30.03.2023 - 18:16

The hands of a clock in some tower are \approx imately 5 m and 3 m in length. How fast is the distance between the tips of the hands changing at 9:00? Write an equation relating the angle between the tw

Question:

The hands of a clock in some tower are \approx imately 5 m and 3 m in length. How fast is the distance between the tips of the hands changing at 9:00? Write an equation relating the angle between the two clock hands,\theta, and the distance between the tips of the two hands, c. Find the rotated rates equation.

Answers (1)
  • Marguerite
    April 5, 2023 в 04:59
    The distance between the tips of the clock hands is the hypotenuse of a right triangle where the lengths of the other two sides are the lengths of the clock hands. At 9:00, the small hand is pointing straight down and the large hand is pointing directly at the 9 on the clock face. This makes the angle between the hands 90 degrees. Using the Pythagorean theorem, we can find the length of the hypotenuse: c^2 = (5m)^2 + (3m)^2 c^2 = 25m^2 + 9m^2 c^2 = 34m^2 c = sqrt(34)m To find how fast the distance between the tips of the hands is changing, we need to find the derivative of c with respect to time. We can use the chain rule to do this: c = sqrt(34)m dc/dt = (1/2)(34m)^(-1/2) * d/dt(34m) dc/dt = (1/2)(34m)^(-1/2) * 34 * dm/dt dc/dt = (17/sqrt(34)) * dm/dt To find dm/dt, we need to use the fact that the small hand moves at a constant rate of 2pi/12 radians per hour and the large hand moves at a constant rate of 2pi/60 radians per minute: dm/dt = 3m * (2pi/12 radians per hour) - 5m * (2pi/60 radians per minute) dm/dt = (pi/2) - (pi/6) dm/dt = (2pi/6) - (pi/6) dm/dt = pi/6 radians per minute Plugging this into the equation we derived earlier, we get: dc/dt = (17/sqrt(34)) * (pi/6) m/minute dc/dt ? 0.788 m/minute So the distance between the tips of the clock hands is changing at a rate of \approx imately 0.788 meters per minute at 9:00.
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