By the divergence theorem, the surface integral taken over
is equivalent to the triple integral of the divergence of
We have
The latter integral is then given by
State the horizontal asymptote of the rational function.
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By the divergence theorem, the surface integral taken over
is equivalent to the triple integral of the divergence of
We have
The latter integral is then given by
The equation of the tangent to the curve at the point corresponding to the given value of the parameter will be x + y = c.
<h3>How to calculate the tangent of the parameter curves at a point?</h3>The curves are given below.
x = t⁵ + 1 and y = t⁶ + t
Then differentiate the functions with respect to t, then we have
...1
and
...2
Divide equation 2 by equation 1, then we have
Then the slope of the equation at t = −1, then we have
dy/dx = [6(-1)⁵ + 1]/[5(-1)⁴]
dy/dx = (-6+1)/5
dy/dx = -5/5
dy/dx = -1
Then the equation of the tangent line will be
y = -x + c
x + y = c
Where c is a constant.
More about the tangent of the parameter curves at a point link is given below.
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