Chapter 8.2, Problem 71E

Solution

(a.)

To Prove: When $a=b=r$ , the given formulas become the familiar formulas for the area and perimeter of a circle.

It has been shown that when $a=b=r$ , the given formulas become the familiar formulas for the area and perimeter of a circle.

Given:

The area of an ellipse is $A=\pi ab$ and the perimeter is $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ .

Concept used:

The area of a circle is given as $A=\pi r^2$ and the perimeter of a circle is given as $p=2\pi r$ .

Calculation:

It is given that the area of an ellipse is $A=\pi ab$ and the perimeter is $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ .

Put $a=b=r$ in $A=\pi ab$ to get,

  $A=\pi r^2$

Similarly, put $a=b=r$ in $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ to get,

  $\begin{array}{c}p\approx \pi \left(r+r\right)\left(3-\frac{\sqrt{\left(3r+r\right)\left(r+3r\right)}}{r+r}\right)\\ \approx \pi \left(2r\right)\left(3-\frac{\sqrt{\left(4r\right)\left(4r\right)}}{2r}\right)\\ \approx 2\pi r\left(3-\frac{4r}{2r}\right)\\ \approx 2\pi r\left(3-2\right)\end{array}$

Solving,

  $p\approx 2\pi r$

This shows that when $a=b=r$ , the given formulas become the familiar formulas for the area and perimeter of a circle.

Conclusion:

It has been shown that when $a=b=r$ , the given formulas become the familiar formulas for the area and perimeter of a circle.

(b.)

A pair of ellipses such that the one with greater area has smaller perimeter.

It has been determined that a pair of ellipses such that the ellipse having greater area has smaller perimeter is given by an ellipse having semi-major axis as $3$ units and semi-minor axis as $2$ units; and the other having semi-major axis as $4$ units and semi-minor axis as $1$ unit.

Given:

The area of an ellipse is $A=\pi ab$ and the perimeter is $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ .

Concept used:

The area of a circle is given as $A=\pi r^2$ and the perimeter of a circle is given as $p=2\pi r$ .

Calculation:

It is given that the area of an ellipse is $A=\pi ab$ and the perimeter is $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ .

Let a pair of ellipses be such that the first ellipse has semi-major axis and semi-minor axis as $a$ and $b$ ; and the second ellipse has semi-major axis and semi-minor axis as $c$ and $d$ .

Then, the area of the first ellipse is $A=\pi ab$ and the area of the second ellipse is $A=\pi cd$ .

Similarly, the perimeter of the first ellipse is $p\approx \pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)$ and the perimeter of the second ellipse is $p\approx \pi \left(c+d\right)\left(3-\frac{\sqrt{\left(3c+d\right)\left(c+3d\right)}}{c+d}\right)$ .

Now, let the first ellipse have greater area but smaller perimeter.

Then,

  $\pi ab>\pi cd$

Simplifying,

  $ab>cd$

Similarly,

  $\pi \left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)<\pi \left(c+d\right)\left(3-\frac{\sqrt{\left(3c+d\right)\left(c+3d\right)}}{c+d}\right)$

Simplifying,

  $\left(a+b\right)\left(3-\frac{\sqrt{\left(3a+b\right)\left(a+3b\right)}}{a+b}\right)<\left(c+d\right)\left(3-\frac{\sqrt{\left(3c+d\right)\left(c+3d\right)}}{c+d}\right)$

On further simplification,

  $3\left(a+b\right)-\sqrt{\left(3a+b\right)\left(a+3b\right)}<3\left(c+d\right)-\sqrt{\left(3c+d\right)\left(c+3d\right)}$

Now consider the values of $a=3$ , $b=2$ , $c=4$ and $d=1$ .

Put these values in $ab>cd$ to get,

  $6>4$

The above inequality obtained is true.

So, the inequality $ab>cd$ holds for the assumed values.

Similarly, put these values in $3\left(a+b\right)-\sqrt{\left(3a+b\right)\left(a+3b\right)}<3\left(c+d\right)-\sqrt{\left(3c+d\right)\left(c+3d\right)}$ to get,

  $\begin{array}{c}3\left(3+2\right)-\sqrt{\left(3\left(3\right)+2\right)\left(3+3\left(2\right)\right)}<3\left(4+1\right)-\sqrt{\left(3\left(4\right)+1\right)\left(4+3\left(1\right)\right)}\\ 3\left(5\right)-\sqrt{\left(9+2\right)\left(3+6\right)}<3\left(5\right)-\sqrt{\left(12+1\right)\left(4+3\right)}\\ 15-\sqrt{\left(11\right)\left(9\right)}<15-\sqrt{\left(13\right)\left(7\right)}\\ -\sqrt{99}<-\sqrt{91}\end{array}$

Solving,

  $\sqrt{99}>\sqrt{91}$

The above inequality obtained is true.

So, the inequality $3\left(a+b\right)-\sqrt{\left(3a+b\right)\left(a+3b\right)}<3\left(c+d\right)-\sqrt{\left(3c+d\right)\left(c+3d\right)}$ holds for the assumed values.

This implies that a pair of ellipses such that the ellipse having greater area has smaller perimeter is given by an ellipse having semi-major axis as $3$ units and semi-minor axis as $2$ units; and the other having semi-major axis as $4$ units and semi-minor axis as $1$ unit.

Conclusion:

It has been determined that a pair of ellipses such that the ellipse having greater area has smaller perimeter is given by an ellipse having semi-major axis as $3$ units and semi-minor axis as $2$ units; and the other having semi-major axis as $4$ units and semi-minor axis as $1$ unit.