Find the area under the curve y=5(cosx) and above y=5(sinx) for 0<x<pi
Sketch a graph of the two functions between 0 and pi.
Note that cos x is above sin x only between x = 0 and x = pi/4
so
we have 5 times integral from 0 to pi/4 of cos x dx - sin x dx
cos x dx --> sin x 0 to pi/4 = .5 sqrt 2
sin x dx = -cos x 0 to pi/4= -.5sqrt 2 +1
cos x dx - sin x dx --> sqrt 2 - 1
times 5 = 5 sqrt 2 - 5
or about 7.07 - 5 = 2.07
Well, since you like to play with curves, let's find some area under them!
To find the area between the curves y = 5(cosx) and y = 5(sinx) for 0 < x < pi, we'll need to integrate. However, the graphs of these functions keep playing their little dance and intersecting each other.
So, let's figure out the points where they intersect. We know that y = 5(cosx) and y = 5(sinx), so when they are equal to each other, we get:
5(cosx) = 5(sinx).
Dividing both sides by 5, we have:
cosx = sinx.
Now, we can solve this equation. But instead of getting too serious with math, let's take a clownish shortcut. We can use our knowledge that sin(pi/4) = cos(pi/4) = sqrt(2)/2.
So, we can say that x = pi/4 is one of the points where the graphs intersect.
Now, we'll need to find the other point of intersection. We know that the graphs repeat every 2*pi, so we can add 2*pi to our previous solution:
x = pi/4 + 2*pi.
Now, let's set up the integral to find the area between the curves:
∫[0, pi/4 + 2*pi] (5(cosx) - 5(sinx)) dx.
But hey, we're still not done clowning around! Let's simplify this a bit:
∫[0, pi/4 + 2*pi] (5cosx - 5sinx) dx.
Now, we can integrate term by term:
∫[0, pi/4 + 2*pi] 5cosx dx - ∫[0, pi/4 + 2*pi] 5sinx dx.
Integrating cosx gives us sinx, and integrating sinx gives us -cosx:
[5sinx] [0, pi/4 + 2*pi] - [-5cosx] [0, pi/4 + 2*pi].
Plugging in the limits of integration:
[5sin(pi/4 + 2*pi) - 5sin(0)] - [-5cos(pi/4 + 2*pi) - (-5cos(0))].
Simplifying a bit:
[5sin(pi/4 + 2*pi) - 5sin(0)] - [-5cos(pi/4 + 2*pi) + 5].
But wait, we're not done yet! Let's appreciate the symmetry of sine and cosine:
[5(sin(pi/4)cos(2*pi) + cos(pi/4)sin(2*pi)) - 5sin(0)] - [-5(cos(pi/4)cos(2*pi) - sin(pi/4)sin(2*pi)) + 5].
Now, let's simplify further:
[5(sin(pi/4)(1) + cos(pi/4)(0)) - 5sin(0)] - [-5(cos(pi/4)(1) - sin(pi/4)(0)) + 5].
Fortunately, sin(0) = 0, so we can simplify even more:
[5(sin(pi/4) - 0)] - [-5(cos(pi/4) - 0) + 5].
Now, let's use our clownish knowledge again: sin(pi/4) = cos(pi/4) = sqrt(2)/2:
[5(sqrt(2)/2) - 0] - [-5(sqrt(2)/2) - 0 + 5].
Simplifying further:
[5sqrt(2)/2] - [-5sqrt(2)/2 + 5].
Finally, let's do the math:
[5sqrt(2)/2] - [-5sqrt(2)/2 + 5] = 5sqrt(2)/2 + 5sqrt(2)/2 - 5 = 10sqrt(2)/2 - 5 = 5sqrt(2) - 5.
And voila! The area under the curve y = 5(cosx) and above y = 5(sinx) for 0 < x < pi is 5sqrt(2) - 5.
Hope you had fun clowning around with this mathematical circus act!
To find the area under the curve y = 5(cosx) and above y = 5(sinx) for 0 < x < pi, we can use definite integration.
Step 1: First, let's visualize the curves and the region we need to find the area of:
Graph:
- y = 5(cosx)
- y = 5(sinx)
Region: The area between the curves above the x-axis for 0 < x < pi.
Step 2: Determine the points of intersection:
Set the two equations equal to each other and solve for x to find the points where the curves intersect.
5(cosx) = 5(sinx)
Divide both sides by 5:
cosx = sinx
Divide both sides by cosx:
1 = tanx
Since tanx = 1 at x = pi/4, the curves intersect at this point.
Step 3: Set up the integral:
To find the area between the curves, we need to integrate the difference between the upper curve (y = 5(cosx)) and the lower curve (y = 5(sinx)) from x = 0 to x = pi.
The integral will be:
∫ [upper curve - lower curve] dx
Step 4: Evaluate the integral:
Integrate the difference between the curves from x = 0 to x = pi.
∫ [5(cosx) - 5(sinx)] dx
= 5∫(cosx - sinx) dx
= 5[sinx + cosx] evaluated from x = 0 to x = pi
Step 5: Substitute the limits of integration:
Evaluate the expression at x = pi and x = 0 and subtract the results.
= 5[(sin(pi) + cos(pi)) - (sin(0) + cos(0))]
= 5[(0 + (-1)) - (0 + 1)]
= 5[(-1) - (1)]
= 5(-2)
= -10
Step 6: Interpret the result:
The area under the curve y = 5(cosx) and above y = 5(sinx) for 0 < x < pi is -10 square units. Since area cannot be negative, we can conclude that there is no area between the curves in this particular region.
To find the area under the curve y = 5cos(x) and above y = 5sin(x) for 0 < x < π, we need to find the definite integral of the difference between the two functions over this interval.
Let's break it down step by step:
Step 1: Determine the points of intersection.
To find the points of intersection between the curves, we need to set y = 5cos(x) equal to y = 5sin(x):
5cos(x) = 5sin(x)
Dividing both sides by 5, we get:
cos(x) = sin(x)
Using trigonometric identities, we can rewrite this equation as:
cos(x) = cos(π/2 - x)
For two angles to be equal, their cosine values must be equal. Therefore, we have:
x = π/2 - x
Simplifying, we get:
2x = π/2
x = π/4
So, the two curves intersect at x = π/4.
Step 2: Set up the integral.
The area under the curve y = 5cos(x) and above y = 5sin(x) can be found by integrating the difference of the two functions over the interval 0 < x < π. So, the integral can be written as:
∫[0, π] (5cos(x) - 5sin(x)) dx
Step 3: Evaluate the integral.
Evaluating this integral gives us the area under the curve:
∫[0, π] (5cos(x) - 5sin(x)) dx = [5sin(x) + 5cos(x)] evaluated from 0 to π
Evaluating at the upper limit (π) gives:
[5sin(π) + 5cos(π)] - [5sin(0) + 5cos(0)]
Simplifying, we get:
[0 + (-5)] - [0 + 5] = -10
Therefore, the area under the curve y = 5cos(x) and above y = 5sin(x) for 0 < x < π is -10 square units.
Note: The negative sign indicates that the area is below the x-axis.