# I think this is hard....

• Apr 25th 2009, 10:55 AM
beq!x
I think this is hard....
In arch $\displaystyle OA\frown$ of the line in the plane , with equation $\displaystyle f(x)=x^3$ where $\displaystyle O(0,0)$ and $\displaystyle A(a,a^3)$ for $\displaystyle a>0$. Find the coordinates of the point $\displaystyle M(X0,Y0)$ in order that $\displaystyle \triangle OAM$ has the biggest area , than find the equation of the tangent and normal line in point $\displaystyle M$
• Apr 25th 2009, 12:01 PM
red_dog
The area of the triangle is

$\displaystyle S=\frac{1}{2}|\Delta|$ where $\displaystyle \Delta=\begin{vmatrix}x_O & y_O & 1\\x_A & y_A & 1\\x_M & y_M & 1\end{vmatrix}$

$\displaystyle \Delta=\begin{vmatrix}0 & 0 & 1\\a & a^3 & 1\\x & x^3 & 1\end{vmatrix}=x^3-a^2x$

Then $\displaystyle S(x)=\frac{a}{2}|x^3-a^2x|=\left\{\begin{array}{ll}\frac{a}{2}(a^2x-x^3), & x\in(-\infty,-a]\cup [0,a]\\ \frac{a}{2}(x^3-a^2x), & x\in(-a,0)\cup(a,\infty)\end{array}\right.$

$\displaystyle S'(x)=\left\{\begin{array}{ll}\frac{a}{2}(a^2-3x^2), & x\in(-\infty,-a)\cup (0,a)\\ \frac{a}{2}(3x^2-a^2), & x\in(-a,0)\cup(a,\infty)\end{array}\right.$

$\displaystyle S'(x)=0\Rightarrow x=\pm\frac{a\sqrt{3}}{3}$ and both are maximum points.

Then, the maximum area is $\displaystyle S\left(\pm\frac{a\sqrt{3}}{3}\right)$
• Apr 26th 2009, 03:23 AM
beq!x
Quote:

Originally Posted by red_dog

Then $\displaystyle S(x)=\frac{a}{2}|x^3-a^2x|=\left\{\begin{array}{ll}\frac{a}{2}(a^2x-x^3), & x\in(-\infty,-a]\cup [0,a]\\ \frac{a}{2}(x^3-a^2x), & x\in(-a,0)\cup(a,\infty)\end{array}\right.$

$\displaystyle S'(x)=\left\{\begin{array}{ll}\frac{a}{2}(a^2-3x^2), & x\in(-\infty,-a)\cup (0,a)\\ \frac{a}{2}(3x^2-a^2), & x\in(-a,0)\cup(a,\infty)\end{array}\right.$

$\displaystyle S'(x)=0\Rightarrow x=\pm\frac{a\sqrt{3}}{3}$ and both are maximum points.

Then, the maximum area is $\displaystyle S\left(\pm\frac{a\sqrt{3}}{3}\right)$

Can you explain me a little bit please, because i can't understand it.(Doh)(Headbang)