2 Part Question: I've done the first bit but I don't understand the second bit

**sweetG** · July 19th 2008, 07:47 AM

a) Find the equation of the normal to y = x^2 / 3 at the point where x = 1

b) This normal cuts the coordinate axis at P and Q. Given that P and Q are adjacent vertices of a rhombus, and that the diagonals of this rhombus intersect at the origin, determine the area of this rhombus
[I don't understand this part and I'm not too sure on what to do can someone please help]

My working out for part a [Please tell me if i made a mistake]

y = x^2 / 3

dy/dx = (x^2 x 0) - (3 x 2x)
_________________

9

= x^2 - 6x
________

9

when x = 1

m(tangent) = -5/9

So m(normal) = 9/5

Therefore Equation of normal = y = 9/5x - 22/15

[Sorry I missed a few lines and skipped straight to the final answer]

**Henderson** · July 19th 2008, 08:09 AM

Yeah, that's not right. Sorry.

Looks like you're trying to use the quotient rule, which is making things difficult. Three problems with what you've done:

The top should be $(3 \cdot 2x) - (x^2 \cdot 0)$ . You've got it backwards.
$x^2 \cdot 0 = 0$ , not $x^2$ .
You've forgotten to square the bottom.

An easier way to do this might be to look at the 'dividing by three' as a coefficient:
$\frac{1}{3}x^2$

**Soroban** · July 19th 2008, 02:33 PM

Hello, sweetG!

a) Find the equation of the normal to $y \:= \:\frac{1}{3}x^2$ where $x = 1$

When $x = 1,\;y = \frac{1}{3}$

We have: . $y' \:=\:\frac{2}{3}x$

When $x = 1$ , the slope of the tangent is: . $m \:=\:\frac{2}{3}$
. . Hence, the slope of the normal is: $-\frac{3}{2}$

The equation of the normal is: . $y - \frac{1}{3}\:=\:-\frac{3}{2}(x-1) \quad\Rightarrow\quad \boxed{y \:=\:-\frac{3}{2}x + \frac{11}{6}}$

b) This normal cuts the coordinate axis at P and Q.
Given that P and Q are adjacent vertices of a rhombus,
and that the diagonals of this rhombus intersect at the origin,
determine the area of this rhombus.

The intercept of the normal are: . $P\left(\frac{11}{9},\:0\right)\,\text{ and }\,Q\left(0,\:\frac{11}{6}\right)$

If the diagonals intersect at the origin, the rhombus looks like this:

Code:

                      |
                    Q o (0,11/6)
                   *  |  *
                *     |     *
          R  *        |        * P
    - - - o - - - - - + - - - - - o - - -
   (-11/9,0) *        |        * (11/9,0)
                *     |     * 
                   *  |  *
                    S o (0,-11/6)
                      |

The area of a rhomus is one-half the product of its diagonals.

The horizontal diagonal has length: . $2 \times \frac{11}{9} \:=\:\frac{22}{9}$

The vertical diagonal has length: . $2 \times \frac{11}{6} \:=\:\frac{11}{3}$

The area is: . $\frac{1}{2} \times \frac{22}{9} \times \frac{11}{3} \;=\;\boxed{\frac{121}{27}\text{ units}^2}$

*

**sweetG** · July 20th 2008, 05:24 AM

Thanks so much guys !
=)
that really helped and wow i made a lot of silly mistakes there
and I just realized how easy this question really was I just wasn't thinking hard enough Thanks once again !

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