Find the point on the curve y = sqrt(x) that is closest to the point (3, 0)

Solution:

Given, the curve is y = √x --- (1)

If D is the distance from (x, y) to the point (3, 0), by Pythagorean distance formula

D² = (x - 3)² + y²

= (x - 3)² + (√x)²

= x² - 6x + 9 + x

D² = x² - 5x + 9

On differentiating,

2D.D’ = 2x - 5

D’ = \(\frac{2x-5}{2D}\)

\(D'=\frac{2x-5}{2\sqrt{x^{2}-5x+9}}\)

\(D'=(2x-5).\frac{1}{2(x^{2}-5x+9)^{\frac{1}{2}}}\)

Now, \((2x-5).\frac{1}{2(x^{2}-5x+9)^{\frac{1}{2}}}=0\)

⇒ 2x - 5 = 0

⇒ 2x = 5

⇒ x = 5/2

This is known as the critical value and it represents the x-value for which the function is minimised.

Put x = 5/2 in (1)

y = √5/2

y = 1.58

Therefore, the point on the curve y = √x closest to the point (3, 0) is (5/2, 1.58).

Find the point on the curve y = sqrt(x) that is closest to the point (3, 0)

Summary:

The point on the curve y = √x that is closest to the point (3, 0) is (5/2, 1.58).