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Chapter 7: Problem 22

Find the point on the parabola \(y=x^{2}\) that has minimal distance from the point \(\left(16, \frac{1}{2}\right).\) [See Fig. 2(b).] [Suggestion: If d denotes the distance from \((x, y)\) to \(\left(16, \frac{1}{2}\right),\) then \(d^{2}=(x-16)^{2}+\left(y-\frac{1}{2}\right)^{2} .\) If \(d^{2}\) is minimized, then \(d\) will be minimized.]

Short Answer

Expert verified
The point on the parabola \( y = x^2 \) that is closest to \((16, \frac{1}{2})\) is \((2, 4)\).

Step by step solution

01

- Define the distance formula

The distance between a point \( (x, y) \) on the parabola and the point \( \left(16, \frac{1}{2}\right) \) is given by the formula \[ d = \sqrt{(x - 16)^2 + \left(y - \frac{1}{2}\right)^2} \]. We will minimize \(d^2\) to find the minimal distance.
02

- Substitute the parabola equation

Since the point is on the parabola \( y = x^2 \), substitute \( y = x^2 \) into the distance formula to get \[ d^2 = (x - 16)^2 + \left(x^2 - \frac{1}{2}\right)^2 \]
03

- Simplify the expression

Expand and simplify the expression: \[ d^2 = (x - 16)^2 + \left(x^2 - \frac{1}{2}\right)^2 \] becomes \[ d^2 = (x - 16)^2 + \left(x^4 - x^2 + \frac{1}{4}\right) \]
04

- Expand the squared terms

Expand the squared terms: \[ d^2 = (x^2 - 32x + 256) + (x^4 - x^2 + \frac{1}{4}) \] becomes \[ d^2 = x^4 + x^2 - x^2 - 32x + 256 + \frac{1}{4} \], which simplifies to \[ d^2 = x^4 - 32x + 256 + \frac{1}{4} \]
05

- Take the derivative

To minimize \( d^2 \), take the derivative with respect to \( x \): \[ \frac{d}{dx}(d^2) = \frac{d}{dx}(x^4 - 32x + 256 + \frac{1}{4}) = 4x^3 - 32 \]
06

- Find the critical points

Set the derivative to zero to find the critical points: \[ 4x^3 - 32 = 0 \] Solving for \( x \), we get: \[ 4x^3 = 32 \rightarrow x^3 = 8 \rightarrow x = 2 \]
07

- Get the corresponding y-coordinate

Plug \( x = 2 \) back into the parabola equation \( y = x^2 \): \[ y = 2^2 = 4 \]
08

- Verify that it's a minimum

Since \( d^2 \) is a polynomial with a positive leading coefficient, so the critical point \(x = 2 \) gives us the minimum distance.

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Key Concepts

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

distance minimization
In calculus, one common problem is to find the minimum distance between a point and a curve. For this exercise, we need to find the point on the parabola defined by the equation \( y = x^2 \) that is closest to another fixed point, \( (16, \frac{1}{2}) \). We begin by using the distance formula to represent the distance \(d\) between any point \((x, y)\) on the parabola and the point \((16, \frac{1}{2})\):
\[ d = \sqrt{(x - 16)^2 + \( y - \frac{1}{2} \) ^2 } \].
Instead of minimizing \(d\) directly, it's easier to minimize \(d^2\) because their minimum points are the same. This approach avoids dealing with the square root, simplifying our calculations.
derivative
To find the minimum value of \(d^2\), we use the concept of the derivative. The derivative helps us understand how a function changes as its input changes. For the function \(d^2 = (x - 16)^2 + (x^2 - \frac{1}{2})^2\), we take the derivative with respect to \(x\). The derivative of a function \(f(x)\) is denoted as \(f'(x)\) or \(\frac{df}{dx}\). By finding where this derivative equals zero, we discover potential minimum or maximum points. In our problem, the derivative of \(d^2\) with respect to \(x\) is \( \frac{d}{dx}(d^2) = 4x^3 - 32\). We then set this equation to zero and solve for \(x\):
\[ 4x^3 - 32 = 0 \rightarrow x^3 = 8 \rightarrow x = 2 \].
critical points
Critical points occur where the derivative of a function is zero or undefined. In the context of our distance minimization problem, we set the derivative \( 4x^3 - 32 \) to zero and solve for \(x\). We found that the critical point is \( x = 2 \). Critical points are useful because they help us identify where a function changes direction, indicating potential minimum or maximum values. However, we should verify whether this critical point leads to a minimum distance. For our polynomial function \( d^2 = x^4 - 32x + 256 + \frac{1}{4} \), the positive leading coefficient means it opens upwards, confirming that \(x = 2\) gives a minimum distance.
parabola
A parabola is a symmetrical, U-shaped curve that can open upwards or downwards. It's generally represented by the equation \( y = ax^2 + bx + c \). In our exercise, the parabola is \( y = x^2 \) which opens upwards. We are tasked to find a point on this parabola closest to the point \( (16, \frac{1}{2}) \). After substituting \( y = x^2 \) into the distance formula and simplifying, we solved for \(x = 2 \) at the critical point. Substituting \( x = 2\) back into the parabola equation gives the corresponding \( y \)-coordinate:

\( y = 2^2 = 4 \). Therefore, the point on the parabola \( y = x^2 \) that is closest to \( (16, \frac{1}{2}) \) is \((2, 4)\).

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