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The hour hand of a clock has length 3. The minute hand has length 4. Find the distance between the tips of the hands when that distance is increasing the most rapidly. Find the precise time on the clock.

d= distance between the hands

m= minute hand

h= hour hand

if angle x; between m and h known then from law of cosines

d^2= m^2+h^2 -2mh cos x

The distance between the tips is sqrt(25 - 24 cos x). I differentiate once and get 12 dx/dt sin x / sqrt(25 - 24 cos x). I differentiate again and find the maximum is when cos x (25 - 24 cos x) = 12 (sinx)^2, rearranging leads to 12 cos2x - 25 cos x + 12 = 0.

Factored to solve x and find one solution of angle x between 0 and 1 is 3/4 so plugged that back into the original distance equation. sqrt(25 - 24cos 3/4) = sqrt 7.

I don't know how to find the time on the clock. Also I am asked to solve using vectors if that is possible but I'm not sure.

Questioner:Mark

Country:Arizona, United States

Category:Calculus

Private:No

Subject:calculus clocks problem

Question:

The hour hand of a clock has length 3. The minute hand has length 4. Find

the distance between the tips of the hands when that distance is increasing

the most rapidly. Find the precise time on the clock.

d= distance between the hands

m= minute hand <<<< Bad idea; if you mean the LENGTH OF THE MINUTE HAND,

it is a constant.

h= hour hand

if angle x; between m and h known then from law of cosines

d^2= m^2+h^2 -2mh cos x

The distance between the tips is sqrt(25 - 24 cos x). I differentiate once

and get 12 dx/dt sin x / sqrt(25 - 24 cos x). I differentiate again and find

the maximum is when cos x (25 - 24 cos x) = 12 (sinx)^2, rearranging leads

to 12 cos2x - 25 cos x + 12 = 0.

>>>>>>>>>. here is where there is a problem. You are differentiating w.r.t.

x, but you should be doing it w.r.t. time.

Factored to solve x and find one solution of angle x between 0 and 1 is 3/4

so plugged that back into the original distance equation. sqrt(25 - 24cos

3/4) = sqrt 7.

I don't know how to find the time on the clock. Also I am asked to solve

using vectors if that is possible but I'm not sure.

..........................................

This is a related rates AND a max-min problem. (Yuk!)

A diagram is always a good idea. Draw a clock with the hour hand length =

3, and minute hand length = 4. (Pick some random time.) Now assume that

the 12:00 position corresponds to t = 0, and you should observe:

The minute hand makes an angle with the vertical (12:00) which we call t1.

The hour hand makes an angle with the vertical (12:00) which we call t2.

(Sorry, I can't make a theta.)

The distance between the TIPS of the hands will be called s.

We can use the LAW OF COSINES to express s:

The angle between the hands is t1 - t2 (or vice versa, but it does not matter)

So :

s^2 = 3^2 + 4^2 - 2(3)(4) cos(t1 - t2)

s^2 = 25 - 24 cos(t1 - t2)

Now THAT is basically what you got, except that I have t1-t2 instead of your x.

Now what are t2, t1?

The hour hand moves 2pi in 12 hours, or pi/6 per hour.

Its angle t2 is t2 = pi/6 t

The minute hand moves 2pi in 1 hour, or 2pi per hour.

t1 = 2pi t

We're getting there.

t1 - t2 = 11pi/6 t

s^2 = 25 - 24 cos(11pi/6 t)

Now you said you want to maximize ds/dt. To do that, let's observe that if s is changing, so is s^2. That is, we can find the derivatives of (s^2) instead of the derivatives of s. [This way, we get rid of those square roots -- I hate them, too.]

(s^2)' = 24 (11pi/6) sin (11pi/6 t) <<<<=== ' means d../dt

(s^2)' = 4 (11pi) sin (11pi/6 t)

(s^2)' = 44 sin (11pi/6 t)

Second derivative:

(s^2)'' = 44 (11pi/6 t) cos (11pi/6 t)

When is that zero? Cosine(pi/2) is zero, so:

11pi/6 t = pi/2

11/6 t = 1/2

t = 6/22 = 3/11 hours

THAT is your time. Now you should be able to determine:

t1 = the (position) angle for the minute hand, and

t2 = the angle hour , and

from that, and a quick call to your third grade teacher, who taught you to tell time, exactly what time that is.

You should be able to finish up from here.

Calculus

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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