Draw a circle with radius of your choice, and draw an ellipse with the same area and a perimeter 10% longer than the perimeter of the circle.

Solution.

Students need to know that:

(1) The equations, x = c*cos(t), and y = d*sin(t), where t changes from 0 to 2π, describe an ellipse with axes having lengths 2c and 2d.

The area of such an ellipse is A = π*c*d.

Its perimeter p is:

From the equation for the area of an ellipse we know that

(2) d = r

So putting together (1) and (2), we conclude that in order to find c and d, for the required ellipse we need to solve for c the following equation,

(In this example, in order to be concrete, we use r = 10 cm.

With our calculator in parametric mode, we enter

\X

\Y

Now place the calculator in Func mode.

Enter

\Y

Use Solver to enter the equation, eqn:0=Y1-.55πR,

Y1-.55πR=0^{ }_{ } |
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·^{ }_{ } |
C=6.9448351745...^{ }_{ } |
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T=0^{ }_{ } |
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R=10^{ }_{ } |
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bound={0,10}^{ }_{ } |
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·^{ }_{ } |
left-rt=0^{ }_{ } |

C^{ }_{ } |
ENTER^{ }_{ } |
6.944835175^{ }_{ } |
||||||

R^{2}/C→D^{ }_{ } |
ENTER^{ }_{ } |
14.39918983^{ }_{ } |

Drawing a circle and an ellipse from their parametric description.

Change MODE to Par (parametric) and set it on Radian.

In the Y= editor, enter

\X1T=Rcos(T)^{ }_{ } |
These two equations describe a circle^{ }_{ } |
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\Y1T=Rsin(T)^{ }_{ } |
with radius R.^{ }_{ } |
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\X2T=Dcos(T)^{ }_{ } |
We switch the position of C and D because it is ^{ }_{ } |
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\Y2T=Csin(T)^{ }_{ } |
better to have the longer axis of the ellipse along the x-axis.^{ }_{ } |

Look at WINDOW, Tmin=0 and Tmax=6.28...

Finally, draw a circle with a compass, and draw the required ellipse with two thumb tacks and a piece of thread.

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Note to those who want to do the problem differently, taking derivatives of trig functions by hand:

The equations, x = c*cos(t), and y = d*sin(t), where t changes from 0 to 2π, describe an ellipse with axes having lengths 2c and 2d.

In order to find c and d, for the required ellipse we need to solve for c the following equation:

(In this example, in order to be concrete, we use r = 10 cm.)

Enter

\Y

Use Solver to enter the equation, eqn:0=Y1-.55πR, the value of R, and to initialize T and C.

Y1-.55πR=0^{ }_{ } |
|||

·^{ }_{ } |
C=6.9448351745...^{ }_{ } |
||

T=0^{ }_{ } |
|||

R=10^{ }_{ } |
|||

bound={0,10} | |||

· |
left-rt=0^{ }_{ } |

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