Question

The minute hand on a watch is 6 mm long and the hour hand is 5 mm long. How fast is the distance between the tips of the hands changing at one o'clock

Answer 1

The distance between the tips of the minute hand and the hour hand at one o'clock is approximately 5.57 mm. The distance is changing at a rate of approximately 450π mm per hour.

Step-by-step explanation:

The distance between the tips of the minute hand and the hour hand of a clock can be found using the Pythagorean theorem. Let the distance be represented by d, the length of the minute hand be 6 mm and the length of the hour hand be 5 mm. At one o'clock, the minute hand points at 12 while the hour hand points at 1. The minute hand has moved 1/12th of the way around the clock dial, which is 360/12 = 30 degrees. Using trigonometry, we can calculate the distance:

d = √((6)^2 + (5)^2 + 2(6)(5)cos(30)) = √(36 + 25 + 60cos(30)) = √(61 - 60cos(30)) = √(61 - 30√3) ≈ √31.01 ≈ 5.57 mm.

To find how fast the distance is changing, we can differentiate the distance with respect to time. At one o'clock, the minute hand completes a full revolution every hour, which is 2π radians. Therefore, the rate of change of the distance is given by:

ddt = dθdt = (5)(6)(60)(2π)sin(30) = 3000πsin(30) ≈ 450π mm/hour.

Answer 2

The distance between the tips of the minute and hour hands at one o'clock can be found using the Pythagorean theorem. It is approximately 7.81 mm. The rate at which this distance is changing can be calculated using the chain rule of calculus and is approximately 360 mm/hour.

Step-by-step explanation:

The distance between the tips of the minute and hour hands of a clock can be calculated using the Pythagorean theorem. Let's call the distance x. We can use the lengths of the minute hand (6 mm) and the hour hand (5 mm), and the given time (one o'clock) to find x. Since the minute hand is pointing directly at the 12, and the hour hand is pointing at the 1, we can consider the distance x as the hypotenuse of a right triangle, with the lengths of the minute and hour hands as the other two sides.

Using the Pythagorean theorem, we have:

x^2 = 6^2 + 5^2 = 36 + 25 = 61

Taking the square root of both sides, we find:

x ≈ 7.81 mm

To find how fast the distance is changing at one o'clock, we can consider the minute and hour hands as two sides of a triangle, with the distance x as the hypotenuse. Since both hands are moving continuously, we can calculate the rate of change of each hand and use the chain rule to find the rate of change of x. The angular speed of the minute hand is 360 degrees per 60 minutes, and the angular speed of the hour hand is 360 degrees per 12 hours. Therefore, the rate of change of the minute hand is 360/60 degrees per minute, and the rate of change of the hour hand is 360/12 degrees per hour. We can use the chain rule to find the rate of change of x:

dx/dt = sqrt((6^2)*((360/60)^2) + (5^2)*((360/12)^2)) = sqrt(129600)

Therefore, at one o'clock, the distance between the tips of the hands is changing at a rate of approximately 360 mm/hour.