mathrm{Nm} occurs 1515 minutes before each whole hour, when the minute hand is pointing directly to the left.\newlineFind the formula of the trigonometric function that models the torque τ \tau applied by the weight t t minutes after Thomas attached the weight. Define the function using radians.\newlineτ(t)= \tau(t)=\square

Full solution

Q. At 1111:5555 p.m., Thomas ties a weight to the minute hand of a clock. The clockwise torque applied by the weight (i.e. the force it applies on the clock's hand to move clockwise) varies in a periodic way that can be modeled by a trigonometric function.\newlineThe torque peaks 1515 minutes after each whole hour, when the minute hand is pointing directly to the right, at 3 3 Nm\mathrm{Nm} (Newton metre, the SI unit for torque). The minimum torque of 3 -3 Nm\mathrm{Nm} occurs 1515 minutes before each whole hour, when the minute hand is pointing directly to the left.\newlineFind the formula of the trigonometric function that models the torque τ \tau applied by the weight t t minutes after Thomas attached the weight. Define the function using radians.\newlineτ(t)= \tau(t)=\square
  1. Identify Trigonometric Function: We need to find a trigonometric function that models the torque. The torque is at its maximum, 3Nm3\,\text{Nm}, at 1515 minutes past the hour and at its minimum, 3Nm-3\,\text{Nm}, at 4545 minutes past the hour. This suggests a sine function because it has a period of 6060 minutes (one hour) and reaches its peak and trough at these times.
  2. General Form of Sine Function: The general form of a sine function is τ(t)=Asin(B(tC))+D\tau(t) = A \cdot \sin(B(t - C)) + D, where AA is the amplitude, BB is the frequency, CC is the phase shift, and DD is the vertical shift. The amplitude is half the distance between the maximum and minimum torque, so A=(3Nm(3Nm))/2=3NmA = (3\,\text{Nm} - (-3\,\text{Nm})) / 2 = 3\,\text{Nm}.
  3. Calculate Amplitude and Frequency: The period of the sine function is the time it takes for the torque to complete one cycle, which is 6060 minutes. The frequency BB is related to the period by B=2πperiodB = \frac{2\pi}{\text{period}}. So B=2π60=π30B = \frac{2\pi}{60} = \frac{\pi}{30}.
  4. Determine Phase Shift: The torque peaks at 1515 minutes past the hour, which means the sine function should be at its maximum at t=15t = 15. To achieve this, we need to shift the sine function to the right by 1515 minutes. Therefore, C=15C = 15.
  5. Calculate Vertical Shift: The vertical shift DD is the average of the maximum and minimum torque, so D=(3Nm+(3Nm))/2=0NmD = (3\,\text{Nm} + (-3\,\text{Nm})) / 2 = 0\,\text{Nm}.
  6. Final Sine Function: Putting it all together, the function is τ(t)=3×sin(π30(t15))\tau(t) = 3 \times \sin\left(\frac{\pi}{30}(t - 15)\right). However, we need to convert time tt from minutes to radians since the trigonometric function is defined using radians. There are 2π2\pi radians in 6060 minutes, so tt radians = (t minutes×2π)/60(t \text{ minutes} \times 2\pi) / 60.

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