The following questions will prepare you for the midterm exam on February 2nd. 1. Let U and V be subspaces of Rn . Define U + V = {x ∈ Rn | x = u + v for some u ∈ U, v ∈ V }, and for a subspace U of Rn let U ⊥ denote the orthogonal complement of U . Show that (U + V )⊥ = U ⊥ ∩ V ⊥ . (Hint: To prove equality of two sets A and B it is usually convenient to show that A contains B and B contains A). Solution: First we show that (U + V )⊥ ⊆ U ⊥ ∩ V ⊥ . Let x ∈ (U + V )⊥ . Since U ⊂ U + V and V ⊂ U + V we have x · u = x · v = 0 for all u ∈ U , v ∈ V . Therefore x ∈ U ⊥ ∩ V ⊥ . Now we show that U ⊥ ∩ V ⊥ ⊆ (U + V )⊥ . Let x ∈ U ⊥ ∩ V ⊥ , then x · u = x · v = 0 for all u ∈ U , v ∈ V . Now let y ∈ U + V . Then y = u + v for some u ∈ U , v ∈ V . Therefore x · y = x · (u + v) = x · u + x · v = 0 and x ∈ (U + V )⊥ . 2. For a set X ⊂ Rn , the boundary of X, denoted ∂X, is the set of points in p ∈ Rn such that every ball around p contains points in X and points not in X. Give a detailed proof from first principles that for any X ⊂ Rn , the set X ∪ ∂X is closed. Solution: To say that X ∪ ∂X is closed is to say that its complement is open. Consider y ∈ Rn with y 6∈ X and y 6∈ ∂X. Since y 6∈ ∂X, by the definition of ∂X some ball around y either does not intersect X or is contained in X. The latter is impossible since y 6∈ X. Thus there is a ball around y that does not intersect X. Since this ball is open, if it interesected ∂X, it would also intesect X: say the ball is B(y, δ) and z ∈ ∂X ∩ B(y, δ). Then a small enough ball around z is contained B(y, δ) and intersects X. But we already showed that B(y, δ) doesn’t intersect X, so it also does not intersect ∂X. Summing up, every point in the complement of X ∪ ∂X has a neighborhood contained entirely in the complement of X ∪ ∂X, so the complement is open and X ∪ ∂X is closed. [You can also make a proof with sequences.] 3. Find the tangent plane to the graph of f (x, y) = 2xey −yex at the point (x, y, z) = (1, 1, e). Solution: The derivative of f is
Df = 2ey − yex 2xey − ex , and Df(1,1) = e e




so the tangent plane has equation (z − e) = e(x − 1) + e(y − 1). [You can also find the equation by treating the graph as a level set of the function z − f (x, y) = z − 2xey + yex .] 1

4. Let f : R2 → R be the function: ( f (x, y) =

x2 y+xy 2 x2 +y 2

if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0)

.

Compute directly from the definition (i.e., as a limit) the directional derivative at f at (x, y) = (0, 0) in the direction of v = (1, 1). Solution: By definition, f (h, h) − f (0, 0) 2h3 = lim 2 = 0. h→0 h→0 2h h 2 5. Consider the path traced out by f (t) = (t, t ), i.e., a parabola. Find the speed and unit tangent vector at each point. Write an expression for the arc length of the traced curve (Hint: arclength=distance travelled). Dfv (0, 0) = lim

Solution: The unnormalized tangent vector (velocity vector) is given by the derivative:   1 0 f (t) = . 2t The speed is the length of this vector, namely (1 + 4t2 )1/2 , and the unit tangent vector is   f 0 (t) (1 + 4t2 )−1/2 = . T(t) = 2t(1 + 4t2 )−1/2 ||f 0 (t)|| The arc length is the integral of the speed: Z s s(t) = (1 + 4t2 )1/2 dt. 0 2

6. Let f : R → R be a differentiable function satisfying f (1, 1) = 1, ∂f /∂x(1, 1) = 3, and ∂f /∂y(1, 1) = −4. a) In what direction should you move away from (1, 1) to make f decrease as rapidly as possible? b) What is the rate of decrease? Say precisely what you mean. Solution: a) The data tells us that the gradient of f at (1, 1) is   3 ∇f (1, 1) = −4   −3 so the greatest rate of decrease is in the direction = 5v where v is the unit vector 4   −3/5 . 4/5 b) Again from the gradient, we deduce that the rate of decrease is 5 in the direction of v. In other words, the directional derivative Dfv = −5. 2

7. You are a chemist at a Very Important Public University in Atlanta, GA and you are trying to find a formula for the concentration C of a certain chemical in terms of two variable x and y. You are certain that C(x, y) is a linear function, but it is hard to measure the constants because you can’t control x and y independently in the lab. However, you can control the pair of them via another variable t. In other words, x = f (t) and y = g(t) for some functions f and g and you can control t. You collect the following data: df dg dC = −1 =1 =1 dt t=1 dt t=1 dt t=1 df dg dC =7 =2 = −1 dt t=2 dt t=2 dt t=2 C|x=1,y=1 = 0 Using this data and the chain rule, find the linear function C. [Here, linear means in the sense of calculus, i.e., a polynomial of degree at most 1 in x and y. It need not satisfy C(0, 0) = 0] Your rival at the Completely Insignificant College in Athens, GA claims that C(0, 0) = −1. Does this agree with your data? Solution: The fact that C is linear and C(1, 1) = 0 means C has the form C(x, y) = a(x − 1) + b(y − 1) for some constants a and b. Using the chain rule, dC df df =a +b dt dt dt so the measurements at t = 1 and t = 2 tell us that a + b = −1 and 2a − b = 7. Solving for a and b yields a = 2 and b = −3. Thus C(x, y) = 2(x − 1) − 3(y − 1). In particular, C(0, 0) = 1 and it appears that your rival has made a sign error.

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