Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $$y=2\sqrt{\ln(x+3) }$$ How do I determine a (x,y) pair satisfying the above relation which is the closest to the origin (0,0)?

share|improve this question
1  
$x^2 + y^2$ is minimized –  Will Jagy Jun 10 '13 at 1:46
    
Using Lagrange multiplier under this constraint. –  Shuhao Cao Jun 10 '13 at 1:46
    
What have you tried so far? –  Loki Clock Jun 10 '13 at 1:47
    
Well, I have no idea how to begin. –  klaufir Jun 10 '13 at 2:04
    
A related problem. –  Mhenni Benghorbal Jun 10 '13 at 2:24
add comment

2 Answers

up vote 7 down vote accepted

Minimize the square of the distance from the curve to the origin (and hence, the distance from the curve to the origin), namely: $$x^2+y^2=x^2+4\ln(x+3).$$

share|improve this answer
    
Thanks, I can start using this. –  klaufir Jun 10 '13 at 2:05
add comment

To minimize the distance from the origin, minimize the square of the distance from the origin, given by $d^2 = M = x^2+y^2$,

$$M = x^2+y^2=x^2+(2\sqrt{\ln(x+3)})^2 = x^2 + 4\ln(x+3)$$

$$\frac{dM}{dx} = 2x + \frac{4}{x+3} = \frac{2x(x+3) + 4}{x+3} = \frac{2x^2 + 6x + 4}{x+3}.$$

This is only (possibly) equal to zero when the numerator is equal to zero,

$$2x^2 + 6x + 4 = 0 \implies x=-2 \ \ \text{and} \ \ x=-1$$

We know that at least one of these must be where the minimum occurs, so just plug the values of $x$ into the given equation and compute the distance from the origin.

$y(-1) = 2\sqrt{\ln(-1+3)} = 2 \sqrt{\ln(2)} = 1.665 \implies d^2 = (-1)^2 + (1.665)^2 = 2.386$

$y(-2) = 2\sqrt{\ln(-2+3)} = 2 \sqrt{\ln(1)} = 0 \implies d^2 = (-1)^2 + (0)^2 = 1$

So the point nearest to the origin is $(-2,0)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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