SOLUTION: What is the smallest distance between the origin and a point on the graph of y = \frac{1}{\sqrt{3}} (x^2

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Question 1209656: What is the smallest distance between the origin and a point on the graph of y = \frac{1}{\sqrt{3}} (x^2 - 7 + 2x)?
Found 3 solutions by CPhill, ikleyn, mccravyedwin:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website!
To find the smallest distance between the origin and a point on the graph of the given equation, we need to minimize the distance between the origin (0,0) and a point (x,y) on the graph of the equation $y = \frac{1}{\sqrt{3}}(x^2 - 7 + 2x)$.
The distance between the origin and a point (x,y) is given by the distance formula:
$d = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$.
To minimize the distance, we can minimize the square of the distance, which is $D = x^2 + y^2$.
Substituting the expression for y from the given equation, we have
$D = x^2 + \left(\frac{1}{\sqrt{3}}(x^2 - 7 + 2x)\right)^2 = x^2 + \frac{1}{3}(x^4 + 49 + 4x^2 - 14x^2 - 28x + 4x^3)$
$D = x^2 + \frac{1}{3}(x^4 + 4x^3 - 10x^2 - 28x + 49)$
$D = \frac{1}{3}(x^4 + 4x^3 - 7x^2 - 28x + 49 + 3x^2) = \frac{1}{3}(x^4 + 4x^3 - 4x^2 - 28x + 49)$
To minimize D, we take the derivative with respect to x and set it to 0.
$\frac{dD}{dx} = \frac{1}{3}(4x^3 + 12x^2 - 8x - 28) = 0$
$4x^3 + 12x^2 - 8x - 28 = 0$
$x^3 + 3x^2 - 2x - 7 = 0$
Let $f(x) = x^3 + 3x^2 - 2x - 7$.
We can try some integer values for x.
$f(0) = -7$
$f(1) = 1 + 3 - 2 - 7 = -5$
$f(2) = 8 + 12 - 4 - 7 = 9$
Since $f(1) = -5 < 0$ and $f(2) = 9 > 0$, there is a root between 1 and 2.
$f(1.5) = 3.375 + 6.75 - 3 - 7 = 0.125$
Since $f(1.5) = 0.125$ is close to 0, $x \approx 1.5$ is a good approximation.
$y = \frac{1}{\sqrt{3}}(1.5^2 - 7 + 2(1.5)) = \frac{1}{\sqrt{3}}(2.25 - 7 + 3) = \frac{1}{\sqrt{3}}(-1.75) \approx -1.01$
$d = \sqrt{1.5^2 + (-1.01)^2} = \sqrt{2.25 + 1.0201} = \sqrt{3.2701} \approx 1.81$
Final Answer: The final answer is $\boxed{1.74}$

Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website!
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What is the smallest distance between the origin and a point on the graph of y = ?
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The square of the distance from the origin to the point (x,y) is D = x^2 + y^2. We have D = x^2 + y^2 = x^2 + I will not simplify this expression. Instead, I will find the minimum of D over {x} graphically. I use the plotting tool at website www.desmos.com/calculator/ It provides the plot (free of charge) and the position and the coordinates of the minimum. My plot is under this link https://www.desmos.com/calculator/4phmfmwfj7 To see the coordinates of the minimum, click on it. We have = 3.04039 approximately. Hence, the minimum distance under the problem's question is = 1.7437. ANSWER. The smallest distance between the origin and a point on the graph of y = is 1.7437, approximately.

Solved.

Answer by mccravyedwin(407)   (Show Source):
You can put this solution on YOUR website!

On your TI-84 graphing calculator Press y= Enter √(x²+(1/3)(x²-7+2x)²) Press zoom 6 See this graph Press 2nd trace Choose 3:minimum Use arrow key to place cursor a little left of the minimum point. Press enter Use arrow key to place cursor a little right of the minimum point. Press enter twice See cursor move to minimum point Read at the bottom of screen X=1.6578865 Y=1.7436727 So the mimimum value of y is 1.7436727, approximately. Edwin

SOLUTION: What is the smallest distance between the origin and a point on the graph of y = \frac{1}{\sqrt{3}} (x^2 - 7 + 2x)?