Find the point on the curve y=cosx closest to the point (0,0)?

1 Answer
May 13, 2018

Answer:

The point is #(0,1)#

Explanation:

By the distance formula, the distance between #(0, 0)# and the graph of #y = cosx# is

#d= sqrt((x -0)^2 + (cosx - 0)^2)#

#d = sqrt(x^2 + cos^2x)#

To find the minimum distance, we need to differentiate.

#d' = (2x - 2cosxsinx)/(2sqrt(x^2 + cos^2x))#

#d' = (x - cosxsinx)/sqrt(x^2 + cos^2x)#

We wish for this to be the smallest possible, thus we need #d' = 0#.

#0 = (x - cosxsinx)/sqrt(x^2 + cos^2x)#

#0 = x - cosxsinx#

Use a graphing application to solve and find that #x = 0# is the only solution.

There derivative is negative when #x < 0#, and positive when #x > 0#, therefore, the minimum distance will occur when #x = 0#. The value of #cosx# at #x = 0# is #1#, therefore the closest point is #(0, 1)#. Let's examine graphically:

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Hopefully this helps!