Question

Find the point on the curve y = x^2 closest to the point (0, 1).

Answer 1

Calculus?

Answer 2

yes

Answer 3

There are two answers, due to the symmetry of the function. For now, we can assume x positive.

Answer 4

okay

Answer 5

Find the distance from (0,1) to the curve, and minimize that function.

Answer 6

ok

Answer 7

We need to set the distance up as a function of x. This is a job for the Pythagorean theorem.\[(d(x))^2=(x-x_1)^2+(y-y_1)^2=x^2+(x^2-1)^2\]

Answer 8

ok

Answer 9

When you multiply the second term, you will get an equation that is quadratic in form (fourth degree polynomial). Take the derivative, set it to zero, and solve for x. Once you have x, you can get y.

Answer 10

okay

Answer 11

Note that we worked with the square of the distance. If we minimize it, the distance will also be minimized.

Answer 12

ok got you

Answer 13

\[Let f(x)=(d(x))^2 \implies f(x)=x^4-x^2+1 \implies f'(x)=4x^3-2x=0\]\[\implies 2x(2x^2-1)=2x(\sqrt2 x+1)(\sqrt2 x-1)=0 \implies critical~values~x=0,\pm \frac{\sqrt2}{2}\]

Answer 14

wow okay

Answer 15

The minimizing critical values are the two nonzero ones.

Answer 16

ok

Answer 17

Do you understand how this worked out?

Answer 18

I'm a little bit confused

Answer 19

About what?

Answer 20

what you wrote the i see that some numbers didn't show

Answer 21

Which line?

Answer 22

the last one where you wrote let f(x) = d(x) and so forth

Answer 23

All I did was define the function, take the derivative, and set it to zero. On the next line, I factored it to get the roots, which are x=0 and x=plus/minus sqrt2/2.

Answer 24

ok

Answer 25

Keep working hard. Do math every day.

Answer 26

ok thanks

Answer 27

The final answers will be ordered pairs (-sqrt2/2,1/2) and (sqrt2/2,1/2).