Problem 27 - Contemporary Precalculus

Question

Calculus can be used to show that the area of the ellipse with equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi\)ab. Use this fact to find the area of each ellipse. $$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$$

Step-by-Step Solution

Answer

Answer: The area of the ellipse is \(8\pi\) square units.

Step 1: Find the semi-major axis a and semi-minor axis b

In the given ellipse equation, the terms \(\frac{x^2}{16}\) and \(\frac{y^2}{4}\) indicate that \(a^2 = 16\) and \(b^2 = 4\). To find the values of \(a\) and \(b\), we can take the square root of these values, giving us: $$a=\sqrt{16}=4$$ $$b=\sqrt{4}=2$$

Step 2: Calculate the area of the ellipse

Now that we know the values of \(a\) and \(b\), we can use the formula for the ellipse area, which is \(A = \pi ab\). Plugging in the values we found in Step 1, we get: $$A=\pi (4)(2)$$

Step 3: Solve

Now we can perform the multiplication to find the area of the ellipse: $$A=8\pi$$ So, the area of the ellipse is \(8\pi\) square units.

Key Concepts

Semi-Major and Semi-Minor Axis

The semi-major and semi-minor axes are fundamental components in the geometry of an ellipse. An ellipse can be envisioned as a stretched circle, with the semi-major axis representing the longest radius, and the semi-minor axis, the shortest. Think of the semi-major axis as the half-length of the ellipse's longest diameter, while the semi-minor axis is the half-length of the shortest diameter. These axes are crucial because not only do they define the shape and size of the ellipse, but they're also integral to calculating the ellipse's area.

When given an equation in the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), the values \(a\) and \(b\) are squared representations of the ellipse’s semi-major and semi-minor axes. For the illustration \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\), the semi-major axis, \(a\), is the square root of 16, while the semi-minor axis, \(b\), is the square root of 4. Therefore, knowing these values is step one in determining the ellipse's area.

Precalculus

Precalculus forms the foundational knowledge necessary to understand concepts in calculus, including the study of ellipses. It's in precalculus that students are first introduced to the conic sections: circles, ellipses, parabolas, and hyperbolas. Here, you learn about the equations that represent each of these shapes. For an ellipse, the standard form equation is a quintessential part of precalculus learning: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).

Students use this equation to explore properties of ellipses, such as the axes length and the coordinates of foci without actually using calculus concepts like integration. Recognizing the values of \(a\) and \(b\) and being comfortable with manipulating square roots and other algebraic concepts is what precalculus is all about. Additionally, understanding the relationship between these values and the shape and size of the ellipse is crucial before one can apply calculus to these shapes.

Calculus Applications

When students transition from precalculus to calculus, they encounter a wide array of applications, and one significant advancement is the use of calculus in determining areas. Whereas the area of a rectangle or triangle can be easily found with simple arithmetic, an ellipse requires a more in-depth approach. Calculus can demonstrate that the area of an ellipse with the equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi ab\).

This result is arrived at using integration, which allows us to add up infinitely small pieces to find the whole. This process involves setting up an integral that determines the sum of all the differential areas and is essential for determining areas of shapes with curved boundaries like ellipses. Thus, calculus turns out to be the perfect tool for handling more complex geometric figures and their properties, extending beyond the capabilities of precalculus.