find the point on the unit circle that is closest to (1,3)
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Note that the unit circle has equation:
x^2 + y^2 = 1.
Suppose that (x, y) lies on the circle. The distance between (x, y) and (1, 3) is:
d = √[(x - 1)^2 + (y - 3)^2].
We want to minimize d with respect to x^2 + y^2 = 1. Since minimizing d and d^2 are the same, we can just minimize:
d^2 = (x - 1)^2 + (y - 3)^2.
Now, we can either get this in terms of one variable and use derivatives to minimize d^2 (and hence minimize d) or use Lagrange multipliers. The first method seems easier.
By solving x^2 + y^2 = 1 for y, we get:
y = ±√(1 - x^2).
Now, you'd expect that the point closest to (1, 3) would probably be in Quadrant I where y > 0, so pick the positive sign.
y = √(1 - x^2).
Substituting this into the expression for d^2 yields:
d^2 = (x - 1)^2 + [√(1 - x^2) - 3]^2
= (x^2 - 2x + 1) + [(1 - x^2) - 6√(1 - x^2) + 9], by expanding
= 11 - 2x - 6√(1 - x^2).
Differentiating yields:
d(d^2)/dx = -2 + 6x/√(1 - x^2).
Setting this equal to zero:
-2 + 6x/√(1 - x^2) = 0
==> 6x - 2√(1 - x^2) = 0, by multiplying both sides by √(1 - x^2)
==> 2√(1 - x^2) = 6x, by isolating the radical
==> 4(1 - x^2) = 36x^2, by squaring both sides
==> 4 - 4x^2 = 36x^2
==> 40x^2 = 4
==> x = ±√10/10.
We know that the required point should be in Quadrant I, so x = √10/10.
(You can use the second derivative to show that this actually produces the minimum.)
Since y = √(1 - x^2), y = √[1 - (√10/10)^2] = √(1 - 1/10) = 3√10/10.
Therefore, the required point is (√10/10, 3√10/10).
I hope this helps!
x^2 + y^2 = 1.
Suppose that (x, y) lies on the circle. The distance between (x, y) and (1, 3) is:
d = √[(x - 1)^2 + (y - 3)^2].
We want to minimize d with respect to x^2 + y^2 = 1. Since minimizing d and d^2 are the same, we can just minimize:
d^2 = (x - 1)^2 + (y - 3)^2.
Now, we can either get this in terms of one variable and use derivatives to minimize d^2 (and hence minimize d) or use Lagrange multipliers. The first method seems easier.
By solving x^2 + y^2 = 1 for y, we get:
y = ±√(1 - x^2).
Now, you'd expect that the point closest to (1, 3) would probably be in Quadrant I where y > 0, so pick the positive sign.
y = √(1 - x^2).
Substituting this into the expression for d^2 yields:
d^2 = (x - 1)^2 + [√(1 - x^2) - 3]^2
= (x^2 - 2x + 1) + [(1 - x^2) - 6√(1 - x^2) + 9], by expanding
= 11 - 2x - 6√(1 - x^2).
Differentiating yields:
d(d^2)/dx = -2 + 6x/√(1 - x^2).
Setting this equal to zero:
-2 + 6x/√(1 - x^2) = 0
==> 6x - 2√(1 - x^2) = 0, by multiplying both sides by √(1 - x^2)
==> 2√(1 - x^2) = 6x, by isolating the radical
==> 4(1 - x^2) = 36x^2, by squaring both sides
==> 4 - 4x^2 = 36x^2
==> 40x^2 = 4
==> x = ±√10/10.
We know that the required point should be in Quadrant I, so x = √10/10.
(You can use the second derivative to show that this actually produces the minimum.)
Since y = √(1 - x^2), y = √[1 - (√10/10)^2] = √(1 - 1/10) = 3√10/10.
Therefore, the required point is (√10/10, 3√10/10).
I hope this helps!