Find the point P on the graph of the function y=sqrt{x} closest to the point (10,0)
the slope of the tangent is 1/(2√x), so the slope of the normal is -2√x
so you want the line with slope -2√x that goes through (10,0) and (x,y) on the curve.
Or, consider the distance z from (x,y) to (10,0).
z = √((10-x)^2+y^2) = √((10-x)^2 + x)
now find where dz/dx = 0
z = distance between^2, minimize z
z = (x-10)^2 + (y-0)^2 = x^2 - 20 x + 100 + y^2 but y^2 = x
z = x^2 - 19 x + 100
dz/dx = 2 x -19
= 0 at min
so
x = 19/2 = 9.5
y = sqrt (9.5)
check my arithmetic !
To find the point on the graph of the function y = √x closest to the point (10, 0), we can use the distance formula.
Let's assume the point P on the graph has the coordinates (x, √x). The distance between the two points P and (10, 0) can be calculated using the distance formula:
Distance = √((x - 10)² + (√x - 0)²)
We want to minimize this distance, so we can take the derivative of the above formula with respect to x and set it equal to zero.
d/dx (Distance) = d/dx ( √((x-10)² + x) ) = 0
Let's simplify this equation:
0.5 * [(x - 10)² + x]⁻⁰.⁵ * [2(x - 10) + 1] = 0
Simplifying further:
[(x - 10)² + x]⁻⁰.⁵ * (2x - 20 + 1) = 0
(2x - 19) * [(x - 10)² + x]⁻⁰.⁵ = 0
Either (2x -19) = 0 or [(x - 10)² + x]⁻⁰.⁵ = 0
For the first case, 2x - 19 = 0, we get:
2x = 19
x = 19/2
For the second case, [(x - 10)² + x]⁻⁰.⁵ = 0, but since the expression inside the square root cannot be zero, this case has no solution.
Therefore, the x-coordinate of the point P on the graph closest to (10, 0) is x = 19/2.
Now, substitute this value into the original equation to find the y-coordinate:
y = √(19/2) = √(38/4) = √(19/2) = √19 / √2 = (√19 * √2) / 2 = (√38) / 2.
So, the coordinates of the point P on the graph closest to (10, 0) are (19/2, √38 / 2).
To find the point P on the graph of the function y = √x closest to the point (10,0), we need to compute the distance between each point on the graph and the point (10,0). The point P will be the one with the shortest distance.
Let's start by writing the distance formula between two points:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the coordinates of the two points are:
P = (x, √x)
Q = (10, 0)
Substituting these values into the distance formula, we get:
d = √((10 - x)^2 + (0 - √x)^2)
To find the point P that minimizes this distance, we need to minimize the function d(x) = √((10 - x)^2 + (√x)^2).
To do this, we can take the derivative of d(x) with respect to x and set it equal to zero to find critical points. Let's differentiate:
d'(x) = [2(10 - x)(-1) + (1/2)(2√x)(1)] / 2√((10 - x)^2 + (√x)^2)
Simplifying, we get:
d'(x) = (-20 + x) / √((10 - x)^2 + x)
Setting d'(x) equal to zero:
(-20 + x) / √((10 - x)^2 + x) = 0
To solve this equation, we multiply both sides by the denominator:
-20 + x = 0
Solving for x, we find:
x = 20
Now, we have a critical point at x = 20. We need to check whether this point gives a minimum distance or not. To do this, we can use the second derivative test.
Taking the second derivative of d(x), we get:
d''(x) = (20 - x) / (√((10 - x)^2 + x))^3
Substituting x = 20, we have:
d''(20) = 0
Since the second derivative is zero at x = 20, the second derivative test is inconclusive.
To determine if x = 20 yields a minimum distance, we can substitute it back into the original distance formula and compare it with the distances at nearby points.
We have:
d(20) = √((10 - 20)^2 + (√20)^2) = √(100 + 20) = √120
Now, we can evaluate d(19) and d(21) to see if d(20) is the minimum distance.
d(19) = √((10 - 19)^2 + (√19)^2) = √(81 + 19) = √100 = 10
d(21) = √((10 - 21)^2 + (√21)^2) = √(121 + 21) = √142
Comparing the distances, we see that d(10) = √120 is the smallest among d(10), d(19), and d(21).
Therefore, the point P on the graph of y = √x closest to the point (10,0) is approximately (20, √20).