Generalizing from Quarter-Circle to Other Curves

[turophile]

This is a 2-part problem about finding the limit of the ratio of two triangles inscribed in a quarter-circle.

Part 1: For each point (x, y) on the quarter-circle x[sup:mwp6lwwv]2[/sup:mwp6lwwv] + y[sup:mwp6lwwv]2[/sup:mwp6lwwv] = 9 (x > 0, y > 0), let A(x) and B(x) be the areas of the triangles A and B. The vertices of triangle A are (0, 0), (1, 0), and (x, y), and the vertices of triangle B are (1, 0), (3, 0), and (x, y). Find the limit of B(x) / A(x) as x approaches 3.

My solution for Part 1: Triangle A has base 1 and altitude x, and triangle B has base 2 and altitude x. Therefore A(x) = x / 2 and B(x) = (2x) / 2 = x, and R(x) = B(x) / A(x) = x / (x / 2) = 2. So the limit of B(x) / A(x) = 2, which approaches 0 as x approaches 3.

Part 2: Generalize this problem to curves other than the quarter-circle.

My questions: Does my solution to Part 1 look right? How should I get started on Part 2? Thanks for any help on this one.

 

[galactus]

So the limit of B(x) / A(x) = 2, which approaches 0 as x approaches 3.[ /quote]

The limit is indeed 2. But, \(\displaystyle \lim_{x\to 3}2 = 2\)

Check some arbitrary triangles, A and B, against one another. The area ratios approach 2 as x-->3.

[quote:1l0tnp7k]Part 2: Generalize this problem to curves other than the quarter-circle.

[/quote:1l0tnp7k]

What sort of curves?. make something up?. Try a whole circle?.

 

[soroban]

Hello, turophile!

You are slightly off . . .

\(\displaystyle \text{(1) For each point }(x, y)\text{ on the quarter-circle: }\:x^2 + y^2 \:=\:9\;\; (x > 0,\:y > 0),\)
. . . \(\displaystyle \text{let }A(x)\text{ and }B(x)\text{ be the areas of the triangles }A\text{ and }B.\)

\(\displaystyle \text{The vertices of triangle }A\text{ are: }\0, 0),\;(1, 0),\;(x, y)\)
\(\displaystyle \text{The vertices of triangle }B\text{ are: }\1, 0),\;(3, 0),\;(x, y)\)

\(\displaystyle \text{Find: }\;\lim_{x\to3} \frac{B(x)}{A(x)}\)

      |
      *  *    P
      |       o(x,y)
      |       : *
      |       :  *
      |       :y
      |       :   *
    - o - o - - - o - -
      O 1 A   2   B

\(\displaystyle \text{Draw }OP,\:AP,\:BP.\)

\(\displaystyle A(x) \:=\: \text{area of }\Delta OPA\)
\(\displaystyle B(x) \:=\:\text{area of }\Delta APB\)

\(\displaystyle \text{Note: the altitude of the triangles is }y.\)


\(\displaystyle \text{We have: }\:x^2+y^2\;+\;9 \quad\Rightarrow\quad y \:=\:\sqrt{9-x^2}\)


\(\displaystyle \begin{array}{c}\text{Triangle }A\text{ has base 1 and altitude }y. \\ \\[-3mm] \text{Triangle }B\text{ has base 2 and altitude }y. \end{array}\)


\(\displaystyle \text{Then: }\;\begin{array}{cccccc}A(x) &=& \frac{1}{2}(1)\sqrt{9-x^2} &=& \frac{1}{2}\sqrt{9-x^2} \\ \\[-3mm] B(x) & =& \frac{1}{2}(2)\sqrt{9-x^2} &=& \sqrt{9-x^2} \end{array}\)

\(\displaystyle \text{Hence: }\;\frac{B(x)}{A(x)} \;=\;\frac{\sqrt{9-x^2}}{\frac{1}{2}\sqrt{9-x^2}} \;=\;2\)


\(\displaystyle \text{Therefore: }\;\lim_{x\to3}(2) \;=\;2\)

(2) Generalize this problem to curves other than the quarter-circle.

\(\displaystyle \text{This result should be the same for }any\text{ function, }\:y = f(x)\)

 

[galactus]

Hey, Soroban. I worked the same way as you but different. I used theta instead for wackiness.

\(\displaystyle A({\theta})=\frac{(1)3sin{\theta}}{2}, \;\ B({\theta})=\frac{2(3sin{\theta})}{2}=3sin{\theta}\)

\(\displaystyle \frac{A({\theta})}{B({\theta})}=\frac{3sin{\theta}}{\frac{3}{2}sin{\theta}}}=2\)

\(\displaystyle \lim_{{\theta}\to 0}2=2\)

I believe turophile thought that the limit was 0 since the 'limitee' was a constant.

:wink: