can you help me with this calculus problem please? its short but i don't get it?

Question

Find the point on the parapolic arc y=x^2, with domain x E [ 0, 1/ k^1/2], that is closest to the point (0,1). Then, find the point on the arc that is most distant from the point (0,1).

Answer 1

Each point on the parabola is given by an ordered pair:

(x, x²)

So the distance from the point to (0, 1) is given by:

d(x) = √((x - 0)² + (x² - 1)²)

Just to make things easier, we can look at the "square of the distance function" instead, without causing a problem:

D(x) = x² + (x^4 - 2x² + 1) = x^4 - x² + 1

Now, to find the minimum and maximum over the interval [0, 1/√k], differentiate and set equal to zero:

D'(x) = 4x³ - 2x = 0
2x(2x² - 1) = 0
x = 0 or x = 1/√2 = √2/2 (throwing out the negative one, since it's not in the range specified)

So it looks to me like x = √2/2 is the x-value to (0, 1), unless 1/√k < √2/2, in which case 1/√k would be the closest.

Answer 2

y = x^2, domain [0, 1/sqrt(k)]

Recall that the distance formula is equal to

D = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 )

In this case, we want to minimize or maximize the distance between (0,1) and the parabola.

We want to find the distance between (0, 1) and (x, y), so

D = sqrt ( (x - 0)^2 + (y - 1)^2 )
D = sqrt (x^2 + (y - 1)^2)

But y = x^2, so

D = sqrt (x^2 + (x^2 - 1)^2)

Which we'll define to be our function, D(x).

D(x) = sqrt(x^2 + (x^2 - 1)^2)

We know that we will eventually make D'(x) = 0, so let's square both sides.

[D(x)]^2 = x^2 + (x^2 - 1)^2

In order to find the min/max, take the derivative with respect to x and then make it 0.

2[D(x)] [D'(x)] = 2x + 2(x^2 - 1)(2x)

Making D'(x) = 0, we get

0 = 2x + 2(x^2 - 1)(2x)
0 = 2x + 4x(x^2 - 1)
0 = 2x + 4x^3 - 4x
0 = 4x^3 - 2x
0 = 2x^3 - x

We can factor this
0 = x(2x^2 - 1)

Therefore,
x = 0
2x^2 - 1 = 0 { which means 2x^2 = 1, x^2 = 1/2, x = +/- 1/sqrt(2) }

So we have three critical points: {0, 1/sqrt(2), -1/sqrt(2)}

What we want are the absolute min and absolute max. We must test these points and the endpoints of the given interval. That is, we want to test D(0), D(1/sqrt(2)), D(-1/sqrt(2)), D(1/sqrt(k)). However, note that -1/sqrt(2) falls out of the domain, so we exclude that one.

D(x) = sqrt(x^2 + (x^2 - 1)^2)

D(0) = sqrt (0 + (0 - 1)^2) = sqrt (0 + 1) = sqrt(1) = 1

D(1/sqrt(2)) = sqrt (1/2 + (1/2 - 1)^2) = sqrt (1/2 + (-1/2)^2)
= sqrt (1/2 + 1/4) = sqrt (3/4) = sqrt(3)/2

D(1/sqrt(k)) = sqrt (1/k + (1/k - 1)^2) = sqrt (1/k + [1-k]/k)
= sqrt ( {1 + 1 - k}/k ) = sqrt ( {2 - k}/k )

What I don't really get is why the domain is given as defined as k. Normally, you would determine which one is the minimum value out of D(0), D(1/sqrt(2)), and D(1/sqrt(k)) to get the mininum, and the same thing goes to get the maximum. What does k represent?

Answer 3

Eva Ivy Kelsey Kelly Kaitlyn Anna Elise Maria Callie Julia Emily Reagan Sophie Brooke Aimee Sarah Ellie Hannah Camille Allie Jessie Gina Sammy Lila Gracie Kylie Katie Kaylee Helen Kiara Kiana Jamie Ruthie Hailey break of day Donna Serena Blaire Julie

Answer 4

The distance between two points? sqrt ((x2-x1)^2 + (y2-y1)^2), right?
one point is 0,1
the other point is (x, x^2) = (x,y) right? plug & chug
you want to minimize distance - sounds like calculus to me!

The other part - I assume sqrt(k) is by convention positive?
If not then you have any number - inf to + inf
in which case you have k as -> +inf sqrt (k^-1) -> 0 and vice versa. No other constaints on k? so if x E (1/k^1/2) where k->0
figure the limit - sounds like calculus to me!
the limiting value for x
Seems to me that will give you a pretty huge distance from 0,1. But it doesn't seem right, does it? Are you sure you've got that right?

Answer 5

Use Pythagoras to get an expression for the distance between the specified point and an arbitrary point on the parabola. As usual, the minimax is obtained by setting the derivative to zero and solving. This will give you either the minimum or maximum, and the other minimax point will be at the edge of the specified domain.