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Chapter 11: Problem 75

A racetrack is in the shape of an ellipse 100 feet long and 50 feet wide. What is the width 10 feet from a vertex?

Short Answer

Expert verified
The width 10 feet from a vertex is 30 feet.

Step by step solution

01

Identify the general form of ellipse equation

An ellipse centered at the origin has the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, a is the semi-major axis and b is the semi-minor axis.
02

Determine the semi-major and semi-minor axes

Given ellipse dimensions are 100 feet long and 50 feet wide. So, the semi-major axis (a) is 50 feet (half of 100) and the semi-minor axis (b) is 25 feet (half of 50).
03

Substitute known values into the ellipse equation

Rewrite the ellipse equation with the given values: \(\frac{x^2}{50^2} + \frac{y^2}{25^2} = 1\)
04

Find the y-coordinate when x is 10 feet from the vertex

The distance 10 feet from the vertex is along the x-axis, hence x = 50 - 10 = 40. Substitute x = 40 into the equation to find y: \(\frac{40^2}{50^2} + \frac{y^2}{25^2} = 1\)
05

Simplify and solve for y

First, compute \( \frac{40^2}{50^2} = \frac{1600}{2500} = \frac{16}{25} \). The equation becomes: \( \frac{16}{25} + \frac{y^2}{625} = 1 \). Then, calculate \( \frac{y^2}{625} = 1 - \frac{16}{25} = \frac{9}{25} \). Thus, \ y^2 = 625 \times \frac{9}{25} = 225\. So, y = \pm \sqrt{225} = \pm 15 foot.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

semi-major axis
The semi-major axis is one of the key elements that define an ellipse. In a standard ellipse equation, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the variable \a\ represents the length of the semi-major axis. The semi-major axis is the longest radius that runs from the center to the edge of the ellipse. It essentially dictates the 'width' of the ellipse along its longest part.
In the given exercise, the total length of the racetrack is 100 feet. Therefore, the semi-major axis is half of this length. So:
  • Given total length (major axis) = 100 feet
  • Semi-major axis (a) = 100 / 2 = 50 feet
This means that from the center of the ellipse to the farthest point on the track, the distance is 50 feet. Understanding the semi-major axis helps in identifying the proportions and dimensions of the ellipse accurately.
semi-minor axis
The semi-minor axis is another critical part of the ellipse. In the ellipse equation, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the \b\ value represents the semi-minor axis. The semi-minor axis is the shorter radius that also extends from the center to the edge of the ellipse, perpendicular to the semi-major axis.
For our racetrack example, the width of the track is 50 feet, so the semi-minor axis is half of this width:
  • Given total width (minor axis) = 50 feet
  • Semi-minor axis (b) = 50 / 2 = 25 feet
This means that from the center of the ellipse to the shortest edge of the track, the distance is 25 feet. Identifying the semi-minor axis is essential for drawing and understanding the shape of the ellipse accurately.
solving for coordinates
Once the semi-major and semi-minor axes are known, we can solve for coordinates on the ellipse using the standard formula. The racetrack problem requires finding the width of the track (i.e., the y-coordinate) 10 feet away from a vertex along the semi-major axis. Here is how you solve it step-by-step:
  • First identify the vertex on the x-axis which is \a = 50\ feet.
  • Since we are looking for the point 10 feet away from this vertex: \x = 50 - 10 = 40\ feet.
Now, substitute \x = 40\ into the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\):
  • \ \frac{40^2}{50^2} + \frac{y^2}{25^2} = 1\
  • Calculate \ \frac{40^2}{50^2} = \frac{1600}{2500} = \frac{16}{25}\
  • The equation now becomes \ \frac{16}{25} + \frac{y^2}{625} = 1\
Next, solve for \ y^2 \:
  • \ \frac{y^2}{625} = 1 - \frac{16}{25} = \frac{9}{25}\
  • \

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Most popular questions from this chapter

Parametric equations of four plane curves are given. Graph each of them, indicating the orientation. \(\begin{array}{ll}C_{1}: & x(t)=t, \quad y(t)=\sqrt{1-t^{2}} ; \quad-1 \leq t \leq 1 \\ C_{2}: & x(t)=\sin t, \quad y(t)=\cos t ; \quad 0 \leq t \leq 2 \pi \\\ C_{3}: & x(t)=\cos t, \quad y(t)=\sin t ; \quad 0 \leq t \leq 2 \pi \\\ C_{4}: & x(t)=\sqrt{1-t^{2}}, \quad y(t)=t ; \quad-1 \leq t \leq 1\end{array}\)

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