Find the measure of the missing angle.
2 answers:
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Answer:
a) f(x)
b) g(x)
c) for all values of x > 1, g(x) has a greater rate of change
Step-by-step explanation:
a) The rate of change of f(x) is the x-coefficient: 4. The average rate of change of g(x) on an interval can be found by dividing the change by the interval width. Here, the width of the interval of interest is 1-0 = 1, so the average rate of change is g(1) -g(0) = 4-2 = 2.
On the interval [0, 1], f(x) has the greater rate of change.
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b) The interval [2, 3] also has a width of 1, so the rate of change of g(x) on that interval is g(3) -g(2) = 20 -10 = 10. This value is greater than 4, so ...
on the interval [2, 3], g(x) has the greater rate of change.
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c) Part (b) demonstrates that g(x) will have a greater rate of change on some intervals. In fact, for any interval whose center is greater than x=1, g(x) will have a greater rate of change than f(x).
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In calculus terms, the rate of change of g(x) is g'(x) = 4x. This will be greater than 4 for any x > 1. The graph shows the rate of change is 4 at x=1, and is higher for x > 1. The average rate of change will be greater than 4 on any interval whose center is greater than x=1.
We can figure the average rate of change on the interval [a, b] as ...
m = (g(b) -g(a))/(b -a)
m = ((2b² +2) -(2a² +2))/(b -a) = 2(b² -a²)/(b-a) = 2(b+a)
For the average rate of change to exceed 4, the sum of the ends of the interval must exceed 2, which is to say the midpoint must exceed 1.
Answer:
Step-by-step explanation:
We have a curve (an ellipse) written as the system of equations
.
And we want to calculate the tangent at the point (3,4,9).
The idea in this problem is to consider two variables as functions of the third. Usually we consider and
as functions of
. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization
.
Recall that the parametric equation of a line has the form
,
where is a point on the line (in this particular case is (3,4,9)) and
is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).
Now, if we have the parametric equation of a curve its tangent line will have direction vector
. So, as we need to calculate the equation of the tangent line at the point
, we must obtain the tangent vector
. This part can be done taking implicit derivatives in the systems that defines the ellipse.
So, let us write the system as
.
Then, taking implicit derivatives:
.
Now we substitute the values and
, and we get the system of linear equations
,
where the unknowns are and
.
The system is
,
and its solutions are
and
.
Then, the direction vector of the tangent is
.
Finally, the tangent line has parametric equation
where .