Related Rates: Calculating Change in Distance Between Clock Hands at 9 o'clock

[rdougie]

Homework Statement

"On a certain clock the minute had is 4in long, and the hour hand is 3in long. How fast is the distance between the tips of the hands changing at 9 o'clock?"

Homework Equations

- a^{2} + b^{2} = c^{2}
- Law of Cosines?

The Attempt at a Solution

Ok i drew a clock, and the 3-4-5 triangle positioned at 9:00. The distance between the 2 tips is 5in, but how would I find the rate of change between them? If the distance is marked by x, then I'm solving for dx/d\Theta, right? For every 6 degrees the minute hand moves, the hour hand moves one. I'm thoroughly confused...

 

[EnumaElish]

I would think of this as two concentric circles with radii R and r (R > r). The inner circle is stationary, but the outer circle is rotating. At t = 0 two points (one on each circle) lie on a straight line through the origin so their distance is d = R - r. At t = \epsilon, the outer circle has rotated some, so the distance between the two points has increased. Find (d(\epsilon) - d(0))/\epsilon as \epsilon --> 0.

 

[hotvette]

Welcome to PF!

What I would do is the following. There are two lines of different lengths (r and R) rotating about a common origin at different frequencies. If the tips of the lines are points a and b, I'd write equations that represent the x,y locations of a and b (with respect to t) using initial conditions such that the postions of the lines are at 9:00 at t=0. I'd then write an expression for the distance between points a and b, take the derivative and evaluate it at t=0. Chain rule involved.

 

[HallsofIvy]

In general, the angle formed by the hands is not a right angle (it happens to be at 9:00 but not immediately after). Your second thought was correct: you can use the cosine law to write the distance between the tips of the hands as a function of the angle between them.

 

[D H]

Why is the chain rule needed here? Defining \hat x as the 12:00 position, \hat y as the 3:00 position, and \theta_m as the angle between the 12:00 position and the minute hand measured clockwise, then the tip minute hand is at

\vec m = 5\text{in}(\cos \theta_m \hat x + \sin \theta_m \hat y)

The minute hand makes one revolution per hour. Thus \dot \theta_m = 2\pi/\text{hr}[/tex]. Differentiating the expression for \vec m,<br /> <br /> \frac{d}{dt}\vec m = 10\pi\text{in}/\text{hr}(-\sin \theta_m \hat x + \cos \theta_m \hat y)<br /> <br /> You can treat the hour hand similarly. Taking the difference between these two velocity vectors yields the tip-to-tip velocity. Apply the specific values for \theta_m and \theta_h, take the magnitude, and voila, you have the distance rate of change.<b>The above is not correct.</b> Correction is imminent.

 

[D H]

I should know better. The magnitude of a velocity vector does not equal to the time derivative of the magnitude underlying displacement vector.

Suppose \vec d is some displacement vector (e.g., the vector from the end of the hour hand to the end of the minute hand). Denote d=||\vec d|| as the magnitude of this displacement vector. For an inner product metric space,

d^2 \equiv \vec d \cdot \vec d[/itex]<br /> <br /> Differentiating with respect to time,<br /> <br /> 2d\dot d = 2\vec d \cdot \dot {\vec d}[/itex]&lt;br /&gt; &lt;br /&gt; Solving for the time derivative of the magnitude of the displacement vector,&lt;br /&gt; &lt;br /&gt; \dot d = \frac{\vec d \cdot \dot {\vec d}}{d}[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The magnitude of the velocity vector, \sqrt{\dot {\vec d} \cdot \dot {\vec d}} is obviously not the same as the time derivative of the magnitude of the displacement vector.