Sketch the region enclosed by the given curves.
y = tan 3x, y = 2 sin 3x, −π/9 ≤ x ≤ π/9
then then find the area.
i can sketch but cant find correct area
To find the region enclosed by the given curves, we need to determine the points where these curves intersect and the limits of integration.
First, let's find the intersection points:
Setting the two curves equal to each other, we have:
tan 3x = 2 sin 3x
Dividing both sides by sin 3x:
tan 3x / sin 3x = 2
Using the identity tan x = sin x / cos x:
(sin 3x / cos 3x) / sin 3x = 2
simplifying further:
1 / cos 3x = 2
cos 3x = 1/2
Taking the inverse cosine (cos^-1) of both sides:
3x = cos^-1(1/2)
Using the symmetry property of cosine, we have two solutions:
3x = π/3 or 3x = -π/3
Dividing by 3 on both sides gives:
x = π/9 or x = -π/9
So the intersection points are (π/9, y) and (-π/9, y).
To find the area, we need to evaluate the integral between these two x-values:
Area = ∫[from -π/9 to π/9] (2 sin 3x - tan 3x) dx
Let's calculate this integral step by step:
Area = ∫[from -π/9 to π/9] (2 sin 3x - tan 3x) dx
= ∫[from -π/9 to π/9] (2 sin 3x - sin 3x / cos 3x) dx
= ∫[from -π/9 to π/9] ((2 cos 3x - 1) / cos 3x) dx
Now, let's evaluate this integral.
To sketch the region enclosed by the given curves, we need to plot both the curves on the same graph and identify the points where they intersect.
First, let's plot the curves y = tan 3x and y = 2 sin 3x on the graph.
Step 1: Plot the curve y = tan 3x:
- Choose a set of x-values within the given range, say x = -π/9, 0, and π/9.
- Calculate the corresponding y-values using the equation y = tan 3x.
- Plot the points (-π/9, y1), (0, y2), and (π/9, y3).
- Connect the points to form the curve of y = tan 3x.
Step 2: Plot the curve y = 2 sin 3x:
- Choose the same set of x-values as before (-π/9, 0, and π/9).
- Calculate the corresponding y-values using the equation y = 2 sin 3x.
- Plot the points (-π/9, y4), (0, y5), and (π/9, y6).
- Connect the points to form the curve of y = 2 sin 3x.
Once the curves are plotted, observe the points of intersection between the two curves. Each point of intersection represents a boundary of the region enclosed by the curves.
To find the area of the enclosed region, integrate the difference between the two curves with respect to x, between the x-values where they intersect.
In this case, the region is bounded by the curves y = tan 3x and y = 2 sin 3x, and the x-range is −π/9 ≤ x ≤ π/9.
To calculate the area, you can use the definite integral:
Area = ∫[a, b] (2 sin 3x - tan 3x) dx,
where a and b are the x-values of the intersection points.
To find these intersection points, set the two equations equal to each other:
tan 3x = 2 sin 3x.
Simplifying further, we get:
tan 3x - 2 sin 3x = 0.
You can solve this equation either algebraically or numerically using a graphing calculator or software. Once you find the x-values of the intersection points, substitute them into the integral formula and evaluate the integral to find the area of the enclosed region.
algebraically, since both functions are symmetric about the origin, the area is zero.
geometrically, using symmetry, you have the area is
2∫[0,pi/9] (tan3x - 2sin3x) dx
= 2/3 (2cos 3x - log cos 3x)[0,pi/9]
= 2/3 (log2 - 1)