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Chapter 9: Problem 54

Find the shortest distance between the curves \(y^{2}=4 x\) and \(x^{2}+(y+12)^{2}=1\)

Short Answer

Expert verified
The most accurate short answer to this question will lie in the particular numerical results of Steps 3 and 4, which rely on some straightforward but potentially lengthy arithmetic and algebra. Proceed through the calculations outlined in the four steps to arrive at the short answer.

Step by step solution

01

Equation for the Distance Between Two Points

The general formula to calculate the distance \(d\) between two points \((x_{1},y_{1})\) and \((x_{2},y_{2})\) in a plane is given by \(d = \sqrt{{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}}\). Now, select an arbitrary point \((x,y)\) on the parabola \(y^{2}=4x\) and another arbitrary point \((x,-12+ \sqrt{1-x^{2}})\) on the circle \(x^{2}+(y+12)^{2}=1\). Calculate the distance between the points.
02

Calculating d(x)

Substitute these points into the distance formula yielding the function \(d(x)=\sqrt{{(x-x)^2 + (\sqrt{4x}+12- \sqrt{1-x^{2}})^2}}\). Simplify the expression to obtain our distance function \(d(x)\) in terms of \(x\).
03

Finding the Minimum Distance

Find the derivative \(d'(x)\) of the distance function \(d(x)\). Set \(d'(x) = 0\) and solve for \(x\) to find critical points of \(d(x)\). Make sure to check the endpoints of the interval of \(x\) as determined by the circle equation. The smallest \(d(x)\) value corresponding to these \(x\) values will be the shortest distance between the curves.
04

Ensuring the Minimum is Positive

The shortest distance must be a positive number. If not, choose a different point on the circle or adjust the calculation. Validate that the point lies on both the parabola and the circle, indicating that the shortest distance is truly the distance between the two curves.

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Key Concepts

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

Parabola Equation
A parabola is a U-shaped curve that can open upwards, downwards, left, or right. It is defined by a quadratic polynomial equation. The equation for a standard parabola that opens rightwards is in the form \( y^2 = 4ax \). Here, each pair \((x, y)\) satisfying this equation lies on the parabola.
  • In our exercise, the parabola is represented by \( y^2 = 4x \), which denotes a standard rightward-opening form, centered at the origin \(0, 0\).
  • The parameter \( a \) helps determine the parabola's width and direction of opening. In this case, \( 4a = 4 \), so \( a = 1 \).
Understanding parabola equations provides clarity on how one part of the initial distance calculation is framed, aiding in comprehending the coordinate system used for finding the shortest distance.
Circle Equation
A circle is defined in two-dimensional space by the equation \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center, and \( r \) is the radius.
  • In our exercise, the circle equation is given as \( x^2 + (y+12)^2 = 1 \). This equation describes a circle centered at \((0, -12)\) with a radius equal to 1.
  • The offset in the \( y \)-coordinate indicates that the circle has been shifted down from the origin by 12 units.
This understanding is critical as it not only shapes the set of points on the circle but also influences the set area within which the shortest distance to the parabola can be sought.
Distance Formula
The distance between two points in Cartesian coordinates is calculated using the distance formula: \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
  • In our exercise, we use this formula to find the distance between points on the given parabola and circle.
  • Choosing important points: One on the parabola, \((x, y)\), and one on the circle, \((x, -12 + \sqrt{1-x^2})\), helps establish the function \( d(x) \).
This function represents how the distance changes as \( x \) varies. It becomes central when determining the minimum distance, showcasing how geometry and algebra intertwine to provide actual solutions.
Derivative and Critical Points
The process of finding the shortest distance involves calculus, particularly derivatives and critical points. To find where a function reaches its minimum or maximum, we find its derivative and solve for zero.
  • Calculating \( d'(x) \) for the distance function allows us to locate critical points where the rate of change is zero.
  • The critical points are potential candidates for minimum distance, but we must also check end conditions set by the circle.
It's also important to ensure that these points satisfy the conditions of being on both the parabola and circle, verifying that the computed shortest distance is legitimate and feasible. This step bridges the gap between calculation and practical application!

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