MATH 260 Homework assignment 8 March 21, px 2yq dy. (c) Where C is the part of the line y x, parametrized any way you like.

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1 MATH 26 Homework assignment 8 March 2, 23. Evaluate the line integral x 2 y dx px 2yq dy (a) Where is the part of the parabola y x 2 from p, q to p, q, parametrized as x t, y t 2 for t. (b) Where is the part of the parabola y x 2 from p, q to p, q, parametrized as x sin t, y sin 2 t for t π{2. (c) Where is the part of the line y x, parametrized any way you like. (d) Why should you have expected the answers to (a) and (b) to be the same, but (c) to be different? (a) If x t and y t 2 then dx dt and dy 2t dt so the integral becomes: x 2 y dx px 2yq dy t 4 dt pt 2t 2 q2t dt t 4 2t 2 4t 3 dt t5 5 2t 3 3 t (b) If x sin t and y sin 2 t, then t goes from to π 2, and we have dx cos t dt and dy 2 sin t cos t dt. The integral becomes: x 2 y dx px 2yq dy sin 4 t cos t dt psin t 2 sin 2 tq2 sin t cos t dt using the substitution u sin t. u 4 u5 5 psin 4 t 2 sin 2 t 4 sin 3 tq cos t dt 2u 2 4u 3 du 2u 3 3 u , (c) The simplest way to parametrize the line is x t, y t for t, so dx dt and dy dt. So we get x 2 y dx px 2yq dy t 3 dt pt 2tq dt t 3 t dt t4 4 t (d) The answers to (a) and (b) are the same because of independence of parametrization (the same line integral parametrized two different ways), but (c) is a different curve, and the vector field rx 2 y, s 2ys is not conservative.

2 2 2. alculate F dr where (a) F yi 2xj and is the circle x 2 y 2, traversed counterclockwise starting from p, q, once around. (b) F yi 2xj and is the circle x 2 y 2 4, traversed counterclockwise starting from p, q, once around. (c) F xi 2yj 2xk and is the part of the semicubical parabola given by x t, y t 2, z t 3 starting from p,, q and ending at p,, q. (a) The simplest parametrization would be x cos t, y sin t for t 2π. Then dx sin t dt and dy cos t dt and so 2π 2π F dr y dx 2x dy sin 2 t 2 cos 2 3 t dt cos 2t dt π, 2 2 using the identities cos 2 t 2 2 cos 2t and sin2 t 2 2 sin 2t. (b) This time we let x 2 cos t and y 2 sin t for t 2π, so dx 2 sin t dt and dy 2 cos t dt and so 2π 2π F dr y dx 2x dy 4 sin 2 t 8 cos 2 t dt 2 6 cos 2t dt 4π. The scaling between this part and part (a) makes sense, since F scales linearly and dr also scales linearly so their dot product scales quadratically (so when we make the circle twice as big, the integral gets multiplied by four). (c) We re given a reasonable parametrization so we use it: x t, y t 2 and z t 3 for t, so dx dt, dy 2t dt and dz 3t 2 dt and so t F dr x dx 2y dy 2x dz t 4t 3 6t 3 2 5t 4 dt Let F be a vector field on R 2 or R 3 whose coefficients are differentiable functions, and let be a smooth curve. Prove that F dr ML, where M is a number such that }Fpxq} M for all x, and L the length of. We can parametrize however we like, so we choose to parametrize by arc length s for a s b. Then dr T ds where T is the unit tangent vector field to and ds is the element of arc length (remember?). Then F dr ds F T }Fpsq}}T} M,

3 3 since }Fpsq} M for all s. But then b F dr F T ds a b F T ds M b a a ds ML. 4. If F xi yj, compute (a) The work done by F on a particle that moves around the circle x 2 y 2 (b) The outward flux of F through the circle x 2 y 2. (c) Is F a conservative field? If so, find its potential function, i.e., the function ϕ such that ϕ F. (a) For practice, we do this directly: Let x cos t and y sin t for t 2π, then dx sin t dt and dy cos t dt and we get ¾ ¾ 2π Work F dr x dx y dy cos t sin t sin t cos t dt. (b) To calculate flux, we first note that the outward pointing normal vector to the curve is xi (calculate this as n ds i j dt) and, using the parametrization from part (a) we get dx dt dy dt ¾ ¾ Flux F nds x 2 y 2 ds length of 2π. yj (c) F is a conservative field, since F 2 px2 y 2 q. We could have used this fact to conclude that the integral in part (a) is zero without computing it. Oh, Wronski! (Read the trig.alt notes before attempting these problems.) 5. Suppose the two functions y pxq and y 2 pxq are both solutions of the ODE y 2 ppxqy qpxqy. p q Recall from class that the Wronskian of y and y 2 is defined to be the function: W pxq y pxq y 2 pxq y pxq y 2 pxq y pxqy2 pxq y 2pxqy pxq. (a) Show that W satisfies the first-order equation W pw. (b) Solve this equation, and conclude that for any two solutions y and y 2 of (*), either W pxq is zero for all values of x or else W pxq is never zero.

4 4 (a) The derivative of W is W y y2 2 y2 y 2, so W pw y py2 2 py 2 q y 2py 2 py q. But because y and y 2 are both solutions of (*), we have y 2 py qy and y2 2 py 2 qy 2. Therefore W pw y p qy 2 q y 2 p qy q. (b) The equation W pw is both a separable and a linear equation for W. If we solve it as a linear equation, we learn that W pxq e ³ p. Since the exponential factor is non-zero for all x, we conclude that if then W pxq is zero for all x and if then W pxq for all x. 6. Suppose ypxq is a solution of equation (*) from the previous problem. Find a function vpxq so that the function zpxq vpxqypxq is a solution of an equation of the form z 2 Qpxqz p q i.e., z satisfies an equation without a first-order term. (Hint: If you do it right, you ll have Qpxq qpxq 4 ppxq2 2 p pxq.) This shows that in order to prove general facts about homogeneous, linear, second-order ODEs, we only need to consider equations of the form (**), since any equation of the form (*) can be converted to (**) by this trick. If zpxq vpxqypxq then z v y vy and z 2 v 2 y 2v y vy 2. Therefore z 2 Qpzqz v 2 y 2v y vy 2 Qvy vy 2 2v y pv 2 Qvqy. So y will satisfy the equation y 2 2v v v y 2 Qv y. v We therefore need to choose Q and v so that 2v {v p and pv 2 Qvq{v q. If we view 2v {v p as a differential equation for v (since p is a known function), we can solve it as a separable equation and learn that 2 ln v ³ p or v e ³ 2 p. Since we already know that v 2 vp we have that v2 2 pv p vp q 2 p 2 vp2 vp q 2 vp 2 p2 p q. To find Q, we solve the equation pv 2 Qvq{v q algebraically and get Q q v 2 {v q 4 p2 2 p, using the expression for v 2 from the preceding paragraph. Therefore, y is a solution of y 2 py qy if (and only if) z vy is a solution of z 2 Qz for Q q 4 p2 2 p, which is what we were trying to show. 7. Now let z be a non-trivial (i.e., not identically zero) solution of (**), where Qpxq for all x. Show (by basic calculus arguments) that there is at most one value of x for which zpxq. If there are no values of x for which zpxq then we are already done. If there is one, say x x, then note that any solution of (**) with zpx q is a constant multiple of the specific one

5 5 that satisfies the initial conditions zpx q, z px q (by the uniqueness theorem for initial-value problems). So we need only prove that the solution of the problem z 2 Qpxqz zpx q z px q is never zero for x x. First, consider the case x x. We ll prove that zpxq. We ll work by contradiction: If there were a value of x bigger than x for which zpxq then there must be a point x x with x x x such that z px q (by the mean-value theorem, which guarantees the existence of such an x satisfying z px q zpxq zpx q x x. Next, since z px q and z px q, the set of points tx x x x and z pxq u is non-empty. This set is also closed and so has a minimum. Let x 2 be the smallest value of x between x and x where z pxq. Therefore we know that z pxq for x x x 2. We can conclude that zpxq for x x x 2 (again by the mean-value theorem) and then the differential equation tells us that z 2 pxq Qpxqzpxq for x x x 2 since Qpxq and zpxq. However, the mean-value theorem also tells us that there would be a point x 3 with x x 3 x 2 with z 2 px 3 q z px 2 q z px q x 2 x x 2 x which contradicts our earlier conclusion that z 2 pxq for all x x x 2. Therefore our assumption was faulty and we have proved that there is no point x x where zpxq. The proof for the case where x x is similar. 8. What happens if Qpxq? More on this next week, but you can see that the situation might be complicated from the following example: Solve the (auchy-euler) equation z 2 k x 2 z and show that every nontrivial solution of has an infinite number of zeroes (like sines and cosines do) if k {4, but only a finite number (how many are possible?) if k {4. What if k {4? Since the coefficient k{x 2 in this problem becomes singular for x, we consider solutions defined only for x. and d 2 z dx 2 To solve auchy-euler equations, let x e u, so u ln x. By the chain rule: d dx dz dx d dx x dz du x 2 dz du dz dz du dz dx du dx x du d x du dz du dz du dx x 2 du x d 2 z d 2 z dz du 2 x x 2 du 2 du

6 6 Substitute these into the equation to get z 2 k x 2 z x 2 d 2 z du 2 dz du k x 2 z d 2 z dz x 2 du 2 du kz. Therefore zpuq satisfies the constant-coefficient equation d 2 z dz du 2 du kz. The characteristic equation for this is r 2 r there are three cases: k, which has solutions r 2 p? 4kq and If k {4 then the the two values of r are real and the solution is For x this is z c e 2 p? 4kqu c 2 e 2 p? 4kqu c x 2 p? 4kq c2 x 2 p? 4kq. whose only positive zero is z x 2 p? 4kq pc x? 4k x? {p 4kq c2 if c and c 2 have opposite signs, and otherwise there are no positive zeros. If k {4 then r {2 is a double root of the characteristic equation and we have c z c e 2 u c 2 ue 2 u c? x c2? x ln x? xpc c 2 ln xq whose single positive zero occurs at x e c{c2. If k {4 then 4k is negative so the roots of the characteristic equation are complex: r 2 p a p4k qq 2 p i? 4k q. So the solution is?4k z c e 2 u cos u c 2 q?4k c 2 e 2 u sin u??4k x c cos ln x?4k c 2 sin ln x Since the u-version of the function has infinitely many zeros, spaced π{? 4k apart, the x version has infinitely many positive zeros at the exponentials of the u zeros.