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# Graphing

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What is the smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 8).

Feb 20, 2023

#1
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I want to find the smallest distance between  $$(x,\frac{x^2-8}{2}) \qquad and \qquad (0,0)$$

The distance between these is

$$d=\sqrt{(x-0)^2+(\frac{x^2-8}{2}-0)^2} \\d= \sqrt{x^2+(\frac{x^2-8}{2})^2}$$

We need to minimize that expression and solve for x

I will get the same result if I minimize

$$x^2+(\frac{x^2-8}{2})^2 \qquad or \qquad 4x^2+(x^2-8)^2$$

Let

$$T=4x^2+(x^2-8)^2\\ \frac{dT}{dx}=8x+2(x^2-8)^1*2x\\ \frac{dT}{dx}=8x+4x(x^2-8)\\ \frac{dT}{dx}=8x+4x^3-32x\\ \frac{dT}{dx}=4x^3-24x\\ \frac{dT}{dx}=4x(x^2-6)\\ \frac{dT}{dx}=4x(x-\sqrt 6)(x+\sqrt 6)\\~\\ \frac{dT}{dx}=0 \;\;\;\;when\;\;\;\;\ x=0,\;\;x=\pm\sqrt6$$

When x=0

$$d= \sqrt{x^2+(\frac{x^2-8}{2})^2}\\ d= \sqrt{16}\\ d=4$$

When   $$x=\pm\sqrt6$$

$$d= \sqrt{x^2+(\frac{x^2-8}{2})^2}\\ d= \sqrt{6+(\frac{6-8}{2})^2}\\ d= \sqrt{6+1}\\ d=\sqrt7\\ d\approx 2.64$$

So it looks to me like the min distance is  when x= $$\sqrt7 \quad \text{and it occurs when }\quad x=\pm\sqrt6$$

I'll look at the second derivative for verification.

$$\frac{dT}{dx}=4x^3-24x\\ \frac{d^2T}{dx^2}=12x^2-24\\ \frac{d^2T}{dx^2}=12(x^2-2)\\$$

When    $$x=\pm \sqrt 6\\ \frac{d^xT}{dx^2}= 48>0 \quad \text{So minimum}$$

When    $$x= 0\\ \frac{d^xT}{dx^2}= -24<0 \quad \text{So maximum}$$

So I get the minimum distance to be    sqrt7

And here is the pic

LaTex:

T=4x^2+(x^2-8)^2\\
\frac{dT}{dx}=8x+2(x^2-8)^1*2x\\
\frac{dT}{dx}=8x+4x(x^2-8)\\
\frac{dT}{dx}=8x+4x^3-32x\\
\frac{dT}{dx}=4x^3-24x\\
\frac{dT}{dx}=4x(x^2-6)\\
\frac{dT}{dx}=4x(x-\sqrt 6)(x+\sqrt 6)\\~\\
\frac{dT}{dx}=0 \;\;\;\;when\;\;\;\;\ x=0,\;\;x=\pm\sqrt6

Feb 21, 2023