Calculus question?

Question

The minute hand on a watch is 8mm long and the hour hand is 4mm long. How fast is the distance between the tips of the hands changing at one o'clock?

Answer 1

Draw a sketch. The hands on the watch are two sides of a triangle - call them x (the minute hand) and y (the hour hand).

The distance between the tips of the hands is z, the third side of the triangle.

Let &alpha be the angle between the hands.

By the rule of the cosines,
z² = x² + y² - 2·x·y·cos&alpha

At 1 o'clock, the angle is 30 degrees and
z² = 8² + 4² - 2·8·4·√3/2
z² = 80 - 32·√3
z = √(80 - 32·√3)
z = 4.957 approx.

OK, now let's think about the speed of the hands. We will be using the angular velocity.

The minute hand makes the full circle in an hour, so that is
360 deg/h

The hour hand makes the full circle in 12 hours, so that is
360 deg/ 12 h = 30 deg/h.

Define &phi := the angle between 12 o'clock and where the minute hand is now and
define &theta := the angle between 12 o'clock and where the hour hand is now .

Then &alpha = &theta - &phi (plus or minus)

&alpha and z are the changing quantities in our triangle - x and y are constants.

so if &alpha = &theta - &phi, then
cos&alpha = cos(&theta - &phi) = cos&theta cos&phi + sin&theta sin&phi

We need this because we will need to plug d&theta/dt and d&phi/dt into an equation later on.

So we are interested in dz/dt.
Therefore we differentiate (implicitly) the equation
z² = x² + y² - 2·x·y·(cos&theta cos&phi + sin&theta sin&phi)
[and remember, x and y are constants]

2·z·dz/dt =
2·x·y·(-sin&theta cos&phi ·d&theta/dt - cos&theta sin&phi·d&phi/dt + cos&theta sin&phi·d&theta/dt + sin&theta cos&phi·d&phi/dt)

Now just plug in the values of x and y and the values of z, &theta and &phi at 1 o'clock:
z = 4.957, &theta = 0, &phi = 30
and, of course, d&theta/dt = 360 deg/h, d&phi/dt = 30 deg/h:

2·4.957·dz/dt =
2·8·4·(-sin0 cos30 ·360 - cos0 sin30·30 + cos0 sin30·360 + sin0 cos30·30)

9.914 ·dz/dt = 64·( 0 - 1·1/2 ·30 + 1· 1/2·360 + 0)
9.914 ·dz/dt = 10560
dz/dt = 1065 mm/hour
dz/dt = 1065 mm/60 min = 17.75 mm/min
or
dz/dt = 0.3 mm/sec

=> at one o'clock, the distance between the tips of the hands is changing (getting smaller) by 0.3 mm/sec.

Hope this helps!

Answer 2

The exact answer is:

- (11π)/(90√(5-2√3)) mm/min

or approximately -0.309827 mm/min

Use Sine & Cosine functions of (2π/60)t and (2π/720)t for the (x,y) positions of the tip of the minute and hour hand, then differentiate the pythagorean distance and let t = 0.

Answer 3

Let r be the distance and ∅ be the angle between the two hands.

r^2 = 8^2+4^2-2*8*4cos∅

At one o'clock, ∅ = 360/12 = 30 degrees, r(30) = 4.9572547 mm

Differentiate r^2 with respect to time,
2rr' = 64sin∅ * ∅'

But
∅' = (360/(12*60) - 360/60)(pi/180) = -0.0959931 rad/min

At r = r(30), ∅ = 30 degrees, we have
r'(30) = 32sin30 * -0.0959931 / r(30) = -.30982668 mm/min
------------
Check:
Initially, r = r(30) = 4.9572547 mm
After 0.01 minute, (0.01)360/60 = .06 degrees has been passed for the minute hand, and (0.01)360/720 = 0.005 degrees has been passed for the hour hand.
∅ = 30-0.06+0.005 = 29.945 degrees
r = √(8^2+4^2-2*8*4cos29.945) = 4.9541580 mm

Approximately, the rate of change of r with time is
r' = (4.9541580 mm - 4.9572547 mm) /.01 min = -.3097 mm/min

Therefore, my answer above should be correct.
---------------
To M,
You have to use rad/time as the unit for the change of the angle between the two hands. That's part of the reason that your answer is much different from mine.

Answer 4

Let x1 denotes the angle between the hour hand and 12 o'clock at any time,
Let x2 denotes the angle between the minute hand and 12 o'clock at any time,
Let x denotes the angle between the minute hand the hour hand at any time,
f is the distance between the tips of the hands
f^2 = (4 sin x)^2 + ( 8- 4cos x )^2
f^2 = 16 sin^2(x) + 64 - 64 cos x + 16 cos^2(x)
f^2 = 16 (sin^2(x) + cos^2(x)) + 64 - 64 cos x
f^2 = 80 - 64 cos x
2 f df/t = 64 sin x (dx/dt)
df/dt = (32 sin x dx/dt)/f
at one o'clock x = 360/12 = 30 degrees
dx/dt = dx1/dt - dx2/dt =2 pi/12 - 2pi/1= -11 pi/12
So d^2 = 80 - 64 cos 30 = 24.57437416
d = 4.9572547
df/dt= 16/d dx/dt = 3.227592885 *( -11 pi/12) mm/hour
= - 9.294800255 mm/hour
= - 0.00258188896 mm/sec