Using Calculus to find a Point on a circle

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November 25th 2010, 06:10 PM
Dudealadude

Using Calculus to find a Point on a circle

Hello, I am trying to find the closest point on the circle x^2+y^2=1 to (1,2) using calculus, not geometry. However, when I try, I get:

x^2+y^2=1
y=sqrt(1-x^2)
trying to find the distance between (x,sqrt(1-x^2)) and (1,2) = (1-x, 2-sqrt(1-x^2))
d=sqrt((1-x)^2+(2-sqrt(1-x^2))^2)
d=sqrt(1-x^2+4-1+x^2)
d=sqrt(5)

Simple geometry tells me this is the distance between the two points, so how do I solve for x and y from here?
November 25th 2010, 06:22 PM
chisigma

You have to minimize the function $f(x,y)= (x-1)^{2} + (y-2)^{2}$ with the constraight $x^{2} + y^{2} = 1$ ... a classical problem involving Lagrange's multipliers...

Kind regards

$\chi$ $\sigma$
November 25th 2010, 06:40 PM
tonio

Quote:

Originally Posted by Dudealadude

Hello, I am trying to find the closest point on the circle x^2+y^2=1 to (1,2) using calculus, not geometry. However, when I try, I get:

x^2+y^2=1
y=sqrt(1-x^2)
trying to find the distance between (x,sqrt(1-x^2)) and (1,2) = (1-x, 2-sqrt(1-x^2))
d=sqrt((1-x)^2+(2-sqrt(1-x^2))^2)
d=sqrt(1-x^2+4-1+x^2)
d=sqrt(5)

Simple geometry tells me this is the distance between the two points, so how do I solve for x and y from here?

Don't write y as a function of x in the beginning. you want the minimum of

$(x-1)^2+(y-2)^2=-2x-4y+6$ , after we substitute $x^2+y^2=1$ . Now you write $y=\sqrt{1-x^2}$ ,

observing that the wanted point must be in the first quadrant and thus both $x,y>0$ :

$-2x-4\sqrt{1-x^2}+6$ . This is a func. of x, derivate and equal to zero and get $\displaystyle{x=\frac{1}{\sqrt{5}}\,,\,y=\frac{2}{ \sqrt{5}}$

Tonio