Find the points on the graph of $\displaystyle y=9-x^2$ that is closest to the point $\displaystyle (0, 3)$.

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- Jan 28th 2010, 09:52 AMshawlioptimization - distance between points on a graph
Find the points on the graph of $\displaystyle y=9-x^2$ that is closest to the point $\displaystyle (0, 3)$.

- Jan 28th 2010, 10:17 AMdrumist
In general, the distance between two points is

$\displaystyle \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Use:

$\displaystyle (x_1,y_1)=(x,9-x^2)$ <-- this is a point on the curve

$\displaystyle (x_2,y_2)=(0,3)$

Then we have the distance between a point on the curve and the point $\displaystyle (0,3)$:

$\displaystyle \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\displaystyle =\sqrt{(x-0)^2+(9-x^2-3)^2}$

$\displaystyle =\sqrt{x^2+(6-x^2)^2}$

$\displaystyle =\sqrt{x^2+36-12x^2+x^4}$

$\displaystyle =\sqrt{x^4-11x^2+36}$

Our goal is to find the minimum value of this, so we find the derivative and set it equal to 0 and solve for x.

Can you take it from here? - Jan 28th 2010, 10:28 AMshawli
Yes, thank you!

One question though, in the general formula for the distance between two lines, does it matter which function i assign to be $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$? - Jan 28th 2010, 10:34 AMdrumist