Calculus Question?

Question

find the point on the parabolic arc y=x^2, with the domain xE [0, 1/(k^1/2), that is clsoest to the point (0,1). Then, find the point that is most distant from th epoint (0, 1)

Answer 1

The question of the closest and most distant points (jointly considered) depends on whether k < 1, or 1 < k < 2, or k > 2.

In general, the distance D between the points (0, 1) and (x,y), where y = x^2 satisfies:

D^2 = x^2 + (y - 1)^2 = y + (y - 1)^2 = y^2 - y + 1.*** ........(1)

[*** NOTE : We can often make life easier for ourselves by choosing the right independent variable. In this case, 'y' is preferable to 'x'. Don't fall into the trap of thinking that "x is always the independent variable" ! ]

d (D^2) / dy = 2 y - 1; so y = 1/2 gives a stationary value for D. That D^2 = 1/4 - 1/2 + 1 = 3/4. Also, d^2 (D^2) / dy^2 is positive. So:

D = 3^(1/2) / 4 is a local minimum.

This is less than 1 (the distance from point (0,1) to the origin, the start of the range for x).

We also need to know the distance at the UPPER END of the range for x. Let that end point be at (x. y) = (1/[k^(1/2)], 1/k), and the distance to it from (0, 1) be E. Let's just use Y for 1/k.

Then by equation (1), E^2 = Y^2 - Y +1. ........(2)

When does E = 1? Only when Y = 0 (our original origin case, in fact) and when Y = 1. So, if Y > 1 (i.e. k < 1), the LARGEST value for the distance is at the upper end of the range. So, to separate out these JOINT cases:

(i) For k < 1, the closest point to (0, 1) is at (1/[2^(1/2)], 1/2) and the furthest point is at (1/[k^1/2], 1/k).

(ii) For 2 > k > 1, the closest point is at (1/[2^1/2], 1/2), and the most distant is at the origin.

(iii) For k > 2, the closest point is at the UPPER-END of the range
(1/[k^1/2], 1/k) and the most distant is at the origin.

(I found that consideration of a sketch of D^2 versus Y was extremely helpful in clarifying the ranges for these joint cases. It cleared up some early confusion that I'd had in summarising my results in words.)

Thank you for an interesting question to consider.

Live long and prosper.

P.S. I thank sahsjing, whose talents I appreciate, for his comment on my original, incomplete, summary claim. On reconsideration, I believe that he is not quite correct : the JOINT question of (closest, least) distances settles down into the THREE cases I've now separated out, which can be viewed as a cooperative blend of both of our earlier conclusions. I find this division into three ranges from the start the most satisfactory way of describing the influence of the extent of the range on the results. Otherwise one has to break the summary conclusions into two ranges for the closest point, one of which has then to be further subdivided to separate out the subsequent, alternative possibilities for the most distant point. Simply dividing the original range for 'k' into the three separate sub-ranges for JOINT consideration seems a bit more aesthetic, to me. But many thanks, anyway, for that comment. It certainly sent me back to be more self-critical. My salutations !

P^2. S. I now see that sahsjing, in his later, more detailed analysis of the problem, effectively breaks his range for k < 2 into two sub ranges ("Therefore the point on the curve [for greatest distance --- ed.] is either (0, 0) or (1/√k, 1/k)"), without explicitly pointing out that this means identifying k = 1 as another value for k where the JOINT behaviour changes. This isn't to my taste --- I prefer to identify both k = 1 and k = 2 as places where the JOINT assignments of the (closest, furthest) pair of points alters. But as I wrote, this summary of the conclusions is perhaps merely a matter of taste, although I think that my way brings out the issues most clearly and explicitly. We're all entitled to our preferences. De gustibus nil disputandum.

Answer 2

Let d be the distance from a point (x, y) on the curve to (0,1).
d^2 = x^2 + (x^2-1)^2......(1) , because y = x^2

Differentiate (1) with respect to x,
2d d' = 2x +2(x^2-1)(2x) ......(2)

Solve d' = 0 for x,
x = 0 or x = 1/√2

If k < 2, then 1/√k > 1/√2.
At x = 0, d = 1,
maximum d = max[1, √[1/k + (1/k - 1)^2]].
Therefore the point on the curve is either (0, 0) or (1/√k, 1/k).

At x = 1/√2, d = √3 / 2,
minimum d = √3 / 2.
Therefore the point on the curve is (1/√2, 1/2)


If k > 2, then 1/√k < 1/√2.
At x = 0, d = 1, maximum d
Therefore the point on the curve is (0, 0)

At x = 1/√k, d = √[1/k + (1/k - 1)^2], minimum d
Therefore the point on the curve is (1/√k, 1/k).
----------------
Dr Spock,
"Depends on whether k is > or < 1" is not right. It should be "Depends on whether k is > 2 or < 2."

P.S. Initially, Dr. Spock just posted this one sentence "Depends on whether k is > or < 1". After I finished my work on a scratch paper, I found that actually the "critical point" should be 2 instead of 1. That's why I made above comment.
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Dr Spock,
The following seems to be the only difference in conclusions made from your approach and my approach:

From your approach:
"(i) For k < 1, ..., the furthest point is at (1/[k^1/2], 1/k).

(ii) For 2 > k > 1, ..., the most distant is at the origin. "

From my approach:
"If k < 2, then 1/√k > 1/√2.
maximum d = max[1, √[1/k + (1/k - 1)^2]]."

Actually, if you read my conclusion carefully, you'll find that the two conclusions are the same.

I don't think using y is better than using x because the domain is related to k directly.

Answer 3

I'm not sure how the right endpoint of your domain will come into play, but you can use the distance formula and some calculus:

The distance formula is derived from the Pythagorean Theorem,
a^2 + b^2 = c^2 (we'll be using the form c = sqrt(a^ + b^2)), where 'a' is the horizontal component and 'b' is the vertical component (or vice versa). Our horizontal component is easy enough: 'x'. The vertical component is the difference of your y-value (which is really x^2) and 1 (from your point (0,1)). So the distance between (0,1) and any point on y=x^2 is given by:
sqrt( (x) + (1 - x^2) ).
You can differentiate this with respect to 'x' to determine the minimum value:
dL/dx = ( 1/2 )( 2x^3 - 3x^2 - x + a )

set dL/dx to zero and solve for 'x'. You'll of course need to test the points to find the minimum.

Feel free to email me for any clarifications