/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the point \(P\) on the line... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 19

Find the point \(P\) on the line \(y=3 x\) that is closest to the point \((50,0) .\) What is the least distance between \(P\) and (50,0)\(?\)

Short Answer

Expert verified
Answer: The point P is (5, 15) and the least distance between P and (50,0) is \(15\sqrt{10}\).

Step by step solution

01

Set up the distance function

Let P(x, y) be the point on the line y = 3x that is closest to the point (50,0). We want to find the coordinates of P and the distance between P and (50,0). Use the distance formula to set up the distance function: Distance, D = \(\sqrt{(x - x_{1})^2 + (y - y_{1})^2}\) where (x, y) are the coordinates of P and \((x_{1}, y_{1})\) are the coordinates of the given point (50,0). So, we have: D = \(\sqrt{(x - 50)^2 + (y - 0)^2}\) Since P is on the line y = 3x, we can substitute y = 3x in our distance function: D = \(\sqrt{(x - 50)^2 + (3x - 0)^2}\)
02

Minimize the distance function

Taking the derivative of the distance function and setting it to 0 can minimize this function. However, to make the calculation easier, we can work with the square of the distance function instead, i.e., \(D^2 = (x - 50)^2 + (3x)^2\). Now, we can differentiate the square of the distance function with respect to x and set it to 0: \(\frac{d}{dx} (D^2) = \frac{d}{dx} [(x - 50)^2 + (3x)^2] = 0\) By differentiating and simplifying, we obtain: \(2(x - 50) + 18x = 0\) Now, solve for x: \(20x = 100\) \(x = 5\)
03

Find the y-coordinate of point P

Now that we have the x-coordinate of point P, we can use the line equation y = 3x to find the y-coordinate: y = 3x y = 3(5) y = 15 The coordinates of point P are (5,15).
04

Calculate the least distance

Now that we know the coordinates of point P, we can plug these values back into the initial distance function D and calculate the least distance between P and (50,0): D = \(\sqrt{(5 - 50)^2 + (15 - 0)^2} = \sqrt{(-45)^2 + 15^2} = \sqrt{2025 + 225} = \sqrt{2250} = 15\sqrt{10}\) The least distance between point P and (50,0) is \(15\sqrt{10}\).

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