Question

Find the point on the curve \[y = \sqrt x \] that is closest to the

point \(\left( {3,0} \right)\).

Short answer

The closest point to the curve is \(\left( {\frac{5}{2},\sqrt {\frac{5}{2}} } \right)\).

Step-by-step solution

Application of Derivative

We can apply the derivative of a function to find the closest point of a curve to a point or the furthest point of a curve to another point. We find these points with the concept of local minimaand local maximarespectively.

We find local minima and maxima of a function by finding critical pointsand the second derivativeof the function.

Critical points are the \(x\) values that satisfies \(f'\left( x \right) = 0\).

The function has a local maximum at a point \(x\) if \(f''\left( x \right) < 0\).

The function has a local minimum at a point \(x\) if \(f''\left( x \right) > 0\).

Finding the between the given point and a point of the curve

The equation of the curve is given by \(y = \sqrt x \).

Let \(\left( {x,\sqrt x } \right)\) be any point of the curve.

Now the distance between \(\left( {3,0} \right)\) and \(\left( {x,\sqrt x } \right)\) will be

\(\begin{aligned}{l}d = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {\sqrt x - 0} \right)}^2}} \\d = \sqrt {{{\left( {x - 3} \right)}^2} + x} \end{aligned}\)

Finding the Local Minima of the square of the distance and hence the closest point

Let \(S = {d^2} \Rightarrow S = {\left( {x - 3} \right)^2} + x\)

Now

\(S' = 2\left( {x - 3} \right) + 1\)

So the critical point will be

\(\begin{aligned}{c}S' = 0\\2\left( {x - 3} \right) + 1 = 0\\2x - 6 + 1 = 0\\2x = 5\\x = \frac{5}{2}\end{aligned}\)

Hence the critical point is \(x = \frac{5}{2}\).

The second derivative will be \(S'' = 2 > 0\).

Hence the minimum value of the function \(S\) will be at the point \(x = \frac{5}{2}\).

Thus the closest point to the curve is \(\left( {\frac{5}{2},\sqrt {\frac{5}{2}} } \right)\).