Minimum value

**chengbin** · July 21st 2009, 05:37 PM

I'm amazed I'm stuck on this question.

I need to find the value of t at which it becomes the minimum value.

$|\vec b|^2-2t\vec ab+t^2|\vec a|^2$

I thought you complete the square to find the minimum value. I can't get what they're getting by completing the square (although I get it when I use calculus)

$|\vec a|^2(t-\frac {\vec a*\vec b} {|\vec a|^2})^2+|\vec b|^2-\frac {(\vec a * \vec b)}{|\vec a|^2})$

How do you get that?

**mr fantastic** · July 21st 2009, 07:41 PM

Originally Posted by chengbin

I'm amazed I'm stuck on this question.

I need to find the value of t at which it becomes the minimum value.

$|\vec b|^2-2t\vec ab+t^2|\vec a|^2$

I thought you complete the square to find the minimum value. I can't get what they're getting by completing the square (although I get it when I use calculus)

$|\vec a|^2(t-\frac {\vec a*\vec b} {|\vec a|^2})^2+|\vec b|^2-\frac {(\vec a * \vec b)}{|\vec a|^2})$

How do you get that?

You have:

$a^2 t^2 - 2c t + b^2$ where $c = \vec{a}\cdot \vec{b}$

$= a^2 \left(t^2 - \frac{2c}{a^2} t + \frac{b^2}{a^2}\right)$

$= a^2 \left( \left [t - \frac{c}{a^2}\right]^2 - \frac{c^2}{a^4} + \frac{b^2}{a^2}\right)$

$= a^2 \left [t - \frac{c}{a^2}\right]^2 - \frac{c^2}{a^2} + b^2$ .

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