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  • Thread starter rdougie
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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[tex]^{2}[/tex] + b[tex]^{2}[/tex] = c[tex]^{2}[/tex]
- 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[tex]\Theta[/tex], right? For every 6 degrees the minute hand moves, the hour hand moves one. I'm thoroughly confused........
 

Answers and Replies

  • #2
EnumaElish
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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 = [itex]\epsilon[/itex], the outer circle has rotated some, so the distance between the two points has increased. Find (d([itex]\epsilon[/itex]) - d(0))/[itex]\epsilon[/itex] as [itex]\epsilon[/itex] --> 0.
 
  • #3
hotvette
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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.
 
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  • #4
HallsofIvy
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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
D H
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Why is the chain rule needed here? Defining [itex]\hat x[/itex] as the 12:00 position, [itex]\hat y[/itex] as the 3:00 position, and [itex]\theta_m[/itex] as the angle between the 12:00 position and the minute hand measured clockwise, then the tip minute hand is at

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

The minute hand makes one revolution per hour. Thus [itex]\dot \theta_m = 2\pi/\text{hr}[/tex]. Differentiating the expression for [itex]\vec m[/itex],

[tex]\frac{d}{dt}\vec m = 10\pi\text{in}/\text{hr}(-\sin \theta_m \hat x + \cos \theta_m \hat y)[/tex]

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 [itex]\theta_m[/itex] and [itex]\theta_h[/itex], take the magnitude, and voila, you have the distance rate of change.


The above is not correct. Correction is imminent.
 
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  • #6
D H
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I should know better. The magnitude of a velocity vector does not equal to the time derivative of the magnitude underlying displacement vector.

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

[tex]d^2 \equiv \vec d \cdot \vec d[/itex]

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}}[/tex] is obviously not the same as the time derivative of the magnitude of the displacement vector.
 
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