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<sub>x</sub> = 3sin(@) <-- as opposed to 3cos(@) from before
H<sub>y</sub> = 3cos(@)
For the minute arm:
let
M<sub>x</sub> = 4sin(12@)
M<sub>y</sub> = 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)
K<sup>2</sup> = 25 - 24cosA
differentiating both side implicitly
2K (<sup>dK</sup>/<sub>dt</sub>) = 24sinA(<sup>dA</sup>/<sub>dt</sub>)
K (<sup>dK</sup>/<sub>dt</sub>) = 12sinA(<sup>dA</sup>/<sub>dt</sub>)
Now, differentiating implicitly again, and using product rule, you get:
(<sup>dK</sup>/<sub>dt</sub>)<sup>2</sup> + K(<sup>d2k</sup>/<sub>dt2</sub>) = 12cosA(<sup>dA</sup>/<sub>dt</sub>)<sup>2</sup>
By moving the square of the first differential to the right and multiplying both side by K<sup>2</sup>, we get
K<sup>3</sup>(<sup>d2k</sup>/<sub>dt2</sub>)
= 12(<sup>dA</sup>/<sub>dt</sub>)<sup>2</sup>(25cosA - 12cos<sup>2</sup>A - 12)
For the second differential to be zero,
25cosA - 12cos<sup>2</sup>A - 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<sup>-1</sup>(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<sup>-1</sup>(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,
(<sup>dK</sup>/<sub>dt</sub>)
= 12sin(cos<sup>-1</sup>(3/4)) (<sup>dA</sup>/<sub>dt</sub>) / 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..
