Find the equation of the tangents to the curve, y = cos(x+y), -2≤ x≤ 2Π, that are parallel to the line x + 2y = 0.
Two parallel lines have the same slope
The given line can be written as-
y = -()∙x
So it has slope:
The slope of tangent line to a curve at point equals at this point. So you need to find the point (x₀,y₀) on the curve for which
The equation of a tangent line is given by point slope form:
y -y₀= m∙(x - x₀) = y₀ - ()∙(x - x₀)
The slope of the tangent can be found by implicit differentiation:
y = cos(x + y)
y'=
= -sin(x + y) ∙(1 + )
=
At the point you're looking for
= -
solving we get,
2sin(x₀ + y₀) = 1 + sin(x₀ + y₀)
sin(x₀ + y₀) = 1
That means
x₀ + y₀ = () ± 2nπ
were n is an arbitrary integer.
More important is the fact that cosine vanishes at points were sine equals zero, i.e.
cos(x₀ + y₀) = 0 if sin(x₀ + y₀) = 1
Hence,
y₀ = cos(x₀ + y₀) = 0
So the condition for the slope leads to
sin(x₀) = 1
which has only one solution on the interval [0,2π], it is:
x₀ =()
So the equation of the tangent line is:
y = y₀ - ()(x - x₀)
y = 0 - ()(x -() )
y =
The given line can be written as-
y = -()∙x
So it has slope:
The slope of tangent line to a curve at point equals at this point. So you need to find the point (x₀,y₀) on the curve for which
The equation of a tangent line is given by point slope form:
y -y₀= m∙(x - x₀) = y₀ - ()∙(x - x₀)
The slope of the tangent can be found by implicit differentiation:
y = cos(x + y)
y'=
= -sin(x + y) ∙(1 + )
=
At the point you're looking for
= -
solving we get,
2sin(x₀ + y₀) = 1 + sin(x₀ + y₀)
sin(x₀ + y₀) = 1
That means
x₀ + y₀ = () ± 2nπ
were n is an arbitrary integer.
More important is the fact that cosine vanishes at points were sine equals zero, i.e.
cos(x₀ + y₀) = 0 if sin(x₀ + y₀) = 1
Hence,
y₀ = cos(x₀ + y₀) = 0
So the condition for the slope leads to
sin(x₀) = 1
which has only one solution on the interval [0,2π], it is:
x₀ =()
So the equation of the tangent line is:
y = y₀ - ()(x - x₀)
y = 0 - ()(x -() )
y =