Problem 17 - Thomas Calculus

Question

Find the point on the parabola \(x=t, y=t^{2},-\infty< t<\infty\) closest to the point \((2,1 / 2) .\) (Hint: Minimize the square of the distance as a function of \(t .\))

Step-by-Step Solution

Answer

The closest point on the parabola to (2, 1/2) is (1, 1).

Step 1: Set up the distance formula

To find the point on the curve that is closest to the point \((2, \frac{1}{2})\), use the distance formula. The point on the curve is given by \((t, t^2)\). The distance \(d\) from \((t, t^2)\) to \((2, \frac{1}{2})\) can be expressed as:\[ d = \sqrt{(t-2)^2 + (t^2 - \frac{1}{2})^2} \]

Step 2: Simplify using the distance squared

Since minimizing \(d\) is equivalent to minimizing \(d^2\), we'll minimize:\[ d^2 = (t-2)^2 + (t^2 - \frac{1}{2})^2 \] This avoids introducing square roots, making differentiation simpler.

Step 3: Expand the distance squared

Expand each part of the expression:- \((t-2)^2 = t^2 - 4t + 4\)- \((t^2 - \frac{1}{2})^2 = t^4 - t^2 + \frac{1}{4}\)Combine these to get:\[ d^2 = t^4 + t^2 - 4t + 4 + \frac{1}{4} \]

Step 4: Differentiate to find critical points

Differentiate \(d^2\) with respect to \(t\):\[ \frac{d}{dt}(d^2) = 4t^3 + 2t - 4 \]Set this derivative equal to zero to find the critical points:\[ 4t^3 + 2t - 4 = 0 \]

Step 5: Solve the derivative equation

Solve the equation \(4t^3 + 2t - 4 = 0\). Divide throughout by 2 to simplify:\[ 2t^3 + t - 2 = 0 \]Using trial and error or the Rational Root Theorem, check that \(t = 1\) is a solution. Quickly factor the cubic equation as:\[ (t-1)(2t^2+2t+2) = 0 \]

Step 6: Check the second factor

The quadratic factor \(2t^2+2t+2 = 0\) does not have real solutions (as its discriminant \(b^2 - 4ac = 4 - 16 = -12\) is negative), so \(t = 1\) is the only real solution.

Step 7: Verify the closest point

With \(t = 1\), the point on the parabola is \((1, 1^2) = (1, 1)\). Calculate the squared distance from \((1, 1)\) to \((2, \frac{1}{2})\):\[ (1-2)^2 + (1-\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4} \]Since there are no other critical points with real \(t\), this point is indeed the closest.

Key Concepts

Minimization

In calculus, minimization involves finding the smallest value of a function within a given range. In this problem, we aim to determine the point on the parabola that is closest to the point \((2, \frac{1}{2})\). To achieve this, we minimize the distance between these two points. However, working directly with the distance formula, which involves square roots, can be complex. Instead, we minimize the distance squared, \(d^2\). The reason behind this is simple. Since we know that the square of a number is always non-negative, the smallest distance squared corresponds to the smallest actual distance. By minimizing \((t-2)^2 + (t^2 - \frac{1}{2})^2\), we find the optimal value for the parameter \(t\) that gives us the point on the parabola which is closest to our target point.

• It simplifies calculations since squares remove square root complexities.
• Derivative calculations become more manageable without square root terms.

By successfully minimizing \(d^2\), students can accurately determine the point of closest approach, harnessing the power of calculus effectively.

Parabola

A parabola is a U-shaped curve that can open upwards or downwards in a two-dimensional plane. In a Cartesian coordinate system, this type of curve is represented by a quadratic equation. In our example, the parabola is given parametrically as \((x = t, y = t^2)\), indicating the relationship between the variables relative to parameter \(t\). This particular parabola opens upwards since the squared term has a positive coefficient (the nature of the quadratic function).

• The vertex of the parabola, given this specific parametric form, is at the origin \((0, 0)\).
• The parabola extends infinitely in the positive and negative directions on the x-axis, specified by \(-\infty < t < \infty\).

Understanding the geometry of the parabola allows students to visualize where points fall on the curve and how minimizing the distance involves shifting along this parabolic trajectory to find the point \((1, 1)\) that is closest to \((2, \frac{1}{2})\). This awareness is vital in grasping why certain points along the parabola represent better solutions than others.

Derivative

A derivative represents the rate at which a function is changing at any given point and is a fundamental tool in calculus. To find the minimum distance in our problem, we focus on the derivative of \(d^2\), denoted as \(\frac{d}{dt}(d^2)\). This derivative helps identify critical points, where the function's rate of change is zero, potentially signaling a minimum or maximum.

• The first derivative \(4t^3 + 2t - 4\) is derived from differentiating the distance squared function to locate such critical points.
• Setting this derivative equal to zero results in an equation needing solutions, offering values that potentially minimize or maximize the distance.

Upon solving \(4t^3 + 2t - 4 = 0\), we determine \(t = 1\) as a valid solution. Substituting this parameter into the parabola defines the precise point \((1, 1)\) on the graph. Because this is the only real solution from the derivative, \(t = 1\) is confirmed as optimal. Derivatives thus act as a crucial calculus element, allowing for precision in identifying minima or maxima within mathematical contexts.