Find Min Segment Length at Point P(1,8)

[leprofece]

determine the minimum segment that relies on the X axis and Y Axis if it passes through the point P (1, 8).

answers
from (5,0) to ( 0,10)

the point is ( x,0) and (0,y)
find the distance
((x-1)²+64)^2 +(1 +(y-8)²)²)

but how must I proceeed?

 

[Evgeny.Makarov]

This is a special case of https://driven2services.com/staging/mh/index.php?threads/10250/ you posted.

 

[leprofece]

This is a special case of https://driven2services.com/staging/mh/index.php?threads/10250/ you posted.

Yeahh the triangle and that is what I don't know to solve

 

[Evgeny.Makarov]

You can solve the problem in this specific case. As you said in the other thread, the equation of the line passing through (1, 8) is
\[
y-8 = m(x-1)
\]
for some $m$. Find the intercepts, i.e., $x$ when $y=0$ and $y$ when $x=0$. Express $x^2+y^2$ through $m$ where $x$ and $y$ are obtained in the previous step. Find the minimum of that expression, or rather prove that $m=-2$ turns its derivative to zero.

 

[leprofece]

You can solve the problem in this specific case. As you said in the other thread, the equation of the line passing through (1, 8) is
\[
y-8 = m(x-1)
\]
for some $m$. Find the intercepts, i.e., $x$ when $y=0$ and $y$ when $x=0$. Express $x^2+y^2$ through $m$ where $x$ and $y$ are obtained in the previous step. Find the minimum of that expression, or rather prove that $m=-2$ turns its derivative to zero.

I could not get the answer
when I found the intercepts I got X = m-8/m and Y = 8- m
maybe I must raise to 2 and derivate?'

 

[Evgeny.Makarov]

when I found the intercepts I got X = m-8/m and Y = 8- m

When you substitute $y=0$ into $y-8=m(x-1)$ and solve for $x$, do you really get $x=m-8/m$? Also, uppercase and lowercase letters are usually distinguished in mathematics, so $x$ and $X$ are different variables. Let's use lowercase letters in this problem.

maybe I must raise to 2 and derivate?

Yes. As I wrote,

Express $x^2+y^2$ through $m$ where $x$ and $y$ are obtained in the previous step. Find the minimum of that expression, or rather prove that $m=−2$ turns its derivative to zero.

 

[leprofece]

I could not get the answer
when I found the intercepts I got X = m-8/m and Y = 8- m
maybe I must raise to 2 and derivate?'

I got x= (m-8)/(m)
do you understand now?

- - - Updated - - -

When you substitute $y=0$ into $y-8=m(x-1)$ and solve for $x$, do you really get $x=m-8/m$? Also, uppercase and lowercase letters are usually distinguished in mathematics, so $x$ and $X$ are different variables. Let's use lowercase letters in this problem.

Yes. As I wrote,

And why x²+ y²?
it is not a circle?

 

[Evgeny.Makarov]

I got x= (m-8)/(m)

Yes, this is better. The formula $x=m-8/m$ means $x=m-\frac{8}{m}$ and is quite a different thing.

And why x²+ y²?
it is not a circle?

The word "circle" was not used up to now in this thread, so no, it is not a circle.

You have a line segment from $(x,0)$ to $(0,y)$. This segment is the hypotenuse in the triangle formed by parts of the axes between $(0,0)$ and $(x,0)$ and between $(0,0)$ and $(0,y)$. The length of the hypotenuse, which you need to minimize (this is why you are interested in it), is found by the Pythagorean theorem. The length of the side between $(0,0)$ and $(x,0)$ is $x$, and the length of the side between $(0,0)$ and $(0,y)$ is $y$. Thus, the length of the hypotenuse is $\sqrt{x^2+y^2}$. It is easier to consider the square of the length because minimizing it also minimizes the length itself.

 

[leprofece]

Yes, this is better. The formula $x=m-8/m$ means $x=m-\frac{8}{m}$ and is quite a different thing.

The word "circle" was not used up to now in this thread, so no, it is not a circle.

You have a line segment from $(x,0)$ to $(0,y)$. This segment is the hypotenuse in the triangle formed by parts of the axes between $(0,0)$ and $(x,0)$ and between $(0,0)$ and $(0,y)$. The length of the hypotenuse, which you need to minimize (this is why you are interested in it), is found by the Pythagorean theorem. The length of the side between $(0,0)$ and $(x,0)$ is $x$, and the length of the side between $(0,0)$ and $(0,y)$ is $y$. Thus, the length of the hypotenuse is $\sqrt{x^2+y^2}$. It is easier to consider the square of the length because minimizing it also minimizes the length itself.

then I must derive $\sqrt{x^2+y^2}$
and substitute , then equate to zero to get m
valuie??
or plug the values i mean sqrt (( m-8)/(m))²+ (m+8)²) solving and derive i think it is easier because i have one variable instead
equate to 0 and get m value?

 

[Evgeny.Makarov]

or plug the values i mean sqrt (( m-8)/(m))²+ (m+8)²) solving and derive i think it is easier because i have one variable instead
equate to 0 and get m value?

Yes. I said it before, I'll say it again:

Find the intercepts, i.e., $x$ when $y=0$ and $y$ when $x=0$. Express $x^2+y^2$ through $m$ where $x$ and $y$ are obtained in the previous step. Find the minimum of that expression, or rather prove that $m=-2$ turns its derivative to zero.

Solving
\[
\frac{d}{dm}\Big((x(m))^2+(y(m))^2\Big)=0
\]
may be a little complicated because it's an equation of the 4th degree, but it is easy to check that $m=-2$ is a root.

 

[leprofece]

Yes. I said it before, I'll say it again:Solving
\[
\frac{d}{dm}\Big((x(m))^2+(y(m))^2\Big)=0
\]
may be a little complicated because it's an equation of the 4th degree, but it is easy to check that $m=-2$ is a root.

again confused
you mean
it is that I got tl derive that
2x+2y
substitute now)))
2(m-8/m)+2 (m+8) = 0
That is that i understood
is that right?

 

[Evgeny.Makarov]

Well, since you refuse to follow instructions...

Find the intercepts, i.e., $x$ when $y=0$ and $y$ when $x=0$.

\[
x=1-8/m,\quad y=8-m\qquad(*)
\]
This part you have done.

Express $x^2+y^2$ through $m$ where $x$ and $y$ are obtained in the previous step.

\begin{align*}
x^2+y^2&=\left(1-\frac{8}{m}\right)^2+(8-m)^2\\
&=1-\frac{16}{m}+\frac{64}{m^2}+64-16m+m^2\qquad(**)
\end{align*}

prove that $m=-2$ turns its derivative to zero.

The derivative of (**):
\[
\frac{16}{m^2}-\frac{128}{m^3}-16+2m\qquad(***)
\]
Now substitute $m=-2$ and check if (***) turns into 0.

Speaking about your previous response:

Solving
\[
\frac{d}{dm}\Big((x(m))^2+(y(m))^2\Big)=0
\]
may be a little complicated because it's an equation of the 4th degree, but it is easy to check that $m=-2$ is a root.

it is that I got tl derive that
2x+2y

The intercepts $x$ and $y$ that you found are functions of $m$. Each slope $m$ determines a new line through (1, 8) with its own intercepts. That's why I wrote $x(m)$ and $y(m)$. Next, you need to know with respect to which variable you take derivative. It should be the same variable that you vary to find a minimum (or maximum) of something. In this case, you vary the slope $m$ and among all resulting lines you choose the one that cuts the shortest segment against the axes. Therefore, you should take derivative with respect to $m$.

Now, $\frac{d}{dm}(x(m))^2\ne 2x(m)$. This is because you have a composition of $x$ and the square function:
\[
m\mapsto x(m)\mapsto (x(m))^2
\]
According to the chain rule,
\[
\frac{d}{dm}(x(m))^2=2x(m)\frac{d}{dm}x(m)
\]
and similarly for $\frac{d}{dm}(y(m))^2$. So another way to find the derivative (***) is to substitute (*) into
\[
2x(m)x'(m)+2y(m)y'(m)
\]

Again, to find an extremum, you need take derivative with respect to the independent variable. So far I've seen problems you posted with a single independent variable ($m$ in this problem). It is possible to consider situation where you can vary two parameters, say $u$ and $v$, and you need to find the minimum of some $f(u,v)$. Then you need to take partial derivatives of $f$ with respect to $u$ and $v$. My guess is that you have not covered that so far. But in a single-variable case, it makes no no sense to say that $(x^2+y^2)'=2x+2y$. Derivative with respect to what: $x$, $y$ or something else?