Math 107. Rumbos Fall Solutions to Review Problems for Exam 1

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1 Math 07. Rumbos Fall 0 Solutions to Review Problems for Exam. Compute the (shortest) distance from the point P (4, 0, 7) in R 3 to the plane given by 4x y 3z =. Solution: The point P o (3, 0, 0) is in the plane. Let w = P o P = The vector n = is orthogonal to the plane. To find the 3 shortest distance, d, from P to the plane, we compute the norm of the orthogonal projection of w onto n; that is, where d = Pˆn (w), ˆn = 4, 6 3 a unit vector in the direction of n, and Pˆn (w) = (w ˆn)ˆn. It then follows that d = w ˆn, where w ˆn = 6 (4 + ) = 5 6. Hence, d = Compute the (shortest) distance from the point P (4, 0, 7) in R 3 to the line given by the parametric equations x = + 4t, y = 7t, z = t.

2 Math 07. Rumbos Fall 0 Solution: The point P o (, 0, ) is on the line. The vector 4 v = 7 gives the direction of the line. Put w = P o P = The vectors v and w determine a parallelogram whose area is the norm of v times the shortest distance, d, from P to the line determined by v at P o. We then have that from which we get that area(p (v, w)) = v d, d = area(p (v, w)). v On the other hand, area(p (v, w)) = v w, where ˆi ˆj ˆk v w = = 63ˆi + 3ˆj + 35ˆk. Thus, v w = (63) + (3) + (35) = 655 and therefore d = Compute the area of the triangle whose vertices in R 3 are the points (,, 0), (, 0, ) and (0, 3, )

3 Math 07. Rumbos Fall 0 3 Solution: Label the points P o (,, 0), P (, 0, ) and P (0, 3, ) and define the vectors v = P o P = and w = P o P =. The area of the triangle determined by the points P o, P and P is then half of the area of the parallelogram determined by the vectors v and w. Thus, area( P o P P ) = v w, where ˆi ˆj ˆk v w = = 3ˆi ˆj + ˆk. Consequently, area( P o P P ) = = Let v and w be two vectors in R 3, and let λ be a scalar. Show that the area of the parallelogram determined by the vectors v and w + λv is the same as that determined by v and w. Solution: The area of the parallelogram determined by v and w +λv is area(p (v, w + λv)) = v (w + λv), where v (w + λv) = v w + λv v = v w. Consequently, area(p (v, w + λv)) = v w = area(p (v, w)). 5. Let ˆu denote a unit vector in R n and Pˆu (v) denote the orthogonal projection of v along the direction of ˆu for any vector v R n. Use the Cauchy Schwarz inequality to prove that the map v Pˆu (v) for all v R n is a continuous map from R n to R n.

4 Math 07. Rumbos Fall 0 4 Solution: Pˆu (v) = (v ˆu)ˆu for all v R n. Consequently, for any w, v R n, It then follows that Pˆu (w) Pˆu (v) = (w ˆu)ˆu (v ˆu)ˆu = (w ˆu v ˆu)ˆu = [(w v) ˆu]ˆu. Pˆu (w) Pˆu (v) = (w v) ˆu, since ˆu =. Hence, by the Cauchy Schwarz inequality, Pˆu (w) Pˆu (v) w v. Applying the Squeeze Theorem we then get that lim Pˆu(w) Pˆu (v) = 0, w v 0 which shows that Pˆu is continuous at every v V. 6. Define f : R R by x y if (x, y) = (0, 0) x + y f(x, y) = 0 if (x, y) = (0, 0). Prove that f is continuous at (0, 0). Solution: For (x, y) = (0, 0) f(x, y) = x y x + y y, since x x + y for all (x, y) R. We then have that, for (x, y) = (0, 0), f(x, y) x + y, which implies that 0 f(x, y) f(0, 0) (x, y) (0, 0),

5 Math 07. Rumbos Fall 0 5 for (x, y) = (0, 0). Thus, by the Squeeze Theorem, lim f(x, y) f(0, 0) = 0, (x,y) (0,0) 0 which shows that f is continuous at (0, 0). 7. Show that is not continuous at (0, 0). x y, (x, y) = (0, 0) f(x, y) = x + y 0, (x, y) = (0, 0) Solution: Let σ (t) = (t, t) for all t R and observe that and It then follows that lim σ (t) = (0, 0) t 0 f(σ(t)) = 0, for all t = 0. lim f(σ (t)) = 0. t 0 Thus, if f were continuous at (0, 0), we would have that f(0, 0) = 0. () On the other hand, if we let σ (t) = (t, 0), we would have that and lim σ (t) = (0, 0) t 0 f(σ(t)) =, for all t = 0. Thus, if f were continuous at (0, 0), we would have that f(0, 0) =, which is in contradiction with (). This contradiction shows that f is not continuous at (0, 0).

6 Math 07. Rumbos Fall Determine the value of L that would make the function ( ) x sin if y = 0; y f(x, y) = L otherwise, continuous at (0, 0). Is f : R R continuous on R? Justify your answer. Solution: Observe that, for y = 0, ( ) f(x, y) = x sin y = x ( ) sin y x x + y. It then follows that, for y = 0, 0 f(x, y) (x, y). Consequently, by the Squeeze Theorem, lim f(x, y) = 0. (x,y) 0 This suggests that we define L = 0. If this is the case, lim f(x, y) f(0, 0) = 0, (x,y) 0 which shows that f is continuous at (0, 0) if L = 0. Next, assume now that L = 0 in the definition of f. Then, for any a = 0, f fails for be continuous at (a, 0). To see why this is case, note that for any y = 0 and the limit of sin ( ) y f(a, y) = a sin ( ) y as y 0 does not exist.

7 Math 07. Rumbos Fall Let σ : R R be the path given by (a) Sketch the image of σ. σ(t) = ( cos t, sin t), for t R. Solution: The image of σ is the set, σ(r) = {(x, y) R x = cos t, y = sin t, for t R}, of points, (x, y), in R such that x = cos t and y = sin t, () for t R. It follows from the equations in (3) that (x, y) σ(r) if and only if x 4 + y =, which shows that σ(r) is the ellipse pictured in Figure y x Figure : Sketch of ellipse (b) Find a tangent vector to the path at t = π/4. Solution: Compute the derivative of the path σ (t) = ( sin t, cos t), for all t R. Then, a tangent vector to the path at t = π/4 is ( v = σ (π/4) = ),. (3)

8 Math 07. Rumbos Fall 0 8 (c) Give the parametric equations to the tangent line to the path at t = π/4. Sketch the line. Solution: The point on the path at time t = π/4 has coordinates ( ), σ(π/4) =. The line through this point in the direction of the vector v in (3) has parametric equations x = ( t π ) ; 4 y = ( + t π ), 4 for t R. Sketch of the tangent line to the path σ and σ(π/4) is shown in Figure Figure y x Figure : Tangent Line at σ(π/4) 0. Let I denote an open interval, and σ : I R n and γ : I R n be differentiable paths on I. Define h(t) = σ(t) γ(t) for all t R. Show that h: I R is differentiable on I and verify that h (t) = σ (t) γ(t) + σ(t) γ (t), for all t I. (4) Solution: Write σ(t) = (x (t), x (t),..., x n (t) and γ(t) = (y (t), y (t),..., y n (t), for all t I, where x i : I R and y i : I R are differentiable on I,

9 Math 07. Rumbos Fall 0 9 for i =,,..., n. Then, h(t) = σ(t) γ(t) = n x i (t)y i (t), for all t I. (5) i= It follows from (5) and the product rule in Calculus I that h is differentiable and h (t) = n [x i(t)y i (t) + x i (t)y i(t)] i= = n x i(t)y i (t) + n x i (t)y i(t) (6) i= i= = σ (t) γ(t) + σ(t) γ (t), for all t I. This establishes (4).. Let I denote an open interval, and σ : I R n be a differentiable path satisfying σ(t) = c, a constant, for all t I. Show that, at any t I, σ(t) is orthogonal to a tangent vector to the path at that t. Solution: Put h(t) = σ(t) = σ(t) σ(t) for all t I. Thus, if σ(t) = c for all t I, it follows that h(t) = c, for all t I. (7) Differentiating on both sides of (7) and using the result of Problem 0in (4) we obtain from (7) that σ (t) σ(t) = 0. for all t I from which we get that σ (t) σ(t) = 0. for all t I; in other words, σ(t) is orthogonal to a tangent vector to the path, σ (t), at any t I.

10 Math 07. Rumbos Fall 0 0. Let σ : R R be given by σ(t) = (t /3, t) for all t R. Show that σ is not differentiable at 0. Solution: For h = 0, compute so that and therefore h [σ(h) σ(0)] = (h /3, ), [σ(h) σ(0)] h = + h 4/3, lim h 0 [σ(h) σ(0)] h = +, which shows that lim [σ(h) σ(0)] does not exist; hence, σ is not h differentiable at 0. h 0