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#### [Calculus] Function regarding distance from (0,0) to another function

8 viewed last edited 4 years ago
Timothy Lee
0
So I picked this (translated) Calculus book by Thomas from library to refresh some math and in here after the first chapter about functions there is this question which I understand and know the answer to (thanks to the solutions part of the book), but I don't know how to do it myself (even a first step would be welcome). So it's not really homework, but I hope this is still the right place to ask. I would be grateful for any help. Thanks. Here a rough translation: You got 2x+4y=5 as a graph with the point (x,y). L(x) is the function that links the distance of any (x,y) to (0,0). Present function L(x). Answer: L = sqrt(20x^2 - 20x + 25) / 4
Krishna
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Given: you got 2x+4y=5 as a graph with the point (x,y). L(x) is the function that links the distance of any (x,y) to (0,0). formula: Distance between two points (0,0), (x, y) S = \sqrt {(x - 0)^2 + (y-0)^2}............(1) 2x+4y = 5 salve this equation for y 4y= 5 - 2x y = \frac{5}{4} - \frac{2x}{4} Subtitude this y value in equation (1) S = \sqrt {(x - 0)^2 + (\frac{5}{4} - \frac{2x}{4}-0)^2} S = \sqrt {(x)^2 + (\frac{25}{16} - 2 *(\frac{5}{4}*\frac{2x}{4}) + \frac{4x^2}{16}} S = \sqrt { \frac{25}{16} - \frac{20x}{16} + \frac{4x^2}{16} + x^2} S = \sqrt {\frac{25}{16} - \frac{20x}{16} + \frac{20x^2}{16}} S = \frac{\sqrt {20 x^2 - 20x + 25}}{\sqrt{16}} S = \frac{\sqrt {20 x^2 - 20x + 25}}{4}