Question

Find the point on the curve y=cosx closest to the point (0,0)?

Answer 1

The point is `(0,1)`

Explanation:

By the distance formula, the distance between `(0, 0)` and the graph of `y = cosx` is

`d= sqrt((x -0)^2 + (cosx - 0)^2)`

`d = sqrt(x^2 + cos^2x)`

To find the minimum distance, we need to differentiate.

`d' = (2x - 2cosxsinx)/(2sqrt(x^2 + cos^2x))`

`d' = (x - cosxsinx)/sqrt(x^2 + cos^2x)`

We wish for this to be the smallest possible, thus we need `d' = 0`.

`0 = (x - cosxsinx)/sqrt(x^2 + cos^2x)`

`0 = x - cosxsinx`

Use a graphing application to solve and find that `x = 0` is the only solution.

There derivative is negative when `x < 0`, and positive when `x > 0`, therefore, the minimum distance will occur when `x = 0`. The value of `cosx` at `x = 0` is `1`, therefore the closest point is `(0, 1)`. Let's examine graphically:

Hopefully this helps!