Question

Consider the ellipse \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(a\) and \(b\) are real numbers. Let \(\theta\) be the angle between the position vector and the \(x\) -axis. a. Show that \(\tan \theta=(b / a) \tan t\) b. Find \(\theta^{\prime}(t)\) c. Note that the area bounded by the polar curve \(r=f(\theta)\) on the interval \([0, \theta]\) is \(A(\theta)=\frac{1}{2} \int_{0}^{\theta}(f(u))^{2} d u\) Letting \(f(\theta(t))=|\mathbf{r}(\theta(t))|,\) show that \(A^{\prime}(t)=\frac{1}{2} a b\) d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

Short answer

Question: Show that equal areas are swept out in equal times for an object moving in an elliptical orbit. Answer: Since the derivative of the area with respect to time, \(A'(t)\), is constant and equal to \(\frac{1}{2}ab\), it means that the rate of area being swept out is the same for any point on the ellipse. Thus, as the object moves around the ellipse, it sweeps out equal areas in equal times.

Step-by-step solution

a. Show that \(\tan \theta=(b / a) \tan t\)

Let \(\mathbf{r}(t) = \langle a \cos t, b \sin t \rangle\). Since \(\theta\) is the angle between the position vector and the x-axis, we can define $$\tan \theta = \frac{y}{x}$$ From the given position vector, $$x = a \cos t \quad \mathrm{and} \quad y = b \sin t$$ Therefore, $$\tan \theta = \frac{b \sin t}{a \cos t} = \frac{b}{a} \cdot \frac{\sin t}{\cos t} = \frac{b}{a} \tan t$$

b. Find \(\theta'(t)\)

To find \(\theta'(t)\), we will first differentiate \(\tan \theta\) and \(\tan t\) with respect to \(t\). $$\frac{d(\tan \theta)}{d\theta} \cdot \frac{d\theta}{dt} = \frac{d (\frac{b}{a} \tan t)}{dt}$$ $$\sec^2 \theta \cdot \theta'(t) = \frac{b}{a} \cdot \sec^2 t$$ Now, we can use the relation we found in part a: $$\tan \theta = \frac{b}{a} \tan t \Rightarrow \tan^2 \theta = \frac{b^2}{a^2}\tan^2 t$$ Since \(\sec^2 x = 1 + \tan^2 x\), we have $$\sec^2 \theta = 1 + \frac{b^2}{a^2} \tan^2 t, \quad \sec^2 t = 1 + \tan^2 t$$ Substitute into the above equation and simplify: $$\theta'(t) (1 + \frac{b^2}{a^2} \tan^2 t) = \frac{b}{a} \cdot (1 + \tan^2 t)$$ $$\theta'(t) = \frac{b}{a}\frac{1 + \tan^2 t}{1 + \frac{b^2}{a^2} \tan^2 t}$$

c. Show that \(A'(t)=\frac{1}{2} ab\)

We are given the formula for the area A as a function of \(\theta\): $$A(\theta) = \frac{1}{2} \int_0^\theta f(u)^2 du$$ Now we need to find the derivative of A with respect to t: $$A'(t) = \frac{d}{dt}(\frac{1}{2} \int_0^{\theta(t)} f(u)^2 du)$$ Using the Fundamental Theorem of Calculus and the Chain Rule, $$A'(t) = \frac{1}{2} f(\theta(t))^2 \theta'(t)$$ We know that the position vector is of the form \(\mathbf{r}(t) = \langle a \cos t, b \sin t \rangle\), so $$f(\theta(t)) = |\mathbf{r}(\theta(t))| = \sqrt{(a \cos t)^2 + (b \sin t)^2} = \sqrt{a^2\cos^2 t + b^2\sin^2 t}$$ Squaring \(f(\theta(t))\), $$f(\theta(t))^2 = a^2\cos^2 t + b^2\sin^2 t$$ Hence, $$A'(t) = \frac{1}{2}(a^2\cos^2 t + b^2\sin^2 t) \cdot \frac{b}{a}\frac{1 + \tan^2 t}{1 + \frac{b^2}{a^2} \tan^2 t}$$ After simplifying, we get $$A'(t) = \frac{1}{2} ab$$

d. Conclude that equal areas are swept out in equal times

Since the derivative of the area with respect to time, \(A'(t)\), is constant and equal to \(\frac{1}{2}ab\), it means that the rate of area being swept out is the same for any point on the ellipse. Thus, as the object moves around the ellipse, it sweeps out equal areas in equal times.

Key concepts

Parametric Equations

Parametric equations allow us to describe curves using a third parameter, often denoted as \(t\). Unlike traditional equations that only describe \(x\) in terms of \(y\) or vice versa, parametric equations provide equations for both \(x\) and \(y\) separately. In the context of an ellipse,

• \(\mathbf{r}(t) = \langle a \cos t, b \sin t \rangle\) represents the position of a point on the ellipse for a given \(t\).
• The parameters \(a\) and \(b\) are the semi-major and semi-minor axes of the ellipse, respectively.
• The parameter \(t\) usually varies from \(0\) to \(2\pi\), covering the entire ellipse.

By manipulating parametric equations, it's possible to analyze various properties of the curve like orientation, curvature, and more. This flexibility is one reason they are so useful in calculus and beyond.

Using parametric equations simplifies the study of ellipses and their geometric properties, transforming complex curves into a set of simple trigonometric relationships.

Differentiation

Differentiation of parametric equations involves finding the rate of change of one coordinate concerning the parameter.

• When differentiating the position vector \(\langle a \cos t, b \sin t \rangle\), we focus on how these positions change as \(t\) changes.
• For an ellipse, differentiating tells us about its angular properties, such as how quickly it moves along the curve.

To find the derivative of the angle \(\theta\) between the position vector and the \(x\)-axis, apply the chain rule.

• Start by noting that \(\tan \theta = \frac{b}{a} \tan t\).
• Differentiate both sides with respect to \(t\) using caution around trigonometric identities and division.

Differentiation thus helps reveal dynamic properties of the ellipse, such as how the tangent angle changes, providing deeper insight into the curve's geometry.

Area Under Curves

In calculus, the area under a curve is often evaluated using integration. For parametric equations, this often involves converting the problem into an integral that spans the range of the parameter.

• The exercise provides that the area \(A(\theta)\) bounded by a polar curve can be expressed as an integral:
• \(A(\theta) = \frac{1}{2} \int_{0}^{\theta} (f(u))^2 \, du\).

In the case of an ellipse, finding the area it sweeps reduces to evaluating such an integral over \(t\).

• Using the chain rule and the Fundamental Theorem of Calculus bridges \(f(\theta(t))\) with its derivative, \(A'(t)\).
• The function \(f(\theta(t))\) describes the magnitude of the vector from the origin to the ellipse's perimeter, \(f(\theta(t)) = |\mathbf{r}(\theta(t))|\).

Thus, calculus allows one to not only understand the geometric area swept out by the curve but also how it develops over time as the point traverses the ellipse, highlighting the uniformity of the area sweep across equal time intervals.