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Magazine Chapter 3: Problem 43
The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the Law of cosines.)
Short Answer
Expert verified
Answer: The distance between the tips of the hour and minute hands is changing at a rate of -蟺/60 m/min at 9:00.
Step by step solution
01
Calculate the position of the hour and minute hands at 9:00.
First, we need to find the angle between both the hour and minute hands at 9:00. Since the clock has 12 hours and makes a full circle of 360 degrees, each hour represents 30 degrees (360/12).
At 9:00, the hour hand is at the 9-hour position and makes a 270-degree angle with the 12-hour position (30 degrees multiplied by 9). The minute hand is at the 12-hour position, making a 0-degree angle with the 12-hour position.
02
Use the Law of Cosines to find the distance between tips.
Let D be the distance between the tips of the hour and minute hands. We can set up a triangle with sides 3, 2.5, and D, with the angle between sides 3 and 2.5 equal to 270 degrees. Using the Law of Cosines,
D^2 = 3^2 + 2.5^2 - 2(3)(2.5)cos 270掳
D^2 = 9 + 6.25 - 15(-1)
D^2 = 9 + 6.25 + 15 = 30.25
Thus, the distance between the tips is D = sqrt(30.25) = 5.5 m.
03
Differentiate the equation with respect to time.
First, we need to express the angle in radians. 270 degrees is equivalent to (3/2)蟺 radians. Let 胃 be the angle between the hour and minute hands. Then, 胃 = (3/2)蟺 when t = 9:00. We have D^2 = 9 + 6.25 - 15cos胃.
Next, differentiate both sides of the equation with respect to time, t.
2D(dD/dt)= 30(-sin胃)(d胃/dt)
04
Calculate the rate of change of the angle with respect to time.
To find d胃/dt, we need to know how fast both the hour and minute hands are moving.
As there are 12 hours on the clock and the hour hand travels 360 degrees, the angular speed of the hour hand is 360/12 = 30 degrees per hour. Since one hour is 60 minutes, the angular speed of the hour hand is 30/60 = 0.5 degrees per minute. In radians, it is (0.5/180)蟺 = (蟺/360) radians per minute.
The minute hand goes through 360 degrees every hour. Therefore, its angular speed is 360/60 = 6 degrees per minute. In radians, it is (6/180)蟺 = (蟺/30) radians per minute.
Finally, the difference in angular speed between the minute and hour hands is (蟺/30) - (蟺/360) = (11蟺/360) radians per minute. This is the value for d胃/dt.
05
Calculate the rate of change of the distance between the tips of the hands.
Substitute the values of D, 胃, and d胃/dt into the equation:
2(5.5)(dD/dt)= 30(-sin(3/2)蟺)((11蟺)/360)
2(5.5)(dD/dt)= 30(-1)((11蟺)/360)
Divide both sides by -30, and we have:
-(11/2)(dD/dt)= (11蟺)/360
dD/dt = -(蟺/60)
Thus, the distance between the tips of the hour and minute hands is changing at a rate of -蟺/60 m/min at 9:00. The negative sign indicates that the distance is decreasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate of Change Calculus
In calculus, the rate of change is a fundamental concept that measures how a quantity varies with respect to another variable, usually time. This can take many practical forms, such as the speed at which a car travels (change in distance over change in time) or the rate at which water flows from a faucet (volume per unit time).
Mathematically, the rate of change is expressed as the derivative of a function. Using the derivative, we can find how fast one variable is changing at a specific instant in relation to another, which is especially helpful in our clock problem where we need to determine the velocity, or the rate of change, of the distance between the two hands.
In the given exercise, the derivative \(dD/dt\) represents the rate at which the distance between the tips of the hands of the clock is changing with respect to time. By differentiating the relationship between the distance and the angle formed by the hands, we obtain a formula that can tell us how quickly this distance changes at any given moment.
Mathematically, the rate of change is expressed as the derivative of a function. Using the derivative, we can find how fast one variable is changing at a specific instant in relation to another, which is especially helpful in our clock problem where we need to determine the velocity, or the rate of change, of the distance between the two hands.
In the given exercise, the derivative \(dD/dt\) represents the rate at which the distance between the tips of the hands of the clock is changing with respect to time. By differentiating the relationship between the distance and the angle formed by the hands, we obtain a formula that can tell us how quickly this distance changes at any given moment.
Related Rates Problems
Related rates problems involve finding the rate at which one quantity changes with respect to time when another related quantity is also changing with respect to time. These problems often require the use of chain rule in differentiation, as one must relate the rates of change of various variables in a multi-variable function.
In these problems, we first identify all the variables involved, establish a relation between them (usually in the form of an equation), and then differentiate that equation with respect to time to relate their rates of change. The challenge usually lies in expressing all variables in terms of a single variable such as time before taking the derivative.
In the clock example, the hour and minute hands move at different rates, and the rate at which the angle \(\theta\) changes is crucial for determining the rate at which the distance between the hands \(D\) changes, making it a classic related rates problem. Through careful differentiation, we are able to find the relationship between \(dD/dt\) and \(d\theta/dt\), which is exactly what's needed to solve the problem.
In these problems, we first identify all the variables involved, establish a relation between them (usually in the form of an equation), and then differentiate that equation with respect to time to relate their rates of change. The challenge usually lies in expressing all variables in terms of a single variable such as time before taking the derivative.
In the clock example, the hour and minute hands move at different rates, and the rate at which the angle \(\theta\) changes is crucial for determining the rate at which the distance between the hands \(D\) changes, making it a classic related rates problem. Through careful differentiation, we are able to find the relationship between \(dD/dt\) and \(d\theta/dt\), which is exactly what's needed to solve the problem.
Trigonometry in Calculus
Trigonometry and calculus frequently intertwine, especially in problems involving rates of change with respect to angles or when dealing with periodic functions. In calculus, trigonometric functions are differentiated and integrated just like any other algebraic function.
In the context of our clock example, we use trigonometry to establish the relationship between the sides of the triangle formed by the clock hands and the angle between them using the Law of Cosines. This law states for any triangle with sides \(a\), \(b\), and \(c\), and the angle \(\gamma\) opposite to side \(c\), the relationship is given by \(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\).
By employing the Law of Cosines, we are able to begin our exploration of the rate of change of the distance between the clock hands. Without this trigonometric step, finding the rate \(dD/dt\) would not be possible, as we need the relationship between the sides and the angle to proceed with the calculus part of the problem.
In the context of our clock example, we use trigonometry to establish the relationship between the sides of the triangle formed by the clock hands and the angle between them using the Law of Cosines. This law states for any triangle with sides \(a\), \(b\), and \(c\), and the angle \(\gamma\) opposite to side \(c\), the relationship is given by \(c^2 = a^2 + b^2 - 2ab\cos(\gamma)\).
By employing the Law of Cosines, we are able to begin our exploration of the rate of change of the distance between the clock hands. Without this trigonometric step, finding the rate \(dD/dt\) would not be possible, as we need the relationship between the sides and the angle to proceed with the calculus part of the problem.
