Find the point on the graph of function that is closest to the point. f(x)=x^2, \ (2,\frac{1}{2})


Find the point on the graph of function that is closest to the point. {eq}f(x)=x^2, \ (2,\frac{1}{2}) {/eq}


The general point on the given curve can be written in the parametric form.

It is easier for us as both the coordinates are now in terms of one variable only.

To minimize the distance, use the distance formula.

We need to minimize this function using the application of derivatives.

Answer and Explanation:

Consider the curve {eq}x^{2} = f \left( x \right) {/eq}

Any point on this curve in parametric form can be written as {eq}\left( t,t^{2} \right) {/eq}

Its distance from the point {eq}\left( 2,\dfrac{1}{2} \right) {/eq} is given by {eq}d = \sqrt {{{\left( {t - 2} \right)}^2} + {{\left( {{t^2} - \dfrac{1}{2}} \right)}^2}} = \sqrt {{t^2} + 4 - 4t + {t^4} + \dfrac{1}{4} - {t^2}} = \sqrt {{t^4} - 4t + \dfrac{{17}}{4}} {/eq}

To minimize {eq}d \left( t \right) {/eq}, we will minimize {eq}s\left( t \right) = {d^2}\left( t \right) = {t^4} - 4t + \dfrac{{17}}{4} {/eq}

Naturally, the points at which {eq}s \left( t \right) {/eq} is minimum will be the same points where {eq}d \left( t \right) {/eq} is minimum.

Let us first find the critical points of {eq}s \left( t \right) {/eq}

Differentiating with respect to {eq}t {/eq}, we get

{eq}4{t^3}-4 {/eq}

Equating to {eq}0 {/eq}, we get

{eq}\eqalign{ & 4{t^3} - 4 = 0 \cr & \Rightarrow {t^3} - 1 = 0 \cr & \Rightarrow t = 1 \cr} {/eq}

Hence, there is only one critical point of this function

Checking for double derivative, we have

{eq}s''\left( t \right) = 12t^{2} {/eq}

We have

{eq}s'' \left( 1 \right) = 12 > 0 {/eq}

which indicates that the value of {eq}s \left( t \right) {/eq} is minimum when {eq}t=1 {/eq} and hence, {eq}d \left( t \right) {/eq} is minimum when {eq}t=1 {/eq}

Corresponding to {eq}t=1 {/eq}, the points on the parabola are {eq}\color{red}{\left( 1 , 1 \right)} {/eq} which is closest to {eq}\left( 2, \dfrac{1}{2} \right) {/eq}

Learn more about this topic:

Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2

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