Mark22's optimization question from another site

[MarkFL]

mark22 wrote:

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.

Note: The OP had shown work, and correctly obtained the distance, but was having trouble finding the corresponding time(s).

My response:

I think I would approach this parametrically. Let 12:00:00 (am or pm) be time $\displaystyle t=0$ in minutes and $\displaystyle m(t)$ represent the position of the tip of the minute hand while $\displaystyle h(t)$ represents the position of the tip of the hour hand both with respect to the center of the clock, which is the origin of our coordinate system. Then:

$\displaystyle m(t)=4\left\langle \cos\left(\dfrac{\pi}{30}t \right),\sin\left(\dfrac{\pi}{30}t \right) \right\rangle$

$\displaystyle h(t)=3\left\langle \cos\left(\dfrac{\pi}{360}t \right),\sin\left(\dfrac{\pi}{360}t \right) \right\rangle$

Let $\displaystyle D(t)$ represent the distance between the two tips, hence:

$\displaystyle D^2(t)=\left(4\cos\left(\dfrac{\pi}{30}t \right)-3\cos\left(\dfrac{\pi}{360}t \right) \right)^2+\left(4\sin\left(\dfrac{\pi}{30}t \right)-3\sin\left(\dfrac{\pi}{360}t \right) \right)^2$

Expanding and simplifying via Pythagorean identities, we find:

$\displaystyle D^2(t)=25-24\left(\cos\left(\dfrac{\pi}{30}t \right)\cos\left(\dfrac{\pi}{360}t \right)+\sin\left(\dfrac{\pi}{30}t \right)\sin\left(\dfrac{\pi}{360}t \right) \right)$

Using the angle-difference identity for cosine, there results:

$\displaystyle D^2(t)=25-24\cos\left(\dfrac{11\pi}{360}t \right)$

Let $\displaystyle \theta=\dfrac{11\pi}{360}t$

$\displaystyle D^2(t)=25-24\cos(\theta)$

Differentiating, we find:

$\displaystyle 2D(t)D'(t)=24\sin(\theta)\dfrac{d\theta}{dt}$

$\displaystyle D'(t)=12\dfrac{d\theta}{dt}\dfrac{\sin(\theta)}{D(t)}$

Let $\displaystyle k_1=12\dfrac{d\theta}{dt}$, differentiate again and equate to zero:

$\displaystyle D''(t)=k_1\dfrac{D(t)\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)D'(t)}{D^2(t)}=0$

This implies:

$\displaystyle D(t)\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)D'(t)=0$

$\displaystyle \sqrt{25-24\cos(\theta)}\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)k_1\dfrac{\sin(\theta)}{D(t)}=0$

$\displaystyle (25-24\cos(\theta))\cos(\theta)\dfrac{d\theta}{dt}-k_1\sin^2(\theta)=0$

$\displaystyle 25\cos(\theta)-24\cos^2(\theta)-12(1-\cos^2(\theta))=0$

$\displaystyle 25\cos(\theta)-24\cos^2(\theta)-12+12\cos^2(\theta)=0$

$\displaystyle 12\cos^2(\theta)-25\cos(\theta)+12=0$

$\displaystyle(4\cos(\theta)-3)(3\cos(\theta)-4)=0$

Discarding the invalid root, we find:

$\displaystyle \cos(\theta)=\dfrac{3}{4}$

Hence, the tips of the hands are moving away from one another at the greatest rate when their distance apart is:

$\displaystyle \sqrt{25-18}=\sqrt{7}$

This coincides with times of:

$\displaystyle \dfrac{11\pi}{360}t=\cos^{-1}\left(\dfrac{3}{4} \right)+2k\pi$ where $\displaystyle k\in\mathbb{Z}$

$\displaystyle t=\dfrac{360}{11\pi}\cos^{-1}\left(\dfrac{3}{4} \right)+\dfrac{720k}{11}$

So, to find the times this corresponds to, use $$0\le k\le 10$$ to find the 11 such times in minutes after 12:00:00.