# Related rates

1. Oct 1, 2007

### rdougie

1. The problem statement, all variables and given/known data
"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?"

2. Relevant equations
- a$$^{2}$$ + b$$^{2}$$ = c$$^{2}$$
- Law of Cosines?

3. 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........

2. Oct 2, 2007

### 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.

3. Oct 3, 2007

### 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.

Last edited: Oct 3, 2007
4. Oct 3, 2007

### HallsofIvy

Staff Emeritus
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.

5. Oct 3, 2007

### D H

Staff Emeritus
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 $\vec m$, $$\frac{d}{dt}\vec m = 10\pi\text{in}/\text{hr}(-\sin \theta_m \hat x + \cos \theta_m \hat y)$$ 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. The above is not correct. Correction is imminent. Last edited: Oct 3, 2007 6. Oct 3, 2007 ### D H Staff Emeritus 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, $$d^2 \equiv \vec d \cdot \vec d$ Differentiating with respect to time, [tex]2d\dot d = 2\vec d \cdot \dot {\vec d}[/itex] Solving for the time derivative of the magnitude of the displacement vector, [tex]\dot d = \frac{\vec d \cdot \dot {\vec d}}{d}[/itex] The magnitude of the velocity vector, [tex]\sqrt{\dot {\vec d} \cdot \dot {\vec d}}$$ is obviously not the same as the time derivative of the magnitude of the displacement vector.

Last edited: Oct 3, 2007