Problem 19 - Calculus Early Transcendentals

Question

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)\(?\)

Step-by-Step Solution

Answer

Answer: The least distance between the point (50,0) and the line y=3x is 15√10.

Step 1: Find the slope of the line perpendicular to y=3x

The given line y=3x, so its slope, m1 = 3. The slope of the line perpendicular to this line, m2, should satisfy the condition m1*m2=-1. Thus, m2 = -\frac{1}{3}.

Step 2: Write the equation of the perpendicular line

We will use the point-slope form, which is y-y1=m(x-x1), where m is the slope and (x1,y1) is the given point. Using the point (50,0) and the perpendicular slope -\frac{1}{3}, the equation becomes y-0=-\frac{1}{3}(x-50). Simplify this equation to obtain y=-\frac{1}{3}(x-50).

Step 3: Find the intersection of the two lines to determine point P

We will find the intersection point of the lines y=3x and y=-\frac{1}{3}(x-50) by solving the system of equations. Equate the two y expressions and solve for x: 3x = -\frac{1}{3}(x-50) => 9x = -(x-50) => 10x = 50 => x = 5 Now substitute this x value back into y=3x to find the corresponding y value: y=3(5) => y=15 So, the point P is (5,15).

Step 4: Calculate the distance between point P and (50,0) using the distance formula

Using the distance formula d=\sqrt{(x2-x1)^2+(y2-y1)^2}, where (x1,y1)=(50,0) and (x2,y2)=(5,15), we get: d = \sqrt{(5-50)^2+(15-0)^2}=\sqrt{(-45)^2+15^2}=\sqrt{2025+225}=\sqrt{2250} = 15\sqrt{10} Thus, the least distance between point P and (50,0) is 15\sqrt{10}.