Given y=(x)^1/2, find the closest point to (3/2,0)
let the closest point be P(a,b)
label the given point A(3/2 , 0)
AP will have a minimum length when AP is perpendicular to the tangent at P
slope of tangent = y' = (1/2)x^(-1/2) or 1/(2√x)
so at P the slope of the tangent is 1/2√a
slope of AP = b/(a-3/2)
b/(a-3/2) = -2√a , (the negative reciprocal of 1/2√a
but also b = √a
√a/(a-3/2) = -2√a
1/(a-3/2) = -2
1 = -2a + 3
2a = 2
a = 1
then b=√1 = 1
the closest point is (1,1)
To find the closest point to the point (3/2, 0) on the curve represented by the equation y = sqrt(x), you can use the distance formula.
The distance formula is given by:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's call the x-coordinate of the closest point on the curve as "a", and the corresponding y-coordinate as "b". We need to find the values of a and b that minimize the distance between (a, b) and (3/2, 0).
Substituting the x-coordinate of the closest point, "a", into the equation y = sqrt(x), we get:
b = sqrt(a)
Now, let's calculate the distance between (a, b) and (3/2, 0).
Distance = sqrt((a - 3/2)^2 + (b - 0)^2)
Substituting the value of b from above, we get:
Distance = sqrt((a - 3/2)^2 + (sqrt(a))^2)
Simplifying further, we have:
Distance = sqrt(a^2 - 3a + 9/4 + a)
Distance = sqrt(2a^2 - 2a + 9/4)
To minimize the distance, we need to find the value of "a" that minimizes the expression inside the square root.
To do this, we can either complete the square or take the derivative and find the critical points. Let's take the derivative approach:
Differentiating the expression with respect to "a", we get:
d(Distance)/da = (1/2) * (2a^2 - 2a + 9/4)^(-1/2) * (4a - 2)
Setting the derivative equal to zero, we have:
(1/2) * (2a^2 - 2a + 9/4)^(-1/2) * (4a - 2) = 0
Simplifying, we get:
4a - 2 = 0
a = 1/2
Now that we have the value of "a", we can substitute it back into the equation y = sqrt(x) to find the corresponding value of "b":
b = sqrt(1/2)
Therefore, the closest point to (3/2, 0) on the curve y = sqrt(x) is approximately (1/2, sqrt(1/2)).