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Chapter 21: Problem 103

The minimum distance of a point on the curve \(\mathrm{y}=\mathrm{x}^{2}-4\) from the origin is: \(\quad\) Online April 9, 2016] (a) \(\frac{\sqrt{15}}{2}\) (b) \(\sqrt{\frac{19}{2}}\) (c) \(\sqrt{\frac{15}{2}}\) (d) \(\frac{\sqrt{19}}{2}\)

Short Answer

Expert verified
The minimum distance from the origin to the curve is \(\frac{\sqrt{15}}{2}\).

Step by step solution

01

Define the Distance Formula

To find the minimum distance of a point \((x, y)\) on the curve \( y = x^2 - 4 \) from the origin \((0,0)\), we'll use the distance formula: \( D = \sqrt{x^2 + y^2} \).Since \(y = x^2 - 4\), substitute \(y\) into the distance formula.
02

Substitute the Curve Equation

Substitute \( y = x^2 - 4 \) into the distance formula. We have:\[D = \sqrt{x^2 + (x^2 - 4)^2}\]Simplify the equation to express \(D\) in terms of \(x\): \[D = \sqrt{x^2 + (x^4 - 8x^2 + 16)}= \sqrt{x^4 - 7x^2 + 16}\]
03

Simplify and Make It Square-free

To minimize \(D\), it may be easier to minimize \(D^2\) (the distance squared), to avoid dealing with the square root:\[D^2 = x^4 - 7x^2 + 16\]This expression is now ready for differentiation to find the minimum point.
04

Differentiate the Function

Differentiate \(D^2 = x^4 - 7x^2 + 16\) with respect to \(x\). We need \( \frac{d}{dx}(D^2)\).\[\frac{d(D^2)}{dx} = 4x^3 - 14x\]
05

Find Critical Points

To find critical points, solve for \(x\) where the derivative is zero:\[4x^3 - 14x = 0\]Factor out \(2x\):\[2x(2x^2 - 7) = 0\]So \(x = 0\) or \(2x^2 - 7 = 0\). Solve \( 2x^2 - 7 = 0 \):\[\ x^2 = \frac{7}{2} \quad \Rightarrow x = \pm\sqrt{\frac{7}{2}}\]
06

Evaluate Critical Points

Consider the candidates: \(x = 0\), \(x = \sqrt{\frac{7}{2}}\), and \(x = -\sqrt{\frac{7}{2}}\).Calculate \(D^2\) for each:- For \(x = 0\), \(D^2 = 0^4 - 7\cdot0^2 + 16 = 16\)- For \(x = \sqrt{\frac{7}{2}}\): \[D^2 = \left(\frac{7}{2}\right)^2 - 7\left(\frac{7}{2}\right) + 16 = \frac{49}{4} - \frac{49}{2} + 16 \]Simplify \(D^2\): - Combine terms: \(D^2 = \frac{49}{4} - \frac{98}{4} + \frac{64}{4} = \frac{15}{4}\)
07

Conclude the Minimum Distance

The minimum value of \(D^2\) found is \(\frac{15}{4}\) at \(x = \pm \sqrt{\frac{7}{2}}\).Thus, the minimum distance \(D\) is:\[D = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}\]Hence, the minimum distance from the origin to the curve is \(\frac{\sqrt{15}}{2}\).

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Key Concepts

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

Distance Formula
The distance formula is a crucial tool used to determine the distance between two points in a plane. It's often written as:
  • \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For a point on the curve to the origin, this simplifies to:
  • \[ D = \sqrt{x^2 + y^2} \]
If you know the equation of the curve, substitute it into the formula. In this example, the curve is given by \(y = x^2 - 4\). By substituting, this forms a new equation, \(D = \sqrt{x^2 + (x^2 - 4)^2}\), which measures the variable-dependent distance.
Differentiation Techniques
Differentiation is a mathematical process to determine how a function changes as its input changes. In this problem, differentiating helps find where the minimum distance occurs. Specifically, we differentiate the expression for the squared distance \(D^2 = x^4 - 7x^2 + 16\), to avoid the complexity of square roots.

The derivative of \(D^2\) with respect to \(x\) is:
  • \[ \frac{d(D^2)}{dx} = 4x^3 - 14x \]
This process, differentiating a polynomial, relies on the power rule, which involves multiplying the coefficient by the power, then reducing the power by one.
Critical Points
Critical points are where the derivative of a function equals zero or is undefined. Finding these points is essential to determine where functions like distances have minimum, maximum, or saddle points.

To find critical points for \(D^2\), the derivative is set to zero:
  • \[ 4x^3 - 14x = 0 \]
Solving, we factor out common terms making it easier:
  • \[ 2x(2x^2 - 7) = 0 \]
Thus, the solutions \(x = 0\) or \(x = \pm \sqrt{\frac{7}{2}}\) are the critical points where potential minimum and maximum distances occur.
Curve Analysis
Curve analysis involves evaluating the behavior and characteristics of the curve or function. After finding the critical points, evaluate the expression \(D^2\) at these points to identify the minimum distance.

For this problem, at \(x = 0\), we articulate:
  • \(D^2 = 16\)
Evaluating at \(x = \pm \sqrt{\frac{7}{2}}\):
  • Calculate: \(D^2 = \frac{15}{4}\)
By comparing, the smaller \(D^2\) indicates a minimum distance. The minimum value is thus \(\frac{\sqrt{15}}{2}\), determined by retaking the square root of the smallest \(D^2\) value.

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Most popular questions from this chapter

The tangent at the point \((2,-2)\) to the curve, \(x^{2} y^{2}-2 x=4(1-y)\) does not pass through the point : [Online April 8, 2017] (a) \(\left(4, \frac{1}{3}\right)\) (b) \((8,5)\) (c) \((-4,-9)\) (d) \((-2,-7)\)

If \(\beta\) is one of the angles between the normals to the ellipse, \(x^{2}+3 y^{2}=9\) at the points \((3 \cos \theta, \sqrt{3} \sin \theta)\) and \((-3 \sin \theta, \sqrt{3} \cos \theta) ; \in\left(0, \frac{\pi}{2}\right)\); then \(\frac{2 \cot \beta}{\sin 2 \theta}\) is equal to [Online April 15, 2018] (a) \(\sqrt{2}\) (b) \(\frac{2}{\sqrt{3}}\) (c) \(\frac{1}{\sqrt{3}}\) (d) \(\frac{\sqrt{3}}{4}\)

If \(x=-1\) and \(x=2\) are extreme points of \(f(x)=\alpha \log |x|+\beta x^{2}+x\) then (a) \(\alpha=2, \beta=-\frac{1}{2}\) (b) \(\alpha=2, \beta=\frac{1}{2}\) (c) \(\alpha=-6, \beta=\frac{1}{2}\) (d) \(\alpha=-6, \beta=-\frac{1}{2}\)

If the tangent to the curve \(y=\frac{{ }^{2}}{x^{2}-3}, x \in R,(x \neq \pm \sqrt{3})\), at a point \((\alpha, \beta)(0,0)\) on it is parallel to the line \(2 x+6 y-11=0\), then: \(\quad\) [April 10, 2019 (II)] (a) \(|6 \alpha+2 \beta|=19\) (b) \(|6 \alpha+2 \beta|=9\) (c) \(|2 \alpha+6 \beta|=19\) (d) \(|2 \alpha+6 \beta|=11\)

For real \(x\), let \(f(x)=x^{3}+5 x+1\), then \([\mathbf{2 0 0 9}]\) (a) \(f\) is onto \(\mathrm{R}\) but not one-one (b) \(f\) is one-one and onto \(\mathrm{R}\) (c) \(f\) is neither one-one nor onto \(\mathrm{R}\) (d) \(f\) is one-one but not onto \(\mathrm{R}\)
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