Prove that (2,0) is the closest point on the curve x^2 + y^2 = 4 on (4,0.
To prove that (2,0) is the closest point on the curve x^2 + y^2 = 4 to the point (4,0), we will use the distance formula. The distance between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distance between (2,0) and (4,0) and compare it with the distance between (2,0) and any other point on the curve x^2 + y^2 = 4.
Distance between (2,0) and (4,0):
d1 = sqrt((4-2)^2 + (0-0)^2)
= sqrt(2^2 + 0^2)
= sqrt(4)
= 2
Now, let's calculate the distance between (2,0) and any other point on the curve x^2 + y^2 = 4, using the equation for the curve.
Let y = f(x) be a point on the curve. Substituting y into the equation, we have:
x^2 + (f(x))^2 = 4
Simplifying, we get:
x^2 + f(x)^2 - 4 = 0
Now, we can use the distance formula to find the distance between (2,0) and (x, f(x)):
d2 = sqrt((x-2)^2 + (f(x) - 0)^2)
= sqrt((x-2)^2 + f(x)^2)
To prove that (2,0) is the closest point on the curve to (4,0), we need to show that d1 < d2 for any other point (x, f(x)) on the curve.
Therefore, we need to show that:
2 < sqrt((x-2)^2 + f(x)^2) for all (x, f(x)) on the curve
Since the equation (x-2)^2 + f(x)^2 represents the square of the distance between (2,0) and any other point on the curve, we want to minimize the expression.
To minimize (x-2)^2 + f(x)^2 subject to the constraint x^2 + f(x)^2 = 4, we can use the method of Lagrange multipliers.
First, set up the Lagrangian L(x, f(x), λ) as follows:
L(x, f(x), λ) = (x-2)^2 + f(x)^2 - λ(x^2 + f(x)^2 - 4)
Now, calculate the partial derivatives of L with respect to x, f(x), and λ, and set them equal to zero:
∂L/∂x = 2(x-2) - 2λx = 0
∂L/∂f(x) = 2f(x) - 2λf(x) = 0
∂L/∂λ = x^2 + f(x)^2 - 4 = 0
From the first equation, we get:
2(x-2) - 2λx = 0
2x - 4 - 2λx = 0
(2-2λ)x = 4
x = 4/(2-2λ)
From the second equation, we get:
2f(x) - 2λf(x) = 0
f(x) - λf(x) = 0
(1-λ)f(x) = 0
f(x) = 0 or λ = 1
If λ = 1, then from the third equation:
x^2 + f(x)^2 - 4 = 0
x^2 = 4
x = ±2
If f(x) = 0, then from the third equation:
x^2 + f(x)^2 - 4 = 0
x^2 = 4
x = ±2
Therefore, the critical points are (x=2, f(2)=0) and (x=-2, f(-2)=0).
Now, substitute these critical points into the expression for d2:
d2 = sqrt((x-2)^2 + f(x)^2)
= sqrt((2-2)^2 + 0^2)
= sqrt(0 + 0)
= 0
Since d2 = 0 at these critical points, and d1 (distance between (2,0) and (4,0)) is positive, we can conclude that (2,0) is the closest point on the curve x^2 + y^2 = 4 to the point (4,0).