1. ## the time

Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 ft. Al is due west of Bob. Al begins walking along the edge of the silo at 6 ft per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is ?
The answer for this question is 48. can anyone solve rthis question for me. thanks

2. Hello, rai2003!

No matter what I do, I get a transcendental equation.
Does anyone see an elementary relationship?

Al and Bob are at opposite ends of a diameter of a silo
in the shape of a right circular cylinder with radius 150 ft.
Al is due west of Bob.
Al begins walking along the edge of the silo at 6 ft/sec.
At the same moment that Bob begins to walk due east at the same speed.
The value closest to the time in seconds when Al first can see Bob is __ ?

Code:
                  P
* * o
6t  *         *o
*               * o
*         .       *    o
o
*           @       *             o
A o         * - - - - o - - - - - - - - o Q
*         O   150   *B      6t

*                 *
*               *
*         *
* * *

We have a circle with center $\,O$ and radius 150.

Al starts at $\,A$.
In the next $\,t$ seconds, he walks $6t$ feet to $\,P.$
. . Draw radius $OP = 150.$

Bob starts at $\,B.$
In the same $\,t$ seconds, he walks $6t$ feet to $\,Q.$

$PQ$ is tangent to the circle: . $\angle OPQ = 90^o$

Let $\theta = \angle POQ$

The circumference of the semicircle is $150\pi$
Then: $\text{arc}(PB) \:=\:150\pi - 6t$

Since arc length is given by: . $s \:=\:r\theta$
. . we have: . $150\theta \:=\:150\pi - 6t \quad\Rightarrow\quad \theta \:=\:\pi - \dfrac{t}{25}$

$\text}In right triangle }OPQ:\;\;\;\cos\theta \:=\:\dfrac{150}{6t+150}$

$\text{So we have: }\;\cos\left(\pi - \frac{t}{25}\right) \;=\;\dfrac{150}{6t+150}$

. . . . . and I'm stuck.

3. Iteration?

$\cos\left(\pi - \frac{t}{25}\right) \;=\;\dfrac{150}{6t+150}$

$-\cos\left(\frac{t}{25}\right) \;=\;\dfrac{150}{6t+150}$

$t_{n+1} \;=\;25\cos^{-1}\left(-\dfrac{150}{6t_{n}+150}\right)$

But then, I get 2265 whatever I do.

I wonder if there is another way besides using graphical method...

I tried another iteration using cos instead of arccos, but the value just jumps randomly...

http://www.wolframalpha.com/input/?i=cos(\pi+-+\frac{t}{25})%3D\frac{150}{6t%2B150}+

EDIT: Wait! My calculator mode was in degrees, my bad. In radians, I get 48.007 which is 48 seconds!

4. Sice the qestion asks for t to the nearest second, I suspect that they do expect you to use an approximation technique. Uisng Newton's method I get an answer near 48.02 seconds.

5. can you guy show me clearly about how to solve this please. Thanks

6. A relatively easy way to estimate a solution to

is to plot these two functions using something like Excel or a graphing calculator and see where they intersect. That should be good enough.

Depending on your proficiency with calculus and derivatives you can apply a more sophisticated technique that involves finding the roots of
$
\cos(\pi - \frac t {25} ) - \frac {150} {6t + 150} = 0
$

First, as Unknown008 pointed out, the function $\cos ( \pi - \frac t {25}) = -\cos(\frac t {25})$ This comes from the identity $\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)$. So you have:
$
f(t) = \cos(- \frac t {25}) - \frac {150} {6t + 150} = 0
$

To use Newton's method, you make a first guess of the root of this equation, calculate the error in your guess, then make a next guess based on the amount of error and the slope of the line evaluated at the first guess. Mathematically it's like this:
$
t_2 = t_1 - f(t_1)/f'(t_1)
$

where $t_1$ is the initial guess, $f(t_1)$ is the function evaluated at $t_1$:
$
f(t_1) = \cos(- \frac {t_1} {25}) - \frac {150} {6t_1 + 150} = 0
$

and $f'(t_1)$ is the first derivative of $f$ evaluated at $t_1$ . This generates a next guess $t_2$. You then plug that new guess a back into the equation to generate a third guess, etc. After 10 or so steps you get a pretty good estimate of the true value of $t$ that satisfies the equation. If you're not familiar with how to find derivatives, don't worry about it - just use the graphical method I outlined above.

Hope this helps!