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Answer to Question #4905 in Calculus for klinc
Question #4905
find the point on the curve y= cosx closest to the point (0,0)
Expert's answer
Distance formula:
d = SQRT( (x1-x2)^2 + (y1-y2)^2 )
Here, x1 = y1 = 0.
d = SQRT( x^2 + y^2 )
y = cos(x) - equation of our curve
d = SQRT( x^2 + (cos(x))^2 )
f(x) = x^2 + (cos(x))^2
f'(x) = 2x - 2cos(x)sin(x) = 2x - sin(2x)
Set f'(x) to zero:
2x - sin(2x) = 0
2x = sin(2x)
2x = 0, or x=0 - the unique solution that minimize the distance
cos(0) = 1
So, the point on the curve y=cosx closest to the point (0,0) is point (0,1).
d = SQRT( (x1-x2)^2 + (y1-y2)^2 )
Here, x1 = y1 = 0.
d = SQRT( x^2 + y^2 )
y = cos(x) - equation of our curve
d = SQRT( x^2 + (cos(x))^2 )
f(x) = x^2 + (cos(x))^2
f'(x) = 2x - 2cos(x)sin(x) = 2x - sin(2x)
Set f'(x) to zero:
2x - sin(2x) = 0
2x = sin(2x)
2x = 0, or x=0 - the unique solution that minimize the distance
cos(0) = 1
So, the point on the curve y=cosx closest to the point (0,0) is point (0,1).
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#340153 on Dec 2023
#340153 on Dec 2023
Comments
Take a look: the distance d = SQRT(x^2 + (cos(x))^2) and its square d^2=x^2 + (cos(x))^2 reaches the minimal value simultaneously. Thus one can find minimum of d^2 because it looks easier.
how did you get rid of the SQRT after the 5th step? you can't just drop that trash....
2x = sin(2x) Applying double angle formula we get 2x = 2sinx cosx or x=sin x cos x This is a transcendental equation which can be solved graphically. This equation has unique solution x=0.
how did you go from 2x = sin(2x) to 2x = 0. How six(2x) =0?
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