Find the the point(s) on the curve y = x^2 that is closest to the point (0,7), what is the minimum distance?
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Use the distance formula:
d = sqrt[(x - 0)^2 + (y - 7)^2]
d = sqrt[(x^2 + (x^2 - 7)^2], (since y = x^2)
d = sqrt(x^2 + x^4 - 14x^2 + 49)
Let D = d^2 = x^4 - 13x^2 + 49
D ' = 2d = 4x^3 - 26x
D ' = d = 2x^3 - 13x.
Let's set our derivative equal to 0 and solve for x.
2x^3 - 13x = 0
x(2x^2 - 13) = 0
x = 0 or,
2x^2 = 13
x^2 = 13/2
x = +/- sqrt(13/2).
Since the point (0,7) lies on the y-axis and x^2 is an even function symmetric about the y-axis, both of the points sqrt(13/2) and -sqrt(13/2) are the same distance away from (0,7). To get the y-coordinate, let's plug one of these points into our original functinon
y = (sqrt(13/2))^2 = 13/2.
Therefore, the points are:
(-sqrt(13/2), 13/2) and (sqrt(13/2), 13/2).
Hope this helped.
d = sqrt[(x - 0)^2 + (y - 7)^2]
d = sqrt[(x^2 + (x^2 - 7)^2], (since y = x^2)
d = sqrt(x^2 + x^4 - 14x^2 + 49)
Let D = d^2 = x^4 - 13x^2 + 49
D ' = 2d = 4x^3 - 26x
D ' = d = 2x^3 - 13x.
Let's set our derivative equal to 0 and solve for x.
2x^3 - 13x = 0
x(2x^2 - 13) = 0
x = 0 or,
2x^2 = 13
x^2 = 13/2
x = +/- sqrt(13/2).
Since the point (0,7) lies on the y-axis and x^2 is an even function symmetric about the y-axis, both of the points sqrt(13/2) and -sqrt(13/2) are the same distance away from (0,7). To get the y-coordinate, let's plug one of these points into our original functinon
y = (sqrt(13/2))^2 = 13/2.
Therefore, the points are:
(-sqrt(13/2), 13/2) and (sqrt(13/2), 13/2).
Hope this helped.
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We want the normal line that contains (0,7)
So y=x^2, y'=2x, and normal slope is -1/(2x)
Since (x,x^2) is on the graph and point (0,7) is on normal line, we get slope is also (7-x^2)/(0-x)
Now solve for x by setting both slopes equal: -1/2x = (x^2-7)/x
Multiply both sides by 2x: -1=14-2x^2 and 2x^2=15 and x=+-sqrt(15/2) = +-sqrt(30)/2
So since x=(+-)sqrt(30)/2 we get y=15/2
So we need the distance between (0,7) and (sqrt30/2,15/2) since if x=-sqrt30/2 the distance is greater
So by distance formula we get D=sqrt[(7-15/2)^2 + (sqrt30/2)^2] = sqrt(31/4) = (sqrt31)/2 = 2.784
So y=x^2, y'=2x, and normal slope is -1/(2x)
Since (x,x^2) is on the graph and point (0,7) is on normal line, we get slope is also (7-x^2)/(0-x)
Now solve for x by setting both slopes equal: -1/2x = (x^2-7)/x
Multiply both sides by 2x: -1=14-2x^2 and 2x^2=15 and x=+-sqrt(15/2) = +-sqrt(30)/2
So since x=(+-)sqrt(30)/2 we get y=15/2
So we need the distance between (0,7) and (sqrt30/2,15/2) since if x=-sqrt30/2 the distance is greater
So by distance formula we get D=sqrt[(7-15/2)^2 + (sqrt30/2)^2] = sqrt(31/4) = (sqrt31)/2 = 2.784
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Any point on the curve y = x^2 can be written as (x, x^2)
We want to minimize the following equation regarding the distance between (x, x^2) and (0, 7) but without the square root.
d(x) = (x)^2 + (x^2 - 7)^2
d '(x) = 2x + 4x(x^2 - 7) = 0
2x + 4x^3 - 28x = 0
4x^3 - 26x = 0
x(4x^2 - 26) = 0
x = 0 or x = +/- √26/2
The point with the minimum distance is (+/- √26/2, 26/4) and the min. distance is 3.64 approx.
We want to minimize the following equation regarding the distance between (x, x^2) and (0, 7) but without the square root.
d(x) = (x)^2 + (x^2 - 7)^2
d '(x) = 2x + 4x(x^2 - 7) = 0
2x + 4x^3 - 28x = 0
4x^3 - 26x = 0
x(4x^2 - 26) = 0
x = 0 or x = +/- √26/2
The point with the minimum distance is (+/- √26/2, 26/4) and the min. distance is 3.64 approx.
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y = x² is the equation for a parabola opening upward with vertex at (0, 0) because there is no constant in the equation. Point (0, 7) is a straight vertical line at x = 0, so the closest point is (0, 0) and the shortest distance is the y-value of the point, or 7 units.
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