Find the points on the curve y=x2+2 closest to the point (0,3). Enter the coordinate with the smallest x-value first and round to the nearest 4 decimal places.
To find the points on the curve y=x^2+2 closest to the point (0,3), we can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is given by:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's plug in the coordinates for the given point (0,3):
distance = sqrt((x - 0)^2 + (y - 3)^2)
Substituting y = x^2 + 2:
distance = sqrt((x - 0)^2 + ((x^2 + 2) - 3)^2)
Simplifying:
distance = sqrt(x^2 + (x^2 - 1)^2)
To find the points with the smallest distance, we need to minimize this distance function. Let's find the derivative of the distance function with respect to x:
distance' = (1/2)(2x)(x^2 - 1)^2 + 2(x^2 - 1)(2x)
Simplifying:
distance' = 2x(x^2 - 1)(x^2 - 1 + 2)
Setting distance' equal to zero and solving for x:
2x(x^2 - 1)(x^2 + 1) = 0
This equation factors into:
2x(x - 1)(x + 1)(x^2 + 1) = 0
Since x^2 + 1 is always positive, the only critical points are x = -1, x = 0, and x = 1.
Evaluating the y-coordinates for these x-values:
- For x = -1: y = (-1)^2 + 2 = 3
- For x = 0: y = (0)^2 + 2 = 2
- For x = 1: y = (1)^2 + 2 = 3
So, the two points on the curve y=x^2+2 closest to the point (0,3) are (-1,3) and (1,3).
To find the points on the curve y = x^2 + 2 closest to the point (0,3), we need to minimize the distance between the two points.
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the point (0,3) is the point (x1, y1), and the points on the curve y = x^2 + 2 are represented by (x, x^2 + 2).
So, we need to minimize the distance between (0,3) and (x, x^2 + 2) using the distance formula.
Let's substitute the values into the distance formula:
d = √((x - 0)^2 + ((x^2 + 2) - 3)^2)
Simplifying the expression inside the square root:
d = √(x^2 + (x^2 - 1)^2)
Now, we need to find the value of x that minimizes this expression.
To minimize the distance, we need to minimize the square of the distance, as square root is a monotonically increasing function.
Let's square the expression inside the square root:
d^2 = x^2 + (x^2 - 1)^2
Expanding the expression:
d^2 = x^2 + (x^4 - 2x^2 + 1)
Combining like terms:
d^2 = x^4 - x^2 + 1
To minimize the distance, we need to find the minimum value of d^2. So, we differentiate d^2 with respect to x and set it equal to zero to find the critical points.
Differentiating the expression:
(d^2)' = (x^4 - x^2 + 1)' = 4x^3 - 2x
Setting the derivative equal to zero:
4x^3 - 2x = 0
Factoring out the common term:
2x(2x^2 - 1) = 0
Setting each factor equal to zero:
2x = 0 or 2x^2 - 1 = 0
Solving the first equation:
2x = 0
x = 0
Solving the second equation:
2x^2 - 1 = 0
2x^2 = 1
x^2 = 1/2
x = ±√(1/2)
x = ±0.7071
Now let's find the corresponding y-values for each x-value we obtained.
For x = 0, y = x^2 + 2 = 0^2 + 2 = 2. So, one point on the curve is (0, 2).
For x = 0.7071, y = x^2 + 2 = 0.7071^2 + 2 ≈ 2.5. So, the second point on the curve is (0.7071, 2.5).
For x = -0.7071, y = x^2 + 2 = (-0.7071)^2 + 2 ≈ 2.5. So, the third point on the curve is (-0.7071, 2.5).
In summary, the two points on the curve y = x^2 + 2 closest to the point (0, 3) are (0, 2) and (-0.7071, 2.5).
To find the points on the curve y = x^2 + 2 closest to the point (0,3), we need to minimize the distance between the given point and the points on the curve.
First, let's write the equation for the distance between two points (x1, y1) and (x2, y2):
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, we have the point (0,3) and we want to find the points on the curve that minimize the distance. Let's call the x-coordinate of the points on the curve as x.
The coordinates of the points on the curve are (x, y) where y = x^2 + 2.
So, the coordinates become (x, x^2 + 2).
Substituting these values into the distance equation, we get:
Distance = sqrt((x - 0)^2 + ((x^2 + 2) - 3)^2)
= sqrt(x^2 + (x^2 - 1)^2)
To find the values of x that minimize the distance, we need to find the derivative of the distance equation with respect to x, set it equal to zero, and solve for x.
Differentiating the distance equation using the chain rule, we have:
d/dx (distance) = d/dx(sqrt(x^2 + (x^2 - 1)^2))
= (1/2) * (x^2 + (x^2 - 1)^2)^(-1/2) * (2x + 2(x^2 - 1)(2x))
= (x + (x^2 - 1)(2x)) / sqrt(x^2 + (x^2 - 1)^2)
Setting this equal to zero and solving for x, we have:
x + (x^2 - 1)(2x) = 0
x + 2x^3 - 2x = 0
2x^3 - 2x + x = 0
2x^3 - x = 0
x(2x^2 - 1) = 0
From this equation, we can see that x = 0 or x^2 = 1/2.
Solving for x^2 = 1/2, we get two possible values for x: x = sqrt(1/2) and x = -sqrt(1/2).
Rounding these values to the nearest 4 decimal places, we get x = 0.7071 and x = -0.7071.
Now, we can substitute these values back into the curve equation y = x^2 + 2 to find the corresponding y-coordinates:
For x = 0.7071: y = (0.7071)^2 + 2 = 2.5 (rounded to 4 decimal places)
For x = -0.7071: y = (-0.7071)^2 + 2 = 2.5 (rounded to 4 decimal places)
Therefore, the points on the curve y = x^2 + 2 closest to the point (0,3) are (0.7071, 2.5) and (-0.7071, 2.5).
y = x^2+2
y' = 2x
So, at (h,k) y'(h) = 2h
So, you want a line with slope -1/(2h) passing through (0,3)
y-3 = -1/(2h) x
h^2+2 - 3 = -1/2
h^2 = 1/2
h = 1/√2
so, (1/√2,5/2) is closest to (0,3)
1/2 +
Or, consider the distance formula. We have
z^2 = (x-0)^2 + (y-3)^2
= x^2 + (x^2+2-3)^2
= x^2 + x^4-2x^2+1
= x^4 - x^2 + 1
2z z' = 4x^3 - 2x
z' = x(2x^2-1)/z
z'=0 at x=0, x = ±1/√2