A 10 meters long ladder is leaned against a wall and a 1 meter times 1 meter big box, as the picture illustrates. The width of the ladder is negligible. How high up on the wall does the ladder reach?
Take hackers equations and turn them into an interative scheme:
$\displaystyle
w_n=\frac{h_n-1}{h_n}
$
$\displaystyle
h_{n+1}=\sqrt{100-w_n^2}
$
Now put $\displaystyle h_0=9.9$, $\displaystyle h_1=9.93794$, $\displaystyle h_2=9.93799 $, $\displaystyle h_3=9.93799 $.
So $\displaystyle h \approx 9.93799 $
(we choose an initial guess of $\displaystyle h=9.9$ by doing a rough drawing of the
arrangement and measuring the approximate height, but a rough guess of 9 will have
added just one more iteration to get this level of accuracy)
The scheme also produces an approimate value for $\displaystyle w\approx 1.11188 $
RonL
I would like to note that there is technically another solution: h = 1.11188 m, which the diagram (sort of) rules out. In this case the ladder is nearly horizontal rather than vertical. (I rarely take diagrams as being "to scale" or in the correct orientation unless I have to.)
-Dan
I solved it without using numerical methods.
It is a bit tricky.
The ladder is 10m long and let at the point of contact with the box, it is divided into x(the upper part) and y(the lower part).{x+y=10}
You see that
$\displaystyle \frac{1}{x^2}+\frac{1}{y^2}=1$
$\displaystyle x^2 +y^2=x^2y^2$
$\displaystyle (x+y)^2=(xy)^2+2xy$
Let xy=b
$\displaystyle 100=b^2+2b$
$\displaystyle b^2+2b-100=0$
since b is positive
$\displaystyle b(=xy)=\sqrt{101}-1$
$\displaystyle x(10-x)=\sqrt{101}-1$
solving for x, we get
$\displaystyle x=5\pm \sqrt{24-\sqrt{101}}$
Since we are looking for maximum height,
we have
$\displaystyle x=5+ \sqrt{24-\sqrt{101}}$
$\displaystyle h=\sqrt{x^2-1}+1$
Keep Smiling
Malay
The cube? I haven't heard of that before, but not strange, regardning what I said before...
I was trying to make the equation more general by puting the sides of the bow to a and b, but I couldn't solve it you'r way since I couldn't get only xy left in the equation and no x or y. But I'm still impressed. Is this the way to solve third order equations?
Yes!
It was first made by Italian mathematicians: Tartalia, Cardano. I think they kept it secret.
I have memorized the original method for solving 3rd order equations . And can up with my own which is simpler but not always works, 2 nights ago when I was falling asleep I was thinking about a techinique that will make my solution always workable did not try it yet.
Maybe, I will post it for thee.