related rates


by rdougie
Tags: rates
rdougie
rdougie is offline
#1
Oct1-07, 10:24 PM
P: 6
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[tex]^{2}[/tex] + b[tex]^{2}[/tex] = c[tex]^{2}[/tex]
- 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[tex]\Theta[/tex], right? For every 6 degrees the minute hand moves, the hour hand moves one. I'm thoroughly confused........
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
EnumaElish
EnumaElish is offline
#2
Oct2-07, 09:38 PM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,483
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.
hotvette
hotvette is offline
#3
Oct3-07, 12:15 AM
HW Helper
P: 930
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
HallsofIvy is offline
#4
Oct3-07, 06:28 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

related rates


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
D H is offline
#5
Oct3-07, 07:10 AM
Mentor
P: 14,459
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.
D H
D H is offline
#6
Oct3-07, 07:31 AM
Mentor
P: 14,459
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.


Register to reply

Related Discussions
Related Rates Calculus & Beyond Homework 1
Related Rates Calculus & Beyond Homework 3
related rates Calculus & Beyond Homework 6
related rates Calculus & Beyond Homework 1