Problem 40 - Thomas Calculus

Question

Find the point on the ellipse \(x=2 \cos t, y=\sin t, 0 \leq t \leq 2 \pi\) closest to the point \((3 / 4,0) .\) (Hint: Minimize the square of the distance as a function of \(t\) .

Step-by-Step Solution

Answer

The point \((2\cos t, \sin t)\) closest to \((3/4, 0)\) occurs at \(t = 0\), which is \((2, 0)\).

Step 1: Write Down the Distance Formula

To find the point on the ellipse closest to \((3/4, 0)\), we start by writing the distance squared between any point on the ellipse \((x,y)\) and \((3/4, 0)\). The distance squared is given by \((x - 3/4)^2 + y^2\).

Step 2: Substitute Parametric Equations

Substitute the parametric equations of the ellipse: \(x = 2 \cos t\) and \(y = \sin t\) into the distance squared formula. This gives us \[(2 \cos t - \frac{3}{4})^2 + (\sin t)^2.\]

Step 3: Simplify the Expression

Expand the expression obtained in Step 2:\[(4 \cos^2 t - 3 \cos t + \frac{9}{16}) + \sin^2 t.\]Combine the trigonometric terms by using the identity \(\cos^2 t + \sin^2 t = 1\), resulting in:\[5 \cos^2 t - 3 \cos t + \frac{9}{16} + 1.\]

Step 4: Differentiate with Respect to t

We differentiate the expression \(5 \cos^2 t - 3 \cos t + \frac{25}{16}\) with respect to \(t\) to find the critical points. Use the derivative formulas: \(\frac{d}{dt}(\cos^2 t) = -2 \cos t \sin t\) and \(\frac{d}{dt}(\cos t) = -\sin t\).

Step 5: Set Derivative Equal to Zero

The derivative of the expression is \[-10 \cos t \sin t + 3 \sin t = 0.\]Set this derivative equal to zero to solve for \(t\), yielding:\[\sin t (3 - 10 \cos t) = 0.\]

Step 6: Solve for t

The equation \(\sin t (3 - 10 \cos t) = 0\) gives us two cases: \(\sin t = 0\) and \(3 - 10 \cos t = 0\). Solve these equations to find the critical points. The first case gives \(t = 0, \pi\). The second case gives \(\cos t = \frac{3}{10}\).

Step 7: Determine the Closest Point

Evaluate the distance squared at each value of \(t\) obtained: \(t = 0, \pi, \cos^{-1}(3/10)\).- At \(t = 0\), the point is \((2, 0)\) and the distance squared is \((2 - 3/4)^2 = 9/4\).- At \(t = \pi\), the point is \((-2, 0)\) and the distance squared is \((5/4)^2 = 25/16\).- At \(\cos t = 3/10\), calculate \(\sin t = \sqrt{1 - (3/10)^2}\) and evaluate the distance.Choose the \(t\) with the smallest distance squared.

Key Concepts

Distance Formula

To tackle problems involving distances between points, the distance formula is a crucial tool. In this exercise, we aim to find the point on an ellipse that is closest to another given point, \(\left(\frac{3}{4}, 0\right)\). For such tasks, minimizing the distance often involves calculating the distance squared, which simplifies manipulation and avoids square roots. The distance squared formula for any two points \(x_1, y_1\) and \(x_2, y_2\) is given by:

• \((x_1 - x_2)^2 + (y_1 - y_2)^2\)

This formula allows us to directly substitute the coordinates of interest. By focusing on the distance squared, we eliminate complications from the derivative and comparison processes. This direct substitution results in an expression entirely in the parameter \(t\), which can be analyzed through calculus to find the minimum distance configuration.

Parametric Equations

Parametric equations allow us to describe geometric figures, such as ellipses, using a parameter—in this case, \(t\). The parametric equations given in the problem are:

• \(x = 2\cos t\)
• \(y = \sin t\)

These equations signify the positions of points on the ellipse as \(t\) varies from \(0\) to \(2\pi\). This is a way to depict curves markedly different from traditional Cartesian coordinates (where x and y are not directly interconnected by a third variable such as \(t\)).

By using these parametric forms, substitutions into the distance squared equation become straightforward. It efficiently transforms the problem into one that revolves around a single variable, simplifying both differentiation and optimization.

Trigonometry

Trigonometry plays a vital role in solving problems involving ellipses and the optimization of functions dependent on angles. In this problem, the ellipse is expressed through sine and cosine functions, which are fundamental trigonometric components. The relationship:

• \(\cos^2 t + \sin^2 t = 1\)

allows for simplification when these trigonometric expressions complicate the formulae.

When we deal with derivatives of trigonometric functions, knowing standard derivations, such as \(\frac{d}{dt}(\cos t) = -\sin t\) and \(\frac{d}{dt}(\sin t) = \cos t\), is indispensable. This knowledge aids in the differentiation needed for identifying critical points in function behavior, allowing us to pinpoint minimum or maximum distances effectively. While this exercise particularly tackles a minimum distance problem, the same principles apply symmetrically to other extremum-related mathematics challenges.