/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Show that an ellipse with semima... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that an ellipse with semimajor axis \(a\) and semiminor axis \(b\) has area \(A=\pi a b\).

Short Answer

Expert verified
The area of an ellipse with semimajor axis \(a\) and semiminor axis \(b\) is \(\pi a b\).

Step by step solution

01

Understand the General Formula for Ellipse Area

An ellipse is a shape that looks like a stretched circle. For any ellipse with semimajor axis \(a\) (the longest diameter) and semiminor axis \(b\) (the shortest diameter), its area \(A\) can be calculated using the formula \(A = \pi a b\). We need to understand where this formula comes from.
02

Recall Circle Area Formula

The area of a circle with radius \(r\) is given by \(A_{circle} = \pi r^2\). This formula will help us understand the area formula for an ellipse, given that an ellipse is essentially a 'stretched' circle.
03

Compare an Ellipse to a Circle

Imagine that a circle with radius \(r\) is transformed into an ellipse. The semimajor axis \(a\) corresponds to the radius stretched in one direction, and the semiminor axis \(b\) to the radius compressed in the perpendicular direction. If you start with a circle \(A_{circle} = \pi r^2\) and stretch/compress it, you transform it into \(A_{ellipse} = \pi a b\).
04

Use Integration to Derive the Ellipse Area

Though not always necessary to show intuitively, you can integrate to find the area of an ellipse using the parametric equations \(x = a \cos(t)\) and \(y = b \sin(t)\). Integrate over the interval from 0 to \(2\pi\) to find the area: \[A = \int_0^{2\pi} ab \sin(t) \cos(t) \, dt \]. This integral evaluates to \(\pi ab\), confirming the formula \(A = \pi a b\).

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Key Concepts

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

Ellipse Geometry
Ellipses are fascinating geometric shapes found frequently in mathematics and nature. Imagine an elongated circle; that’s essentially what an ellipse is. It has two axes - the longer one, called the **semimajor axis** (denoted as \(a\)), and the shorter one, known as the **semiminor axis** (denoted as \(b\)). These axes intersect at the center of the ellipse.
By understanding ellipses as stretched circles, you can appreciate how their areas are calculated. The geometric properties of ellipses differ from circles, as ellipses have eccentricity, a measure of how much they deviate from being a perfect circle. A circle, with radius \(r\), is merely a special case of an ellipse where \(a = b = r\).
To calculate the area of an ellipse, you simply multiply \(\pi\) (a mathematical constant) by both the semimajor and semiminor axes: \(A = \pi a b\).
This formula reflects the concept of stretching a circle to form an ellipse. An important thing to remember is, no matter how much you stretch or compress a circle into an ellipse, the formula retains its simplicity, thanks to its elegant derivation.
Integration in Calculus
Integration is a powerful tool in calculus that allows us to calculate areas under curves, among many other applications. In the case of ellipses, integration helps us derive the area formula \(A = \pi a b\). When dealing with ellipses, we often look at their parametric equations: \(x = a \cos(t)\) and \(y = b \sin(t)\).
Imagine drawing an ellipse by plotting these equations from \(t = 0\) to \(t = 2\pi\). These parametric forms represent how the points are traced along the ellipse's perimeter. By setting up an integral to evaluate the area enclosed by these curves, you apply the principles of calculus to a geometrical figure. This integral: \[A = \int_0^{2\pi} ab \sin(t) \cos(t) \, dt\], helps us accumulate the total area as you integrate over one complete cycle of the ellipse.
While the mathematical steps for solving the integral might look complex, the result - \(\pi ab\) - simplifies beautifully to confirm the traditional formula. This shows how calculus serves to both prove and provide insight into geometric properties.
Circle Area Formula
The area of a circle is a cornerstone concept in geometry and serves as a reference point for understanding more complex shapes like ellipses. The formula \(A_{circle} = \pi r^2\) tells us how much space a circle occupies based on its radius \(r\).
Think of a circle as a perfect shape with all points at an equal distance from its center. The simplicity of a circle's structure makes its area formula easy to grasp and enjoyable to use in various mathematical contexts.
The transition from a circle to an ellipse involves transforming the circle through stretching or compressing. This process reflects in the formulas for their areas: just as \(A_{circle} = \pi r^2\) represents a perfect circle, \(A = \pi a b\) extends this concept to account for the stretching and compressing inherent in forming an ellipse. Understanding this connection helps reinforce why the ellipse’s area formula holds true as an adaptation of the circle’s geometry.

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Most popular questions from this chapter

In the late seventeenth century the Italian astronomer Giovanni Domenico Cassini (1625-1712) introduced the family of curves $$ \left(x^{2}+y^{2}+a^{2}\right)^{2}-b^{4}-4 a^{2} x^{2}=0 \quad(a>0, b>0) $$ in his studies of the relative motions of the Earth and the Sun. These curves, which are called \mathrm{\\{} C a s s i n i ~ o v a l s , ~ h a v e ~ one of the three basic shapes shown in the accompanying figure. (a) Show that if \(a=b\), then the polar equation of the Cassini oval is \(r^{2}=2 a^{2} \cos 2 \theta\), which is a lemniscate. (b) Use the formula in Exercise 71 to show that the lemniscate in part (a) is the curve traced by a point that moves in such a way that the product of its distances from the polar points \((a, 0)\) and \((a, \pi)\) is \(a^{2}\).

Find the area of the region described. The region enclosed by the inner loop of the limaçon \(r=1+2 \cos \theta .\) [Hint: \(r \leq 0\) over the interval of integration.]

(a) Use a graphing utility to generate the graph of the parametric curve $$ x=\cos ^{3} t, \quad y=\sin ^{3} t \quad(0 \leq t \leq 2 \pi) $$ and make a conjecture about the values of \(t\) at which singular points occur. (b) Confirm your conjecture in part (a) by calculating appropriate derivatives.

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. $$ \begin{aligned} &\text { Find the area of the surface generated by revolving the curve }\\\ &x=\cos ^{2} t, y=\sin ^{2} t(0 \leq t \leq \pi / 2) \text { about the } y \text { -axis. } \end{aligned} $$

Sketch the curve in polar coordinates. \(r=-2 \cos 2 \theta\)

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