Chapter 12, Problem 14PS

a.

To determine

To write: the perimeter of each triangle in term x .

Answer to Problem 14PS

Thus perimeter of $AOP=1+x\sqrt{1+x^2}+\sqrt{x^4-x^2+1}.$

Thus perimeter of $BOP=1+x\sqrt{1+x^2}+\sqrt{2x^2-2x+1}.$

Explanation of Solution

Given information: Let P(x,y) be a point on the parabola $y=x^2$ in the first quadrant.

Consider the triangle PAO formed by P,A (0,1),and the origin O (0,0), and the triangle PBO formed by P,B (1,0),and the origin O (0,0).

Calculation:

Let P be the point $(x,x^2)asy=x^2.$

Then

  $\begin{array}{l}OA=1unit.\\ OP=\sqrt{x^2+x^4}=x\sqrt{1+x^2}.\\ AP=\sqrt{x^2+{(x^2-1)}^2}=\sqrt{x^2+x^4+1-2x^2}=\sqrt{x^4-x^2+1}\end{array}$

Thus perimeter of $AOP=1+x\sqrt{1+x^2}+\sqrt{x^4-x^2+1}.$

  $\begin{array}{l}OB=1unit.\\ OP=\sqrt{x^2+x^4}=x\sqrt{1+x^2}.\\ BP=\sqrt{{(x-1)}^2+x^2}=\sqrt{x^2-2x+1+x^2}=\sqrt{2x^2-2x+1}\end{array}$

Thus perimeter of $BOP=1+x\sqrt{1+x^2}+\sqrt{2x^2-2x+1}.$

b.

To determine

To complete: the table let r(x) be the ratio of the perimeter of the two triangles.

  $r(x)=\frac{Perimeter△PAO}{Perimeter△PBO}.$

x	4	2	1	0.1	0.01
$Perimeter△PAO$
$Perimeter△PBO$
r(x)

Answer to Problem 14PS

Explanation of Solution

Given information: Let P(x,y) be a point on the parabola $y=x^2$ in the first quadrant.

Consider the triangle PAO formed by P,A (0,1),and the origin O (0,0), and the triangle PBO formed by P,B (1,0),and the origin O (0,0).

Calculation:

  $r(x)=\frac{Perimeter△PAO}{Perimeter△PBO}=\frac{1+x\sqrt{1+x^2}+\sqrt{x^4-x^2+1}}{1+x\sqrt{1+x^2}+\sqrt{2x^2-2x+1}}.$

Completed table shown below.

x	4	2	1	0.1	0.01
$Perimeter△PAO$	33.06	9.07	3.41	2.00	2.00
$Perimeter△PBO$	22.49	7.71	3.41	2.00	2.00
r(x)	1.47	1.18	1	1	1

c.

To determine

To find: $\underset{x\to 0^{+}}{\lim }r(x)$ .

Answer to Problem 14PS

  $\underset{x\to 0^{+}}{\lim }r(x)=1.$

Explanation of Solution

Given information: Let P(x,y) be a point on the parabola $y=x^2$ in the first quadrant.

Consider the triangle PAO formed by P,A (0,1),and the origin O (0,0), and the triangle PBO formed by P,B (1,0),and the origin O (0,0).

Calculation:

  $\begin{array}{l}\underset{x\to 0^{+}}{\lim }r(x)=\underset{x\to 0^{+}}{\lim }\frac{1+x\sqrt{1+x^2}+\sqrt{x^4-x^2+1}}{1+x\sqrt{1+x^2}+\sqrt{2x^2-2x+1}}=\frac{1+\sqrt{0}+\sqrt{1}}{1+\sqrt{0}+\sqrt{1}}=\frac{2}{2}=1.\\ \underset{x\to 0^{+}}{\lim }r(x)=1.\end{array}$