Solutions to Midterm 2 | Calculus III | MATH 226, Exams of Mathematics

Download Solutions to Midterm 2 | Calculus III | MATH 226 and more Exams Mathematics in PDF only on Docsity! Solutions to Midterm #2 Math 226, Fall 1996 Given Monday, 21 October 1996 Problem 1. If h = (h1, h2), show that h1h2 is o(h) as h→ 0. Solution. I did this example in class. I gave ten points if you wrote “it’s of higher degree than one.” That’s the right philosphy, but a little vague, unless you explained further. You were being asked to show that h1h2 ‖h‖ → 0 as h→ 0, where of course ‖(h1, h2)‖ = √ h21 + h22. In class I worked this by noting that |h1h2| ≤ 12(h 2 1 + h22) = 1 2 ‖h‖2, thus h1h2/‖h‖ is bounded (in absolute value) by 1/2‖h‖, which goes to zero. But you don’t need any fancy inequalities to do this. You just need to use the fairly obvious facts that |h1|√ h21 + h22 ≤ 1, |h2|√ h21 + h22 ≤ 1. This says that if h → 0, then h1 → 0 and h2 → 0 at comparable rates. In fact, we explicitly have the estimates ‖h1‖ ≤ ‖h‖, |h2| ≤ ‖h‖. Thus anytime we have a factor of h1 or h2 in an expression, we pick up an extra factor of ‖h‖ on the right-hand side of the inequalities. If there are two or more, and we then divide by ‖h‖, we’re left with some power of ‖h‖ on the right-hand-side: |h1h2| ‖h‖ ≤ ‖h‖. And as h→ 0 — which means exactly that ‖h‖ → 0 — we see that h1h2 ‖h‖ → 0 as h→ 0. When h = (h1, h2, . . . , hn), Each hi is of the same order of magnitude as ‖h‖. Typeset byAMS-TEX 2 Problem 2. Show that u = 1√ x2 +y2 + z2 satisfies Laplace’s equation for three variables: ∂2u ∂x2 + ∂ 2u ∂y2 + ∂ 2u ∂z2 = 0. Solution. There are ugly ways, and there are pretty ways, to do this. The ugliest is to differentiate u with respect to x, then to differentiate that with respect to x; do the same for y ; do the same for z; and add. A slightly better way is to differentiate u with respect to x twice, then use the symmetry of u in x, y and z to write down the other two formulas for ∂2u ∂y2 and ∂2u ∂z2 , and add. Most people did the problem this way, and under the time-pressure of an exam that’s reasonable. But the best way is to introduce a variable r by r = √ x2 +y2 + z2, to note that r is differentiable as a function of x (or y , or z) at every point except (0,0,0); and that r2 = x2 +y2 + z2, hence 2r ∂r ∂x = 2x, i.e. ∂r ∂x = x r , with similar formulas for ∂r/∂y and ∂r/∂z. Then u = 1/r , so ∂u ∂x = − 1 r 2 x r = − x r 3 . This is the chain rule. Since u is a function of r , which is a function of x,y, z, the only dependency path from u to x is u→ r → x, and so ∂u ∂x = du dr ∂r ∂x . Anyway, we therefore get ∂2u ∂x2 = ∂ ∂x ( −xr−3 ) = −r−3 + (−x)∂r −3 ∂x = −r−3 + 3r−4 ∂r ∂x = −r−3 + 3x 2 r 5 . 5 where the last line was obtained by substituting for r′(t) using the defining equation r′(t) = −∇v(r(t)). Since∇u(x,y)·∇v(x,y) = 0, this means q′(t) ≡ 0, which means q is a constant function. That is, r(t) remains on the same level curve of u. To illustrate this, we have drawn the level curves for the two functions u(x,y) = x4 − 6x2y2 +y4, v(x,y) = 4x3y − 4xy3 on the same graph. (These functions satisfy the Cauchy-Riemann equations, since they are the real and imaginary parts of (x+ iy)4.) The level curves of u are always orthogonal to the level curves of v : -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 Problem 6. Suppose f is a differentiable function of two variables (x,y). Show that the directional derivative of f in the direction of (−∂f/∂y, ∂f/∂x) is zero. Solution. This is trivial. The gradient of ϕ is ( ∂ϕ ∂x , ∂ϕ ∂y ), and the dot product of this with the given direction is ( ∂ϕ ∂x , ∂ϕ ∂y ) · (−∂f/∂y, ∂f/∂x) = 0. There is the technical problem that when the book defines directional derivative in the direction of u, it means that u should be a unit vector; so we should divide the vector (−∂ϕ/∂y, ∂ϕ/∂x) by its length. We’ll still get zero for the dot product. 6 The purpose of this problem is to indicate a possible strategy for drawing the level curves of a function, ϕ(x,y) = c. Namely, we take a curve r(t) = (x(t),y(t)) so that its derivative is pointing in the direction of (−∂ϕ ∂y , ∂ϕ ∂x ): x′(t) = −∂ϕ ∂y (x(t),y(t)) y ′(t) = ∂ϕ ∂x (x(t),y(t)). This is a system of differential equations (there are two unknown functions, x(t) andy(t), and two equations). It will usually have to be solved numerically. This problem shows that the derivative of ϕ(x(t),y(t)) will remain zero, i.e., ϕ(x(t),y(t)) will remain constant, i.e. r(t) will remain on the same level curve. This is a very fast way of drawing level curves, if they’re smooth.