A clock has hands 1 and 1 3/5 inches long respectively. At what rate are the ends of the hands approaching each other when the time is 2 o'clock?
with the center of the clock at (0,0), the distance z between the tips at t minutes after 12:00 is
z^2 = m^2+h^2
z dz/dt = m dm/dt + h dh/dt
m = 8/5 cos(π/30 t)
h = cos(π/360 t)
Now just plug in the numbers at t=120
sorry about z. It is really found using the law of cosines
z^2 = (8/5)^2 + 1^2 - 2(8/5)cosθ
where θ is the angle between the hands. I'm sure you can figure that out.
Well, let's see. As the time moves from 12 o'clock to 2 o'clock, the minute hand would have moved 120 degrees (2/12 of a full revolution). Assuming we're using radians, this would be 2π/3 radians.
Now, we need to find the length of the arc described by the tip of the hour hand in the same time period. From 12 o'clock to 2 o'clock, the hour hand covers 1/6 of the clock dial. So, the arc length covered by the tip of the hour hand would be (1/6)(2π), which simplifies to π/3.
We know that the ends of the hands are approaching each other, so the rate at which they are getting closer is simply the difference in their arc lengths divided by the time.
Thus, we can calculate the rate as (π/3 - 2π/3) / (2/12), which simplifies to -π/6 inches per hour.
So, when the time is 2 o'clock, the ends of the clock hands are approaching each other at a rate of -π/6 inches per hour. I hope that makes you tickle with laughter!
To find the rate at which the ends of the clock hands are approaching each other, we need to find the rates of change of the distances covered by each hand.
Let's start by finding the distance covered by each hand in a given time, t.
The hour hand completes a full revolution every 12 hours, which is equivalent to 360 degrees or the circumference of a circle with a radius of 1 inch. Therefore, the distance covered by the hour hand in t hours is 2π(1) = 2πt inches.
The minute hand completes a full revolution every 60 minutes, which is also equivalent to 360 degrees or the circumference of a circle with a radius of 1 3/5 inches. First, we need to convert the radius to an improper fraction: 1 3/5 = 8/5 inches. Therefore, the distance covered by the minute hand in t minutes is given by 2π(8/5) = 16π/5t inches.
Now, let's differentiate both expressions to find the rates at which each hand is moving.
The derivative of the distance covered by the hour hand with respect to time, t, is 2π inches/hour.
The derivative of the distance covered by the minute hand with respect to time, t, is (16π/5) inches/minute.
At 2 o'clock, the time is 2 hours, so we can substitute t = 2 into the derivatives to find the rates at that time.
The rate at which the ends of the hands are approaching each other at 2 o'clock is 2π inches/hour - (16π/5) inches/minute. To compare these units, we need to convert inches/minute to inches/hour by multiplying by 60:
2π inches/hour - (16π/5) inches/minute = 2π - 192π/5 = 192π/5 - 32π/5 = 160π/5 = 32π inches/hour.
Therefore, the ends of the hands are approaching each other at a rate of 32π inches/hour when the time is 2 o'clock.
To find the rate at which the ends of the clock hands are approaching each other, we need to use the concept of related rates.
Let's consider the clock as a circle with a radius of 1 inch. The longer hand (minute hand) has a length of 1 3/5 inches, which is equivalent to 8/5 inches.
At 2 o'clock, the minute hand is pointing directly at the 12, while the hour hand is pointing directly at the 2. We can imagine this as an isosceles triangle with the clock center as the vertex angle, and the two hands as the legs of the triangle.
Now, let's calculate the distance between the ends of the hands using the Pythagorean theorem. The length of one side of the triangle is the shorter hand (hour hand), which is the radius of the clock, 1 inch. The length of the other side is the difference between the lengths of the two hands.
Using the Pythagorean theorem, the distance between the ends of the hands (d) is given by:
d^2 = (8/5)^2 - 1^2
d^2 = 64/25 - 1
d^2 = 64/25 - 25/25
d^2 = 39/25
Taking the square root of both sides, we get:
d = √(39/25)
Now, let's find the rate at which the ends of the hands are approaching each other. We'll differentiate the equation with respect to time.
d/dt * d = d/dt * (√(39/25))
Now, let's differentiate both sides of the equation:
d/dt * d = d/dt * (√(39/25))
(d/dt) * d = (d/dt) * (39/25)^(1/2)
To find the rate at which the ends of the hands are approaching each other, we need to find d/dt, the derivative of d with respect to time. We can differentiate the equation using the chain rule:
(d/dt) * d = (1/2) * (39/25)^(-1/2) * d/dt (39/25)
Now, let's calculate d/dt (39/25):
d/dt (39/25) = 0
Since the hands of the clock are fixed and don't move independently, the distance d remains constant. Therefore, the rate at which the ends of the hands are approaching each other at 2 o'clock is 0.