The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
The generalized coordinate for the pendulum is the angle of the arm off vertical, theta. Theta is 0 when the pendulum arm is down and pi when the arm is up. M = mass of pendulum L = length of pendulum arm g = acceleration of gravity \theta = angle of pendulum arm off vertical \dot{\theta} = time derivative of \theta What are the kinetic and potential energies? Kinetic energy: T = (1/2)*M*(L*\dot{\theta})^2 Potential energy: V' = MLg(1-cos(\theta)) V = -MLg*cos(\theta) --note: we can shift the potential by any constant, so lets choose to drop the MLg The Lagrangian is L=T-V: L = (1/2)ML^2\dot{\theta}^2 + MLg*cos(\theta)
If it's referring to music it's a bass clef, not a treble clef!
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
There are three of them. Granted this means that there are different variations of all three. I'll show you the variations as well. This is coming straight from my Math 1060 (Trigonometry) notebook. Sorry there is no key to represent the angle; Theta.1. Sin2 (of Theta) + Cos2 (of Theta)= 1Variations: Sin2 (of Theta) = 1- Cos2 (of Theta)AND: Cos2 (of Theta) = 1-Sin2 (of Theta)2. Tan2 (of Theta) + 1 = sec2 (of Theta)Variations: Tan2 (of Theta) = Sec2 (of Theta) -13. 1 + Cot2 (of Theta) = Csc2 (of Theta)Variations: Cot2 (of Theta) = Csc2 (of Theta) -1
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
Tan theta is a function of the number theta.