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

1. ## 2 Part Question: I've done the first bit but I don't understand the second bit

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]

2. 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 $\displaystyle (3 \cdot 2x) - (x^2 \cdot 0)$. You've got it backwards.
• $\displaystyle x^2 \cdot 0 = 0$, not $\displaystyle x^2$.
• 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:
$\displaystyle \frac{1}{3}x^2$

3. Hello, sweetG!

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

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

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

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

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

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: .$\displaystyle P\left(\frac{11}{9},\:0\right)\,\text{ and }\,Q\left(0,\:\frac{11}{6}\right)$

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: .$\displaystyle 2 \times \frac{11}{9} \:=\:\frac{22}{9}$

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

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

*

4. 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 !