Question

Find the point P on the graph of the function y=√x closest to the point (9,0)

Answer 1

Answer:

$\displaystyle \frac{17}{2}$.

Step-by-step explanation:

Let the $x$-coordinate of $P$ be $t$. For $P\!$ to be on the graph of the function $y = \sqrt{x}$, the $y$-coordinate of $\! P$ would need to be $\sqrt{t}$. Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$.

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$.

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$. Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$, would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$.

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$:

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$.

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$.

Set the first derivative, $(2\, t - 17)$, to $0$ and solve for $t$:

$2\, t - 17 = 0$.

$\displaystyle t = \frac{17}{2}$.

Notice that the second derivative is greater than $0$ for this $t$. Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$. This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$, the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$.

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$-coordinate of $P\!$ is $\displaystyle \frac{17}{2}$.