The shortest distance between the point (32, 0) and the curve y = √x, (x > 0) is

Let points (32, 0), (t², t) t > 0
Distance = √(t² - (32)²)² + (t - 0)² = √(t² - 32)² + t²
= √t⁴ - 64t² + 1024 + t² = √t⁴ - 63t² + 1024
Let f(t) = t⁴ - 63t² + 1024
f'(t) = 4t³ - 126t = 0
t²(4t² - 126) = 0
t² = 0 or t² = 63/2
t = √(63/2)
Distance = √((63/2)² - 63(63/2) + 1024) = √(3969/4 - 3969/2 + 4096/4)
= √(8192 - 7938)/4 = √(254)/4 = √(63.5)
Let's find the distance by another method.
Let (x,y) be a point on the curve y=√x
Then the distance from (32,0) to (x,y) is
D² = (x-32)² + y² = (x-32)² + x
D² = x² - 64x + 1024 + x = x² - 63x + 1024
To minimize D², we find the derivative and set it to zero
2x - 63 = 0
x = 63/2
y = √(63/2)
D² = (63/2)² - 63(63/2) + 1024 = (63²/4) - (2*63²/4) + 1024
D² = 1024 - 63²/4 = 1024 - 3969/4 = 4096/4 - 3969/4 = 127/4
D = √(127/4) ≈ 5.63
Let’s use the distance formula:
Distance = √((x-32)² + x) = √(x² - 63x + 1024)
Taking the derivative and setting it to 0, we get
2x - 63 = 0 => x = 63/2
Distance = √((63/2 - 32)² + 63/2) = √( ( -1/2)² + 63/2 ) = √(1/4 + 63/2) = √(127/4) = √52