The minute and hour hands on the face of a school clock are 8 inches and 6 inches long, respectively. Find the rate of change, in inches per minute, of the distance between the tips of the hands at 9:00.
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To find the rate of change of the distance between the tips of the hands at 9:00, we need to first determine the positions of the hands and then differentiate the distance between them with respect to time.
Let's start by finding the positions of the minute and hour hands at 9:00. At exactly 9:00, the minute hand points directly at the 12 while the hour hand points directly at the 9. This means that the hour hand has moved 1/4th of the way from 9 to 10.
Next, we need to find the coordinates of the tips of the hands. We can represent the position of each hand as a vector in polar coordinates, where the length of the vector represents the distance from the origin (center of the clock) and the angle represents the position of the hand. Assuming the origin is the center of the clock face, the coordinates of the tips of the hands are given by:
Minute hand: (8, 90°)
Hour hand: (6, 270°)
Now, let's find the distance between the tips of the hands. We can use the distance formula between two points in Cartesian coordinates:
Distance = √[(x2 - x1)² + (y2 - y1)²]
Plugging in the values:
Distance = √[(6 - 8)² + (270° - 90°)²]
= √[(-2)² + (180°)²]
= √[4 + 32400]
= √32404
= 180.01 inches
Now, to find the rate of change of the distance between the tips of the hands, we differentiate the distance formula with respect to time. Since the rate of change is with respect to minutes, we differentiate both sides with respect to minutes:
d(distance)/dt = d/dt[180.01]
= 0
Therefore, the rate of change of the distance between the tips of the hands at 9:00 is 0 inches per minute. This means that the distance between the tips of the hands remains constant at 180.01 inches throughout the minute.