1
$\begingroup$

I need to find the value of $w$ using a numerical method for finding function's root (like Newton's method).

By the Pythagorean theorem we know that $BF = \sqrt {100 -w^2}$ and $AD = \sqrt {144 - w^2}$.

I'm not so sure how to utilize $OE$ to create a function so that it's root is the answer for $w$.

I'd be glad for help

Thanks!

enter image description here

$\endgroup$
0
2
$\begingroup$

Put the origin at $D$, $A=(0,a)$, $B=(w,b)$. Then the diagonal lines have the equations $y=a·(1-\frac xw)$ and $y=b·\frac xw$. They intersect at $x=w·\frac{a}{a+b}$ at height $y=\frac{ab}{a+b}$.

The three unknowns $a,b,w$ are thus connected to the 3 given lengths via $$ a^2+w^2=144\\ b^2+w^2=100\\ ab=5(a+b) $$ leading to $$ \sqrt{(144-w^2)(100-w^2)}=5(\sqrt{144-w^2}+\sqrt{100-w^2}). $$ This can now be solved using your favorite univariate solver, like bisection, secant method, Newton method and others to find the root close to $w=4.29732800472$.

$\endgroup$
0
1
$\begingroup$

From triangle $ADF$ and triangle $ODF$: $$\frac{AD}{DF}=\frac{OE}{EF}$$

$$\frac{AD}{w}=\frac{5}{EF}$$

$$EF=\frac{5w}{AD}$$

From triangle $BFD$ and $OED$:

$$\frac{BF}{DF}=\frac{OE}{DE}$$

$$\frac{BF}{w}=\frac{5}{DE}$$

$$DE=\frac{5w}{BF}$$

Since $DE+EF=w$

$$\frac{5}{AD}+\frac{5}{BF}=1$$

Hence,

$$\frac{5}{\sqrt{144-w^2}}+\frac{5}{\sqrt{100-w^2}}=1$$

$\endgroup$
0
0
$\begingroup$

Starting from @Siong Thye Goh's answer, $$\frac{5}{\sqrt{144-w^2}}+\frac{5}{\sqrt{100-w^2}}=1$$ let $x=w^2$ and consider that you look for the zero of function $$f(x)=\frac{5}{\sqrt{144-x}}+\frac{5}{\sqrt{100-x}}-1$$ If you plot it, it looks very linear and this is very good for a root-finding algorithm.

Being very lazy, using $x_0=0$, the first iteration of Newton, Halley and Householder methods will be $$\frac{7200}{341} \qquad \frac{89280}{4823}\qquad \frac{173628000}{9399763}$$ Converted to decimals, th first estimate of $w$ is $4.59504$, $4.30248$ and $4.29785$ while the exact solution is $4.29733$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.