Question

What point on the graph of y = sin x is the closest to (0,1)?

Our topics include derivatives and optimization

Answer 1

`x approx 0.4787`

Explanation:

We can find a distance function to the point via Pythagorean theorem:
`d(x)^2 = Delta x^2 + Delta y^2 = x^2 + (1 - sin(x))^2`

We want to minimize `d(x)`. Now, we could take a squareroot and then find the min, but minimizing a number is the same as minimizing its squareroot, so we can just minimize `d(x)^2`:

`d/dx(d(x)^2) = 2x + 2(1-sin(x))(-cos(x)) = 0`
`0 = x -cos(x)(1-sin(x))`

This equation is transcendental, i.e. we cannot actually solve it analytically. We have to solve numerically.

The plot of the above clearly only has one zero:

graph{x -cos(x)(1-sin(x)) [-3, 3, -3, 3]}

This is at `x approx 0.4787`, where the distance is minimized.

Answer 2

`x_0 approx 0.4787`

Explanation:

There's a nifty way to do this with any curve with a geometric property.

The line connecting (0,1) to (x, sin(x)) will have distance which peaks (is minimized) at a number `x_0` which we want to find. We know that this line will be perpendicular to the curve at that point.

`m_{text(line)} = (Delta y)/(Delta x) = (1-sin(x_0))/(0 - x_0) = (sin(x_0) - 1)/x_0`
`m_{text(curve)} = d/dx(sin(x)) |_(x = x_0) = cos(x_0)`

Perpendicular lines are negative inverses, i.e.
`m_(text(curve)) = - 1/(m_(text(line))`
`cos(x_0) = x_0 / (1-sin(x_0)) implies x_0 = cos(x_0)(1-sin(x_0))`

This must be solved numerically, not analytically. We find that `x_0 approx 0.4787`