How do i do this? It's for calculus. My math hw is due in an hour :o
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The distance from origin to a point x,y is SQRT(x^2 + y^2)
Here y must be on the line 3x + 5 so the distance is: SQRT( x^2, (3x+5)^2) =
SQRT(x^2 + 9x^2 + 30x + 25). So factoring common terms, distance = SQRT(5( 2x^2 + 6x + 5))
This is a perabola, so you know the smallest y value occurs at its base.
To find the base you get the first derivative and set it to zero.
dy/dx y = 5 (2x^2 + 6x + 5) = 5 (4x + 6);
Solving this for dy/dx = 0 use 5(4x+6) = 0 so x = -6/4 = -1.5
Now find the distance from origin to the point on the curve where x = -1.5.
Distance = SQRT( x^2, (3x+5)^2) where x = -1.5
So distance =
SQRT(5( 2 x (-1.5)^2 + 6x-1.5 + 5) = 5 x (2.25 x 2 - 9 + 5) = SQRT(2.5) = SQRT(5/2).
You could double check by using x = -1.6 and x = -1.4 - they both should produce larger distance values than SQRT(5/2).
Here y must be on the line 3x + 5 so the distance is: SQRT( x^2, (3x+5)^2) =
SQRT(x^2 + 9x^2 + 30x + 25). So factoring common terms, distance = SQRT(5( 2x^2 + 6x + 5))
This is a perabola, so you know the smallest y value occurs at its base.
To find the base you get the first derivative and set it to zero.
dy/dx y = 5 (2x^2 + 6x + 5) = 5 (4x + 6);
Solving this for dy/dx = 0 use 5(4x+6) = 0 so x = -6/4 = -1.5
Now find the distance from origin to the point on the curve where x = -1.5.
Distance = SQRT( x^2, (3x+5)^2) where x = -1.5
So distance =
SQRT(5( 2 x (-1.5)^2 + 6x-1.5 + 5) = 5 x (2.25 x 2 - 9 + 5) = SQRT(2.5) = SQRT(5/2).
You could double check by using x = -1.6 and x = -1.4 - they both should produce larger distance values than SQRT(5/2).