Why Is My Integral Result Incorrect for Cosine of 2 Theta?

  • Thread starter Thread starter PieOperator
  • Start date Start date
  • Tags Tags
    Cosine Theta
AI Thread Summary
The integral of cosine of 2 theta from pi minus theta1 to pi plus theta1 was initially evaluated incorrectly, resulting in a positive one half sine of 2 theta1 instead of the expected negative value. The error was traced back to the orientation of the vectors used in the calculation. By reversing the direction of the vector, the interval also changes, leading to the correct result. The discussion highlights the importance of vector orientation in integral evaluations. Understanding these concepts is crucial for accurate calculations in trigonometric integrals.
PieOperator
Messages
15
Reaction score
0

Homework Statement



From angles pi minus theta1 to pi plus theta1, what is the integral of cosine of 2 theta times d theta?

Homework Equations





The Attempt at a Solution



When I evaluate this definite integral from pi minus theta1 to pi plus theta1, I get positive one half sine of 2 theta1.
It should be NEGATIVE one half sine of 2 theta1.

Is 2pi + 2theta1 and 2pi - 2theta1 somehow wrong?
 
Physics news on Phys.org
Hi PieOperator! :smile:

(have a pi: π and a theta: θ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

you should have got [1/2 sin2θ]π-θ1π+θ1 …

what happened then? :confused:​
 
Thanks. I found my mistake. It has to do with vectors and the correct orientation. It is positive one half sine of 2 theta in the incorrect vector orientation. I just have to reverse the direction of the vector, and the interval reverses as well.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Vectors Directions: Where is this Resultant Vector Pointing?
Replies
14
Views
1K
Block sliding on vertical track with friction (Solved problem)
Replies
4
Views
2K
Torque on a Circular Current Loop under a Misaligned Magnetic Field
Replies
1
Views
995
2D Steady-state Temperature in a Circular Plate - Bessel Function
Replies
2
Views
2K
Displacement Function of an Idealized Circular Loop Coaster
Replies
21
Views
2K
Circular motion of a mass on a string on an inclined plane
Replies
14
Views
4K
I How Does the Biot-Savart Law Apply to Angles in an Infinite Wire Calculation?
Replies
5
Views
3K
Equilibrium Position of a Hanging Candy Cane
Replies
17
Views
886
Why is the elastic potential energy in position 2 zero?
Replies
6
Views
2K
Finding Resultant Force Using Cosine and Sine Law
Replies
2
Views
7K

Hot Threads

Recent Insights

Back
Top