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The curve y = x/(1 + x^2) is called a serpentine. Find an equation of the tangent line to this curve at the point (2, 0.40). (R...Asked by Andy
The curve y = x/(1 + x^2) is called a serpentine. Find an equation of the tangent line to this curve at the point(2, 0.40).(Round the slope and y-intercept to two decimal places.)
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Answered by
Reiny
use the quotient rule to get
dy/dx = [(1+x^2)(1) - x(2x)]/(1+x^2)^2
when x = 2,
dy/dx = [ 5 - 8]/25 = -3/25
equation of tangent:
y - .40 = (-3/25)(x-2)
25y - 10 = -3x + 6
25y = -3x + 16
y = (-3/25)x + 16/25
or
y = -.12x + .64 or 3x + 25y = 16
(why would you round anything? The numbers are exact)
dy/dx = [(1+x^2)(1) - x(2x)]/(1+x^2)^2
when x = 2,
dy/dx = [ 5 - 8]/25 = -3/25
equation of tangent:
y - .40 = (-3/25)(x-2)
25y - 10 = -3x + 6
25y = -3x + 16
y = (-3/25)x + 16/25
or
y = -.12x + .64 or 3x + 25y = 16
(why would you round anything? The numbers are exact)
Answered by
Andy
I guess whoever wrote this problem didn't test it out first.
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