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. Recall 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 along an ellipse, as described by the position vector r(t) = . Answer: Using the provided position vector r(t) = and the area formula for polar curves, it was shown that the derivative A'(t) = (1/2)ab, which is a constant independent of time t. This constant rate of area change indicates that equal areas are swept out in equal time intervals, consistent with Kepler's second law of planetary motion.

Step-by-step solution

Show that tan(θ) = (b/a)tan(t)

The ellipse is given by the position vector r(t) = . We can find the tangent of the angle θ between the position vector r and the x-axis using trigonometric ratios. To do so, let's find the ratios sin(θ) and cos(θ): sin(θ) = b*sin(t)/|r|, cos(θ) = a*cos(t)/|r|, where |r| = √(a^2*cos^2(t) + b^2*sin^2(t)), is the magnitude of the position vector r(t). Now we can find tan(θ): tan(θ) = sin(θ)/cos(θ) = (b*sin(t))/(a*cos(t)) = (b/a)tan(t).

Find θ'(t)

To find θ'(t), we take the derivative of tan(θ) = (b/a)tan(t) with respect to t: tan'(θ) * θ'(t) = d/dt[(b/a)tan(t)]. Using the chain rule, the derivative of tan(θ) with respect to t is θ'(t)*sec^2(θ) So, θ'(t) * sec²(θ) = (b/a)sec²(t). Since sec(θ) = |r| / √(a^2 * cos^2(t)), we can rearrange the equation to find θ'(t): θ'(t) = (b/a)sec²(t) * (√(a^2 * cos^2(t))/(|r|)).

Show that A'(t) = (1/2)ab

The formula for area in polar coordinates is A(θ) = (1/2) ∫(f(u))^2 du, where f(θ) = |r(θ)|. The position vector is given in Cartesian coordinates, and we have established that tan(θ) = (b/a)tan(t). By letting f(θ(t)) = |r|(θ(t)), we can apply the chain rule to obtain A'(t): A'(t) = (1/2) * d/dt [∫(f(u))^2 du] = (1/2) * (f(θ(t)))^2 * θ'(t). Recall from our previous calculations: f(θ(t))=|r|= √(a^2*cos^2(t) + b^2*sin^2(t)), θ'(t) = (b/a)sec²(t) * (√(a^2 * cos^2(t))/(|r|)). Now, substitute the expressions of f(θ(t)) and θ'(t) into A'(t) equation: A'(t) = (1/2) * (a^2*cos^2(t) + b^2*sin^2(t)) * (b/a)sec²(t) * (√(a^2 * cos^2(t))/(|r|)). Cancelling out some terms, we have: A'(t) = (1/2) * ab.

Conclude that equal areas are swept out in equal times

Since the derivative A'(t) = (1/2)ab is a constant with respect to t, we can conclude that the ellipse sweeps out equal areas in equal times. This is also known as Kepler's second law of planetary motion. The area swept out by the object is proportional to the time spent moving around the ellipse.

Key concepts

Tangent of an Angle

Understanding the tangent of an angle is crucial in trigonometry, and it plays a significant role in the study of different shapes, such as circles and ellipses. In the context of polar coordinates, the tangent represents the ratio of the y-coordinate to the x-coordinate of a point.

For the ellipse described by the position vector \r(t) = \, the angle \(\theta\) made with the x-axis can be expressed in terms of the tangent function as \(\tan \theta = \frac{b}{a} \tan t\). This relationship is derived by dividing the y-component by the x-component of the position vector.

Derivative of Trigonometric Functions

The derivative of trigonometric functions is a cornerstone in calculus, especially in physics and geometry, where motion and change are analyzed. When it comes to the derivative of the tangent function, we use the fact that \(\frac{d}{dt} \tan t = \sec^2 t\).

Applying this knowledge to our problem, the derivative of \(\theta\) with respect to time, denoted as \(\theta'(t)\), can be found using the chain rule and understanding that the secant square of an angle is the square of the hypotenuse over the square of the adjacent side in a right triangle. In the given ellipse problem, this leads to the relation \(\theta'(t) \sec^2(\theta) = \frac{b}{a}\sec^2(t)\).

Area Calculation in Polar Coordinates

Calculating area in polar coordinates can be visualized as summing up a series of infinitesimally small sectors of circles, each with a different radius. For any polar curve described by \(r = f(\theta)\), the area from the origin to the curve between angles 0 and \(\theta\) is given by the integral \(A(\theta) = \frac{1}{2} \int_{0}^{\theta}(f(u))^2 du\).

Within our ellipse problem, we've exploited this formula to demonstrate that the rate of change of the area with respect to time, \(A'(t)\), is constant. This constant rate, which we've found to be \(\frac{1}{2} ab\), suggests a uniformity in area 'swept' over any given time, connecting neatly with physical principles such as the conservation of angular momentum.

Kepler's Second Law

Kepler's Second Law, also known as the law of equal areas, states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This law is a result of angular momentum conservation in celestial mechanics.

The implication of this for our problem with the ellipse is profound. Since the area derivative \(A'(t)\) is constant, it implies that an object moving along the path of our ellipse will cover equal areas in equal times. This mirrors the behavior of planets orbiting the Sun, where the speed of the planet increases as it approaches the Sun and decreases as it moves away, resulting in equal areas being covered in the same amount of time regardless of the planet's position in its orbit.