harder rate of change (1 Viewer)

[OLDMAN]

A clock's minute and hour hands are lengths 4 and 3 respectively.
At the moment when the distance between the two tips is increasing most rapidly:
i) what is the distance?
ii) what is the speed?
iii) what is the time after 3 o'clock does this first happen?

 

[underthesun]

For the hour arm:

let
H_x = 3cos(@)
H_y = 3sin(@)

For the minute arm:

let
M_x = 4cos(12@)
M_y = 4sin(12@)

D² = (4cos(12@) - 3cos(@))² + (4sin(12@) - 3sin(@))²
where D is distance between the two tips.
Now, I only have one problem. I don't seem to be able to think of a way to simplify them. My poor algebra..

To find the time where the two tips is increasing most rapidly, I'd find a zero of the second differential, right? and see if it is a maximum of the first differential..

 

[Affinity]

EDIT: was asleep

 

[Affinity]

Underthesun:

expand and collect terms, see what you get.

read the question carefully, and think about what you are maximizing.

 

[underthesun]

Is the distance between them sqrt(7) cm?
rate of change 1.237 per second?

 

[underthesun]

Originally posted by Affinity
Underthesun:

expand and collect terms, see what you get.

read the question carefully, and think about what you are maximizing.

Yeah, i was thinking some advanced trig..

did you get 25-24 cos(11@) as D²?

 

[Affinity]

Hint: you're maximizing speed, not distance.

 

[Affinity]

also, your units.... what units of time are you using? seconds? mins? hours?

EDIT: I see, but it would be easier if you used a unit of time instead of the angle directly

 

[underthesun]

I did a mistake on the rate of change, it is 12 times the value of that, i was thinking minute arrow instead of hours

Anwyays, aren't you supposed to maximise the speed?

"At the moment when the distance between the two tips is increasing most rapidly"

edit 2: For the second part, my answer is now 0.00567 cm / second..

 

[Affinity]

hm...

 

[underthesun]

The derivatives are very very complicated., those thetas hurt my eyes!

How much marks would this question worth? I hope i'll never get one of these questions in a real HSC

reminds me of the CSSA..

 

[OLDMAN]

Try the following.
Let @= angle between minute and hour hands, measured positive clockwise (of course!)

Then D=sqrt(25-24cos@)

and proceed to differentiate wrt time.

There are valuable lessons in this question for students who seriously try.

 

[underthesun]

This is my full solution to the three questions:

Let @ be the angle between the hour clock and the 12:00 position, with clockwise direction as positive.

From then, for the hour arm,

H_x = 3sin(@) <-- as opposed to 3cos(@) from before
H_y = 3cos(@)

For the minute arm:

let
M_x = 4sin(12@)
M_y = 4cos(12@)

let A be the angle between the minute and hour hands

A = 12@ - @ = 11@

Now, the angle between the two tips is 11@. Using cosine rule, the distance between the two tips is given by the equation:

K = sqrt(25 - 24cos_A)
K² = 25 - 24cosA

differentiating both side implicitly

2K (^dK/_dt) = 24sinA(^dA/_dt)
K (^dK/_dt) = 12sinA(^dA/_dt)

Now, differentiating implicitly again, and using product rule, you get:

(^dK/_dt)² + K(^d2k/_dt2) = 12cosA(^dA/_dt)²

By moving the square of the first differential to the right and multiplying both side by K², we get

K³(^d2k/_dt2)

= 12(^dA/_dt)²(25cosA - 12cos²A - 12)

For the second differential to be zero,

25cosA - 12cos²A - 12 = 0

using quadratic formula we get cosA = 3/4 or 4/3

however, cosA = 4/3 has no solution, hence cosA = 3/4 is the only zero for the second derivative.

Because there must be a maximum to the rate of change of the first derivative, cos^-1(3/4) must be a maximum.

What is the best method to convince the marker that it is a minimum here?

Now, we substitute the value of A = cos^-1(3/4), and put it down the equation for distance.

D = sqrt(25 - 24cosA)
D = sqrt(25 - 24(3/4))
D = sqrt(7)

It happens to be the same answer as my previous answer, but different method..

Question 1 answer :: [ sqrt(7) cm ]

Now, off to the next question:

A = 11@
dA/dt = 11d@/dt.
however, d@/dt is the rate of change of the hourly clock, in radians.

In one hour, @ changes by 30 degrees, which is pi/6

d@/dt = (pi/6)/3600

hence dA/dt = 11pi / 21600

Anyways, progressing with the next question, we have, by substituting the values in the first differential equation,

(^dK/_dt)
= 12sin(cos^-1(3/4)) (^dA/_dt) / K

= 0.0048 cm/s, or should be, unless I screwed up on the units..

Question 2 answer :: [ 0.0048 cm/s ]

Third question, I did by using the general form of cos-1(x)

Question 3 answer :: [ I got 3:08:50 ]

is that the right answer?

I guess setting up the question neatly can get you somewhere..

 

[underthesun]

Here's a simulator for this problem : http://www.utslive.com/clockclock.html

It suggests an answer close to mine. What do you think?

 

[Affinity]

EDIT:Never mind, was half asleep when I did the question.

 

[OLDMAN]

underthesun: you got the distance and speed right, but the time might be worth 1/2 pt. The time of 3:08 is the most rapid decrease in distance, the time 3:24 gives the most rapid increase in distance. Otherwise, well done.

A slightly simpler approach :

D=sqrt(25-24cos@) where @ is the angle between the 2 hands.
dD/dt=12sin@/sqrt(25-24cos)*d@/dt ,implicit diff.
noting that d@/dt is constant since @=m-h where m,h are minute and hour hands' angles respectively. Thus
d@/dt=dm/dt-dh/dt=.1047-.0087=.096 rad/min.
differentiating implicitly again,
d^2(D)/dt^2=.096^2[sqrt(25-24cos)*12cos@-12sin@*12sin@/sqrt(25-24cos)]/(25-24cos)
equating to zero to get maximum, we get
(12cos@-16)(4cos@-3)=0, ignore cos@=16/12, therefore
cos@=3/4. Thus the angle @=41.40 degrees or .723 rad.

Interesting to note that D exhibits "asymetrical" harmonic motion in that it reaches maximum positive speed and maximum negative speed within 16 minutes of each other, straddling the hour hand.