Question

Find the point on the line \(y = 2x + 3\) that is closest to the origin.

Short answer

The closest point to the curve is \(\left( { - \frac{6}{5},\frac{3}{5}} \right)\).

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 satisfy\(f'\left( x \right) = 0\).

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\(y = 2x + 3\).

Let\(\left( {x,2x + 3} \right)\)be any point of the curve.

Now the distance between\(\left( {0,0} \right)\)and\(\left( {x,2x + 3} \right)\)will be

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

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

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

Now, the derivative of S will be:

\(\begin{aligned}{c}S' = 2\left( {2x + 3} \right)\left( 2 \right) + 2x\\ = 10x + 12\end{aligned}\)

So, the critical point will be,

\(\begin{aligned}{c}S' = 0\\10x + 12 = 0\\10x = - 12\\x = \frac{{ - 12}}{{10}}\\x = - \frac{6}{5}\end{aligned}\)

Hence the critical point is\(x = - \frac{6}{5}\).

The second derivative will be \(S'' = 10 > 0\).

Hence the minimum value of the function\(S\)will be at the point\(x = - \frac{6}{5}\). The y value for function will be

\(\begin{aligned}{c}y = 2\left( { - \frac{6}{5}} \right) + 3\\ = \frac{3}{5}\end{aligned}\)

Thus, the closest point to the curve is \(\left( { - \frac{6}{5},\frac{3}{5}} \right)\).