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What point on the curve y =\sqrt{x+1} is closest to the point (2, 0)?

Question:

What point on the curve {eq}y =\sqrt{x+1} {/eq} is closest to the point {eq}(2, 0) {/eq}?

Distance between the two Points:

First we have to understand the distance between the two points to solve this problem:

Suppose {eq}(x_1, y_1) {/eq} and {eq}(x_2, y_2) {/eq} are two points on the line, then distance between these two point is mathematically defined as: {eq}D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 } {/eq}

Answer and Explanation:


Given:

{eq}y = \sqrt{x + 1} \text{ and point } (2, 0) {/eq}

Let {eq}P(2, 0) {/eq} and {eq}Q {/eq} be a point on the curve {eq}y = \sqrt{x + 1} {/eq}.

Assume the x-coordinate of the point Q is t, then the y-coordinate is:

{eq}y = \sqrt{t + 1} {/eq}

Now, we have point Q in the terms of t : {eq}Q(t, \sqrt{t + 1}) {/eq}


We will compute the distance between the P and Q.

As we know the distance formula is:

$$\begin{align*} \displaystyle D &= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 } &\text{(Where } (x_1, y_1) = (2, 0) \text{ and } (x_2, y_2) = \left (t, \sqrt{t + 1} \right ) \text{)}\\ &= \sqrt{(2 - t)^2 + (0 - \sqrt{t + 1})^2 } &\text{(Plugging in the points)}\\ &= \sqrt{4 + t^2 - 4 t + t + 1}\\ D^2 &= t^2 - 3 t + 5 &\text{(Taking square both sides )}\\ &\text{ If we minimize the } D^2 \text{ that means we are minimize } D \\ \frac{d}{dt} [D^2] &= \frac{d}{dx} [ t^2 - 3 t + 5 ] &\text{(Taking derivative with respect to t)}\\ &= 2 t - 3 \times 1 + 0 &\text{(Differentiating using the formula } \frac{d}{dx} [x^n] = n x^{n - 1} \text{)}\\ &= 2 t - 3 \\ \frac{d}{dt} [D^2] &= 0 &\text{(For getting the critical point because the critical point will give minimum } D^2 \text{)}\\ 2 t - 3 &= 0\\ 2 t &= 3 \\ t &= \frac{3}{2} &\text{(Dividing both sides by 2)}\\ \end{align*} $$

Plug the value of t in the point Q

{eq}\displaystyle Q \left (\frac{3}{2}, \ \sqrt{\frac{3}{2} + 1} \right ) = \left ( \frac{3}{2}, \ \sqrt{\frac{5}{2}} \right ) {/eq}

Thus the closest to (2, 0) is {eq}\displaystyle \left ( \frac{3}{2}, \ \sqrt{\frac{5}{2}} \right ) {/eq}.


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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