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?
At time t minutes after 12:00
minute hand: x = 1.6 sin(π/30 t), y = 1.6 cos(π/30 t)
hour hand: x = sin(π/360 t), y = cos(π/360 t)
the distance z between the tips is thus
z^2 = ((1.6 sin(π/30 t) - sin(π/360 t))^2 + (1.6 cos(π/30 t) - cos(π/360 t))^2)
= 3.56 - 3.2cos(11π/360 t)
2z dz/dt = 3.2 * 11π/360 sin(11π/360 t)
so find z(120) and then dz/dt there
walang dulot
To find the rate at which the ends of the clock hands are approaching each other at 2 o'clock, we can use the concept of trigonometry and related rates.
Let's denote:
- The length of the first hand (hour hand) as "h" (h = 1 inch)
- The length of the second hand (minute hand) as "m" (m = 1 3/5 inches)
We can determine the rates of change by considering the angles formed by the hands with the vertical axis.
At 2 o'clock, the minute hand will be pointing directly at the 12, while the hour hand will be pointing at the 2. This forms a right triangle, where the minute hand is the hypotenuse, and the hour hand is the adjacent side.
To find the rate at which the ends of the hands are approaching each other, we need to find the rate of change of the distance between the ends of the hands. This can be determined by taking the derivative of the distance with respect to time.
Let's assume:
- "x" as the distance between the ends of the hands
- "θ" as the angle between the minute hand and the vertical axis
We can write the relationship as:
x^2 = h^2 + m^2 - 2hm * cos(θ)
Now, let's differentiate both sides of the equation with respect to time (t):
2x * dx/dt = -2hm * d(θ)/dt * sin(θ)
To find the rate at which the ends of the hands are approaching each other, we need to find dx/dt when t = 2 o'clock.
At 2 o'clock, the hour hand is at the 2, creating an angle of θ with the vertical axis. We know that the minute hand completes a full circle in 60 minutes, which means it completes 360 degrees of rotation in 60 minutes. Therefore, we can determine the rate at which the minute hand moves as d(θ)/dt = (360/60) degrees per minute.
Substituting the given values:
h = 1 inch
m = 1 3/5 inches
d(θ)/dt = (360/60) degrees per minute
We can solve for dx/dt using the equation:
2x * dx/dt = -2hm * d(θ)/dt * sin(θ)
Substituting the values:
2x * dx/dt = -2 * 1 * (8/5) * sin(θ)
We need to find the value of sin(θ) at 2 o'clock. We know that at 2 o'clock, the hour hand is at the 2, which forms a 60-degree angle with the vertical axis. Therefore, sin(60) = (√3)/2.
Substituting this value:
2x * dx/dt = -2 * 1 * (8/5) * (√3)/2
Simplifying the equation:
2x * dx/dt = -8√3/5
Finally, solving for dx/dt:
dx/dt = (-8√3/5) / (2x)
Now, we need to determine the value of x when the time is 2 o'clock. We can use the trigonometric relations to calculate this.
In a right triangle, the sine of an angle is equal to the length of the side opposite the angle divided by the hypotenuse.
Let's denote the distance from the center of the clock to the end of the hour hand as "a."
sin(60) = a / h
Substituting the values:
sin(60) = a / 1
√3/2 = a
Therefore, a = √3/2.
Since x is the sum of "a" and "m," we can find the value of x:
x = (√3/2) + (8/5)
Now, substituting the value of x into the equation for dx/dt:
dx/dt = (-8√3/5) / (2 * (√3/2) + (8/5))
Simplifying:
dx/dt = (-8√3/5) / (√3 + (8/5))
Therefore, the rate at which the ends of the hands approach each other when the time is 2 o'clock is given by dx/dt = (-8√3/5) / (√3 + (8/5)).
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 two hands of the clock as line segments. The shorter hand can be represented by the length 'a' (1 inch) and the longer hand by the length 'b' (1 3/5 inches).
We need to find the rate at which these line segments are approaching each other, which can be represented by the derivative da/dt. To find this, we first need to determine the relationship between a, b, and t (time).
The longer hand completes a full rotation every 12 hours (720 minutes), while the shorter hand completes a full rotation every hour (60 minutes). At 2 o'clock, the time is 2 hours.
Let's convert these times into minutes to work with a common unit:
Longer hand = (2/12) * 720 = 2 * 60 = 120 minutes
Shorter hand = 2 * 60 = 120 minutes
Now, let's set up a right triangle to represent the position of the hands. The distance between the ends of the hands will be the hypotenuse of the triangle, which we can find using the Pythagorean theorem:
a^2 + b^2 = c^2
Substituting the values:
1^2 + (8/5)^2 = c^2
1 + 64/25 = c^2
25/25 + 64/25 = c^2
89/25 = c^2
c = sqrt(89)/5 inches
Differentiating both sides of the equation with respect to time (t), we get:
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
Since the hands are approaching each other, their distance (c) is decreasing, so dc/dt is negative.
At 2 o'clock, we are given a = 1 inch, b = 8/5 inches, c = sqrt(89)/5 inches, and we need to find da/dt when c = sqrt(89)/5 inches.
Substituting these values into the equation:
2(1)(da/dt) + 2(8/5)(db/dt) = 2(sqrt(89)/5)(dc/dt)
Simplifying:
2(da/dt) + 16/5(db/dt) = 2(sqrt(89)/5)(dc/dt)
We need to find da/dt, so we need to solve for it. Rearranging the equation:
2(da/dt) = 2(sqrt(89)/5)(dc/dt) - 16/5(db/dt)
da/dt = (sqrt(89)/5)(dc/dt) - 8/5(db/dt)
To find da/dt, we need to know the rates of change of c and b with respect to time (dc/dt and db/dt). Unfortunately, this information is not provided in the question. Without the rates of change, we cannot calculate da/dt accurately.
Therefore, to find the rate at which the ends of the hands are approaching each other at 2 o'clock, we need to know the rates of change for both the longer and shorter hand lengths.