Question

Average distance on a triangle Consider the right triangle with vertices (0,0) ,(0,b) , and (a,0) , where a \u003e 0 and b \u003e 0. Show that the average vertical distance from points on the 𝓍-axis to the hypotenuse is b/2 , for all a \u003e 0 .

Accepted Answer

First, identify the hypotenuse line connecting the points (0, b) and (a, 0). Find its equation by calculating the slope and using point-slope form. The slope $$ m  is $$given by $$ m = \frac{0 - b}{a - 0} = -\frac{b}{a} . $$Using point (0, b), the line equation is $$ y = -\frac{b}{a} x + b . $$Next, consider a point on the x-axis, which has coordinates $$ (x, 0) $$ where $$ x $$ ranges from 0 to $$ a . We $$want to find the vertical distance from this point to the hypotenuse. Since the point is on the x-axis, the vertical distance to the hypotenuse is simply the y-value of the hypotenuse at $$ x $$, which is $$ y = -\frac{b}{a} x + b . To $$find the average vertical distance from all points on the x-axis between 0 and $$ a  to $$the hypotenuse, set up the integral of the vertical distance function over the interval $$ [0, a] $$ and divide by the length of the interval $$ a . $$This gives the average distance $$ D_{avg} = \frac{1}{a} \int_0^a \left(-\frac{b}{a} x + b\right) dx . $$Evaluate the integral $$ \int_0^a \left(-\frac{b}{a} x + b\right) dx  by $$integrating term-by-term: $$ \int_0^a -\frac{b}{a} x \, dx + \int_0^a b \, dx . $$Compute each integral separately. After integrating, simplify the expression and divide by $$ a  to $$find the average vertical distance. You should find that the average distance simplifies to $$ \frac{b}{2} $$, which is independent of $$ a $$, as required.

Average distance on a triangle Consider the right triangle with vertices (0,0) ,(0,b) , and (a,0) , where a > 0 and b > 0. Show that the average vertical distance from points on the 𝓍-axis to the hypotenuse is b/2 , for all a > 0 .

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Equation of a Line (Hypotenuse)

Vertical Distance from a Point to a Line

Average Value of a Function over an Interval