Area Under Curve: Find Intersection Points & Area

In summary: If in [a,b] f(x)<g(x) then you integrate g-f over that range. Within each subrange, figure out which function is the greater. If in the range [a,b] f(x)>g(x) then you integrate f-g over that range. If in [a,b] f(x)<g(x) then you integrate g-f over that range. In summary, the student was having trouble finding intersections between the cosine and sine functions and was trying to solve for the area using the greater function. However, the student ran into problems when trying to determine which function was greater within each subrange
  • #1
bonfire09
249
0

Homework Statement


Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

Homework Equations



The sketch of the curve is given too which I uploaded.

The Attempt at a Solution


Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
0=2sin(6x)-1
sin(6x)=1/2
6x=sin^-1(1/2)
x=(sin^-1(1/2))/6

Then I am not sure what to do from here.
 
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  • #2
bonfire09 said:

Homework Statement


Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

Homework Equations



The sketch of the curve is given too which I uploaded.

The Attempt at a Solution


Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
Good up to here.

0=2sin(6x)-1
either 2sin(6x)- 1= 0 or cos(x)= 0

sin(6x)=1/2
Have you never learned that [itex]sin(\pi/6)= 1/2[/itex]?

Draw an equilateral triangle with side length s. All three angles have measure [itex]\pi/3[/itex]. Draw a perpendicular from one vertex to the opposite side forming a right triangle. That perpendicular bisects the angle and the opposite side. The angle at that vertex is [itex](\pi/3)/2= \pi/6[/itex]. The hypotenuse is a side of the original triangle and has length s. The "opposite side" has length s/2 so [itex]sin(\pi/6)= (s/2)/s= 1/2[/itex].

6x=sin^-1(1/2)
x=(sin^-1(1/2))/6

Then I am not sure what to do from here.
[itex]6x= \pi/6[/itex] so [itex]x= \pi/36[/itex]

Of course, cos(x)= 0 for [itex]x= \pi/2[/itex].

Since sine and cosine are periodic, there are many solutions to these equations. Your graph should show which one you need to use.
 
  • #3
bonfire09 said:
Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.
Area or areas?
The sketch of the curve is given too which I uploaded.
Where?
Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
0=2sin(6x)-1
There's another solution, but I'll assume you already covered this in your sketch.
sin(6x)=1/2
What angle has sine equal to 0.5?
Assuming the idea is to find the total area enclosed, treating each individual area as positive, you need to break the range of integration at each intersection. (If you don't do that some areas will come out negative and cancel against others.)
Within each subrange, figure out which function is the greater. If in the range [a,b] f(x)>g(x) then you integrate f-g over that range.
 

1. What is the Area Under Curve?

The Area Under Curve (AUC) is a measure of the total area underneath a line or curve on a graph. It is often used to calculate the overall performance of a model or to compare different models.

2. How is the Area Under Curve calculated?

The AUC is calculated by finding the integral of the function that represents the curve or line. This can be done using mathematical calculations or with the help of software programs, such as Excel or Python.

3. Why is it important to find the intersection points of a curve?

Finding the intersection points of a curve can provide valuable information about the relationship between two variables. It can also help identify the optimal point where the curve reaches its maximum or minimum value.

4. How do you find the intersection points of a curve?

To find the intersection points of a curve, you can set the two equations that represent the curves equal to each other and solve for the common variable. This will give you the x-coordinate of the intersection point. You can then substitute this value into one of the original equations to find the corresponding y-coordinate.

5. Can the Area Under Curve be negative?

No, the AUC cannot be negative. It represents the total area underneath the curve, which is always a positive value. If the curve dips below the x-axis, the area below the x-axis is subtracted from the total AUC to give the final positive value.

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