Find the point P on the parabola y=x² closest to the point (3,0).?

Question

Calculus I - Using Applied Optimization
- P is approximately (0.59, 0.35)
BUT, I need the systematic pathway to go the answer.

Answer 1

use the distance formula to determine the distance between (3,0) and each other point on the function

d = √[(x - 3)² + (y - 0)²]
d = √[(x - 3)² + (x² - 0)²]
d = √[(x - 3)² + (x²)²]
d = √[x² - 6x +9 + x^4]

Now, find the minimum value 'd'

This can be done by finding the local point that has a tangent with a slope of zero... or, in other words, the value where the first derivative is equal to zero

Find the derivative of d over x... ∂d/∂x... and set it equal to 0

d = √[x² - 6x +9 + x^4]
∂d/∂x = ∂/∂x √[x² - 6x +9 + x^4]
∂d/∂x = ∂/∂x [x² - 6x +9 + x^4]^½
∂d/∂x = ½[x² - 6x +9 + x^4]^-½ ⋅ ∂/∂x [x² - 6x +9 + x^4]
∂d/∂x = 1/√[x² - 6x +9 + x^4] ⋅ [2x - 6 + 4x³]
∂d/∂x = [2x - 6 + 4x³] / √[x² - 6x +9 + x^4]

set ∂d/∂x = 0

[2x - 6 + 4x³] / √[x² - 6x +9 + x^4] = 0

Solve for x

When the numerator is zero, the expression is zero. The numerator is zero when x = 1


===

An alternative route you could take would be to find the slope at a point on the function, a tangent, whose perpendicular passes through (3,0)

Answer 2

Let P(x,y) be the point on the parabola. The distance, d, satisfies
d² = (x-3)² + (y-0)²
d² = x² - 6x + 9 + y²
d² = x² - 6x + 9 + (x²)²
d² = x^4 + x² - 6x + 9
d is minimum when d² is minimum, so find derivative of x^4 + x² - 6x + 9
and set it to 0:
4x³ + 2x - 6 = 0
2x³ + x - 3 = 0
x = 1 is clearly a solution, so (1,1) is the point.

probably less work to use y = 2x as tangent to parabola, and y = (-1/2)(x-3) as perpendicular to tangent through (3,0) and (x,y), and solve that way.

Answer 3

actually the min distance is for the point ( 1, 1)

d = sqrt ( ( x - 3 )^2 + y^2 ), where y = x^2

so d = sqrt ( ( x - 3)^2 + x^4 )

take d ', to get [ 2 ( x - 3 ) + 4 x^3 ] / { 2 ( sqrt ( ( x - 3)^2 + x^4 ) ) } = 0,
x = 1 is the critical # .. { cn }

thus ( 1, 1 ) is the point on the parabola to give the min distance to ( 3, 0 ) ......

the first deriv test... x < 1, d ' < 0, and x > 1, then d ' > 0, so x = 1 is cn

Answer 4

given y = x^2
get the slope of this curve,
slope = dy/dx = 2x
now, there is a line (say AB) joining (3,0) with the curve y=x^2,
this line is the mirror image of the slope (dy/dx=2x),
the slope of AB = -(2^-1)x = -0.5x
need to find the eqn for line AB,
point (3,0) falls on line AB,
y = mx + c
=> 0 = -0.5(3) + c => 1.5
thus, AB is given by y = -0.5x + 1.5

now, AB intersects with y=x^2 to give shortest length,
so, let
y = x^2 --- (1)
y = -0.5x + 1.5 --- (2)

subtract (2) from (1),
x^2 + 0.5x - 1.5 = 0
=> 2x^2 + x - 3 = 0
=> (2x +3) (x - 1) = 0
solving gives x = 1 (ignore x = -3/2 since y = 0 when x = 0)
subs. into (1) to get y,
y = (1)^2 = 1

so the point P is (1,1)