# Can you help with this problems????

• May 29th 2006, 01:47 PM
Susie38
Can you help with this problems????
Let T be a triangle witht vertices (0,0), (0,c^2), (c,c^2) and let R be the region between y =cx and y = x^2 where c>0. Fidn the ratio area T/ area R.
• May 29th 2006, 05:40 PM
Soroban
Hello, Susie38!

It's a straight-forward area problem.
Exactly where does your difficulty lie?

Quote:

Let $\displaystyle T$ be a triangle with vertices $\displaystyle (0,0),\;(0,c^2),\;(c,c^2)$
and let $\displaystyle R$ be the region between $\displaystyle y = cx$ and $\displaystyle y = x^2,\; c > 0$
Find the ratio $\displaystyle \frac{area\:T}{area\:R}$

Code:

             |       (0,cē)* - - - - - * (c,cē)             |        /:             |      /::*             |    /::::             |  /::::*             | /:::*       - - - **- - - - - -           (0,0)

The base of the triangle is $\displaystyle c$, its height is $\displaystyle c^2$.
. . Its area is: $\displaystyle A_T \:=\:\frac{1}{2}(c)(c^2) \:=\:\frac{1}{2}c^3$

The area of the parabolic segment is:
. . $\displaystyle \displaystyle{A_R \;= \;\int^c_0(cx - x^2)\ dx \;=\;\frac{c}{2}x^2 - \frac{1}{3}x^3\bigg|^c_0 \;= \;\frac{1}{6}c^3$

Therefore: $\displaystyle \displaystyle{\frac{A_T}{A_R}\;=\;\frac{\frac{1}{2 }c^3}{\frac{1}{6}c^3} \;= \;3$