perplexus.info :: Calculus : I can see the origin from here.

I can see the origin from here. (Posted on 2012-03-11)

Consider the function f(x)=x²+c
Let P=(x,y) be a point on the graph of y=x²+c
Let d(x) = the distance from P to the origin.

[1] Find the value(s) of x, in terms of c, that minimizes d(x).

[2] Prove d(x) ≥ f(x) for all x and c.

[3] Find, in terms of c, the limit as x goes to infinity of d(x)-f(x)

[4] Find, in terms of c, and if it exists, the integral from negative infinity to positive infinity of (d(x)-f(x)-the above limit) dx

Submitted by Jer

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Solution: (Hide)

[1]
If c ≥ -1/4 then x=0
If c < -1/4 then x = ą√((2c+1)/-2)

[2]
Connect P to O. The hypotenuse is always the longest.
[3]
The answer is 1/2 for any c.
We want √(x²+(x²+c)²) - (x²+c)
Multiply by the conjugate and its reciprocal: (x²+(x²+c)²-(x²+c)²) / √(x²+(x²+c)²) + (x²+c)
x² / [x²((1/x²)√(x²+(x²+c)²) + (1/x²)(x²+c))]
1 / [√(1/x² + 1 + 2c/x² + c²/x²) + 1 + c/x²]
The limit as x goes to infinity is 1 / [√(0+1+0+0) + 1+0] = 1/2
[4]
Don't know yet, but if c = -4 the answer seems to be about 23.75 I dare you to put sqrt(x^2+(x^2+c)^2) - (x+c)^2 - 1/2 into the wolfram integrator. When c=0 it's not hard to show the answer is -2/3.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Part 4 still unsolved	Jer	2012-10-05 10:57:06
Parts 1-3	Brian Smith	2012-03-13 20:18:30

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