Monday, 12 August 2013

`(0,0) , (8,15)` Find the distance between the two points using integration.

 Given the equation of a line `y = mx + b,`


=> slope = `dy/dx = m` . Thus, the distance is:


`L = int_a^b sqrt(1+(dy/dx)^2) dx  , a<=x<=b` 



we know the two points` (x_1,y_1)=(0,0)`


`(x_2,y_2)=(8,15)`


`m = (y_2- y_1)/(x_2-x_1) = (15-0)/(8-0) = 15/8`


so now the length is `L = int_0^8 sqrt(1+(15/8)^2) dx`


 `= int_0^8 sqrt(1+(225/64)) dx`


 =` int_0^8 sqrt((64+225)/64)) dx`


= `int_0^8 sqrt((289)/64)) dx`


= `int_0^8 (17/8) dx`


= `(17/8) int_0^8 1...

 Given the equation of a line `y = mx + b,`


=> slope = `dy/dx = m` . Thus, the distance is:


`L = int_a^b sqrt(1+(dy/dx)^2) dx  , a<=x<=b` 



we know the two points` (x_1,y_1)=(0,0)`


`(x_2,y_2)=(8,15)`


`m = (y_2- y_1)/(x_2-x_1) = (15-0)/(8-0) = 15/8`


so now the length is `L = int_0^8 sqrt(1+(15/8)^2) dx`


 `= int_0^8 sqrt(1+(225/64)) dx`


 =` int_0^8 sqrt((64+225)/64)) dx`


= `int_0^8 sqrt((289)/64)) dx`


= `int_0^8 (17/8) dx`


= `(17/8) int_0^8 1 dx`  


= `(17/8) |_0^8 x` 


=` (17/8 )[8-0]`


= `17 `


so the distance between the two points = 17

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