# Math Help - Need helping starting - (sorry for bad thread name)

Hi,

Sorry for the thread title i really didn't how to describe the question,

I really need help starting this off - i've been heere for ages and i'm clearly missing a trick! Pulling hair almost!

Question:

A Point R has coordinates (x, y). Let D be the distance from (0, 0) to R so that

$D^2 = x^2 + y^2$

R Lies on the curve $y = (1+ x^2)^-1$

Find the coordinates of R such that D^2 is a minimum.

Thanks in advance if you can help!!

2. Originally Posted by aceband
Hi,

Sorry for the thread title i really didn't how to describe the question,

I really need help starting this off - i've been heere for ages and i'm clearly missing a trick! Pulling hair almost!

Question:

A Point R has coordinates (x, y). Let D be the distance from (0, 0) to R so that

$D^2 = x^2 + y^2$

R Lies on the curve $y = (1+ x^2)^-1$

Find the coordinates of R such that D^2 is a minimum.

Thanks in advance if you can help!!
Plug in the term of y into the equation

$(D(x))^2 = x^2+\dfrac1{(1+x^2)^2}$

Differentiate (D(x))² and solve the equation for x:

((D(x))²)' = 0

I've got $((D(x))^2)' = 2x\cdot \left(1-\dfrac{2}{(x^2+1)^3}\right)$

This equation has 3 real solutions.

You must check if the 3 corresponding points R actually produce a minimum at D².

3. Thank you so much!

Just to check, i'm getting these values for x which seem Very odd to me:

$
x = 0 and x= \sqrt[2]{\sqrt[3]{2}-1}
$

:s

4. Originally Posted by aceband
Thank you so much!

Just to check, i'm getting these values for x which seem Very odd to me:

$
x = 0 ~\vee~ x= \sqrt[2]{\sqrt[3]{2}-1}
$

:s
Your results are correct. I've got the points R as:

$R_1(0,1),\ R_2\left(\sqrt{\sqrt[3]{2} - 1},\ \frac32 \sqrt[3]{2} - 1\right),\ R_3\left(-\sqrt{\sqrt[3]{2} - 1},\ \frac32 \sqrt[3]{2} - 1\right)$

Btw:
1. There are 3 x-values.
2. The solutions are connected by an or-sign.