This preview shows page 1. Sign up to view the full content.
Unformatted text preview: WEEK 6 HOMEWORK SOLUTIONS 3 We will calculate the rate that the volume of the sweater is changing by using the total diﬀerential of the volume. The volume is given by V (r, h) = π r2 h The total diﬀerential of the volume is then ∂V ∂V dV = dr + dh ∂r ∂h We are given information about the circumference but it will be more useful for us to work in terms of the radius. So, we use C = 2π r to get that C =r 2π dC =⇒ = dr 2π Now, we can compute the requisite partial derivatives: ∂V ∂r ∂V ∂h = 2π rh = π r2 We are given that at t = t0 , h = 3in and C = 18in =⇒ r = 18 in. The circumference π is shrinking at a rate of 0.2in/min and the height is increasing at a rate of 0.1in/min. this means that the corresponding total diﬀerentials are dC = −0.2in/min and dh = 0.1 in/min. Putting this all together we get that dV = 27 3 in /min. 10π 4 a. Don’t worry about the absolute value too much. The labeling of particle 1 and 2 are arbitrary so at any give time we can choose the labeling such that x1 > x2 and
1 2 WEEK 6 HOMEWORK SOLUTIONS we can drop the absolute values. This ends up being equivalent to getting a ±1 if you were to evaluate the derivative. Now, the Lagrangian is L ( x 1 , x2 , x 1 , x 2 ) = ˙˙ m1 2 m2 2 k x1 + ˙ x2 − ˙ 2 2 | x1 − x2 | So, assuming mass is not changing in time, d ∂L dt ∂ x1 ˙ d ∂L dt ∂ x2 ˙ ∂L ∂ x1 ∂L ∂ x2 = Now, the Euler-Lagrange equations give us that d m1 x1 = m1 x1 ˙ ¨ dt d = m2 x2 = m2 x2 ˙ ¨ dt ∂V −k =− =− ∂ x1 | x1 − x2 | 2 ∂V −k =− =− ∂ x2 | x1 − x2 | 2 d ∂L ∂L = dt ∂ xj ˙ ∂ xj and we have found that d ∂L dt ∂ xj ˙ = Fj , we must have that Fj = ∂V k = ∂ xj | x1 − x2 | 2 b. We want to show that 2 d dt
j =1 xj ( t ) ˙ ∂L − L(x1 (t), x2 (t), x1 (t), x2 (t)) ˙ ˙ ∂ xj ˙ Let us deal with the time derivative of the terms in the summation ﬁrst. Using this product rule: 2 2 d ∂L ∂L d ∂L xj ( t ) ˙ = xj ( t ) ¨ + xj ( t ) ˙ dt j =1 ∂ xj ˙ ∂ xj ˙ dt ∂ xj ˙ j =1
d ∂L ∂L By the Euler-Lagrange equations we know that dt ∂ xj = ∂ xj , so our expression can ˙ be written as: 2 2 d ∂L ∂L ∂L xj ( t ) ˙ = xj ( t ) ¨ + xj ( t ) ˙ . dt j =1 ∂ xj ˙ ∂ xj ˙ ∂ xj j =1 WEEK 6 HOMEWORK SOLUTIONS 3 To compute the time derivative of the Lagrangian we need to use the chain rule because it does not explicitly depend on time, just on variables that do. So, we will get 4 terms out of this total derivative1. d ∂L ∂L ∂L ∂L L(x1 (t), x2 (t), x1 (t), x2 (t)) = ˙ ˙ x1 + ˙ x2 + ˙ x1 + ¨ x2 ¨ dt ∂ x1 ∂ x2 ∂ x1 ˙ ∂ x2 ˙ Now, we can subtract these two expressions: 2 ∂L ∂L ∂L ∂L ∂L ∂L xj ( t ) ¨ + xj ( t ) ˙ − x1 + ˙ x2 + ˙ x1 + ¨ x2 = 0 ¨ ∂ xj ˙ ∂ xj ∂ x1 ∂ x2 ∂ x1 ˙ ∂ x2 ˙ j =1 c. Now, we deﬁne2 the momentum of the ith particle to be pj = mj xj . Now, to show ˙ ∂L ∂L that pj = ∂ xj , we just need to work out what ∂ xj is. We have already compute this. ˙ ˙ ∂L = m1 x1 = p1 ˙ ∂ x1 ˙ ∂L = m2 x2 = p2 ˙ ∂ x2 ˙ Now, to show that pj = ˙
∂L ∂ xj we should compute the time derivative of pj pj = ˙ d ( mj xj ) = mj xj ˙ ¨ dt ∂V ∂L = . ∂ xj ∂ xj From working with the Euler-Lagrange equations, we know that mj xj = ¨
∂L So, we have that pj = ∂ xj . ˙ We deﬁne the Hamiltonian as H = p1 x1 + p2 x2 − ˙ ˙
1The m1 2 m2 2 k x1 + ˙ x2 − ˙ 2 2 | x1 − x2 | chain rule should remind you of the total diﬀerential! is a more general way to construct the momentum. Usually it will work out that it is equivalent to the mv deﬁnition we use here. It is this equality that later lets us interpret the Hamiltionian as the total energy. So, in general it is not true that the Hamiltonian can be intepreted as energy, but 99% of the time it is.
2There 4 WEEK 6 HOMEWORK SOLUTIONS We can eliminate the dependence on xj in this equation by using xj = pj /mj . We ˙ ˙ then get: p2 p2 k 1 H= + 2+ 2 m1 2 m2 | x1 − x2 | = T +V Since dH = 0 and we have determined that the Hamiltonian is the total energy in dt the system, we have shown that energy is conserved. e. To show that Hamilton’s equations of motion are true, let us compute the relevant derivatives of the Hamiltonian. ∂H pj = = xj ˙ ∂ pj mj So, the ﬁrst equations is true. ∂H ∂V − =− = mj xj = pj . ¨ ˙ ∂ xj ∂ xj So, the second of Hamilton’s equations is true. This is the one that gives us back Newton’s second law because it is saying that the change in momentum (mx is mass ¨ is constant) is equal to the force. ...
View Full Document
- Fall '07