DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§2. Calculus Review 109EXERCISES A.1♯ *1.Suppose {v 1 , v 2 } gives a basis for R 2 . Given vectors x, y ∈ R 2 , prove that x = y if and only ifx · v i = y · v i , i =1, 2.*2. The geometric-arithmetic mean inequality states that√ab ≤a + b2for positive numbers a and b,with equality holding if and only if a = b. Give a one-line proof using the Cauchy-Schwarzinequality:|u · v| ≤‖u‖‖v‖ for vectors u and v ∈ R n ,with equality holding if and only if one is a scalar multiple of the other.3. Let w, x, y, z ∈ R 3 . Prove that(w × x) · (y × z) =(w · y)(x · z) − (w · z)(x · y).(Hint: Both sides are linear in each of the four variables, so it suffices to check the result onbasis vectors.)♯ 4. Suppose A(t) isadifferentiable family of 3 × 3 orthogonal matrices. Prove that A(t) −1 A ′ (t) isalways skew-symmetric.[ ] a b5. If A = is a symmetric 2 × 2 matrix, set f(x) =Ax · x and check that ∇f(x) =2Ax.b c2. Calculus ReviewRecall that a function f : U → R defined on an open subset U ⊂ R n is C k (k =0, 1, 2,...,∞)if all its partial derivatives of order ≤ k exist and are continuous on U. We will use the notation∂f∂u and f ∂ 2 fu interchangeably, and similarly with higher order derivatives:∂v∂u = ∂ ( ∂f)is the∂v ∂usame as f uv , and so on.One of the extremely important results for differential geometry is the followingTheorem 2.1. If f is a C 2 function, then∂2 f∂u∂v = ∂2 f∂v∂u (or f uv = f vu ).The same results apply to vector-valued functions, working with component functions separately.If f : U → R is C 1 we can form its gradient by taking the vector ∇f = ( )f x1 ,f x2 ,...,f xn of itspartial derivatives. One of the most fundamental formulas in differential calculus is the chain rule:Theorem 2.2. Suppose f : R n → R and α: R → R n are differentiable. Then (f ◦α) ′ (t) =∇f(α(t)) · α ′ (t).