6.5 Q-8

Question Statement

An arc is shaped like a semi-ellipse. The base of the arc is 90 meters wide, and the height at the center is 30 meters. Determine the distance from the center at which the height is $20\sqrt{2} , \text{m}$.

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Background and Explanation

A semi-ellipse can be described using the standard equation of an ellipse:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$

where $a$ is the semi-major axis (horizontal radius) and $b$ is the semi-minor axis (vertical radius). For this problem:

• The base width corresponds to $2a = 90$, giving $a = 45$.
• The height of the semi-ellipse corresponds to $b = 30$.
• The task is to find the horizontal distance $x$ from the center when the height $y = 20\sqrt{2}$.

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Solution

Step 1: Write the equation of the ellipse

Using the given dimensions:

$$\frac{x^2}{45^2} + \frac{y^2}{30^2} = 1.$$

Step 2: Substitute $y = 20\sqrt{2}$ into the equation

$$\frac{x^2}{45^2} + \frac{(20\sqrt{2})^2}{30^2} = 1.$$

Step 3: Simplify the terms

1. Calculate $(20\sqrt{2})^2$:

$$(20\sqrt{2})^2 = 800.$$

2. Substitute into the equation:

$$\frac{x^2}{45^2} + \frac{800}{900} = 1.$$

3. Simplify $\frac{800}{900}$:

$$\frac{800}{900} = \frac{8}{9}.$$

The equation becomes:

$$\frac{x^2}{45^2} + \frac{8}{9} = 1.$$

Step 4: Solve for $x^2$

1. Subtract $\frac{8}{9}$ from both sides:

$$\frac{x^2}{45^2} = 1 - \frac{8}{9}.$$

2. Simplify $1 - \frac{8}{9}$:

$$1 - \frac{8}{9} = \frac{9}{9} - \frac{8}{9} = \frac{1}{9}.$$

Thus:

$$\frac{x^2}{45^2} = \frac{1}{9}.$$

3. Multiply through by $45^2$ to isolate $x^2$:

$$x^2 = \frac{45^2}{9}.$$

Step 5: Calculate $x$

1. Simplify $\frac{45^2}{9}$:

$$\frac{45^2}{9} = \frac{2025}{9} = 225.$$

2. Take the square root:

$$x = \sqrt{225} = 15 , \text{m}.$$

Thus, the horizontal distance from the center when the height is $20\sqrt{2}$ is $\mathbf{15 , \text{m}}$.

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Key Formulas or Methods Used

1. Equation of an Ellipse:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

2. Semi-Major and Semi-Minor Axes:

$$a = \frac{\text{base width}}{2}, \quad b = \text{height at center}.$$

3. Substitution and Simplification to solve for $x$.

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Summary of Steps

1. Identify the semi-major axis $a = 45$ and semi-minor axis $b = 30$.
2. Write the standard equation of the ellipse:

$$\frac{x^2}{45^2} + \frac{y^2}{30^2} = 1.$$

3. Substitute $y = 20\sqrt{2}$ into the equation and solve for $x^2$.
4. Calculate $x = 15 , \text{m}$.

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Final Answer

The horizontal distance is:

$$\mathbf{x = 15 , \text{m}}$$