Math 211, Sections 02,04, Solutions to Practice Problems 2

Math 211, Sections 02,04, Fall 2019 Solutions to Practice Problems 2 1. Each of the following limits converges. Compute them. ey + x sin y x4 − 3y 2 x4 + 2x2 + x2 y 2 + 2y 2 lim lim lim x2 + y 2 (x,y)→(0,0) ln(1 + x) + cos2 (xy) (x,y)→(1,2) x2 + y 2 (x,y)→(0,0) ey + x sin y e0 + 0 1 Answer. (a): lim = = = 1, by limit laws. 2 2 ln(1 + x) + cos (xy) ln(1) + cos (0) 1 (x,y)→(0,0) 1−3·4 11 x4 − 3y 2 = = − , by limit laws. (b): lim 2 2 1+4 5 (x,y)→(1,2) x + y 4 2 2 2 2 x + 2x + x y + 2y (x2 + y 2 )(x2 + 2) (c): lim = lim = lim x2 + 2 = 2, x2 + y 2 x2 + y 2 (x,y)→(0,0) (x,y)→(0,0) (x,y)→(0,0) by limit laws. 2. Prove that each of the following limits diverges. 2x2 − y 2 x2 y + y 3 + xy 4x2 y 3 + 3xy 4 lim lim lim x2 + y 2 x8 + 5y 4 (x,y)→(0,0) x2 + 2y 2 (x,y)→(0,0) (x,y)→(0,0) Answer. (a): 1 −y 2 2x2 − y 2 , along the line x = 0 we have lim = − , but along y = 0 we have For lim 2 2 2 y→0 x + 2y 2y 2 (x,y)→(0,0) 2 2x 1 lim = 2 6= − , so the limit diverges. x→0 x2 2 x2 y + y 3 + xy 0 (b): For lim , along the line y = 0 we have lim 2 = 0, but along y = x we have 2 2 x→0 x +y x (x,y)→(0,0) 1 1 2x3 + x2 = lim x + = 6= 0, so the limit diverges. lim x→0 x→0 2x2 2 2 4x2 y 3 + 3xy 4 0 (c): For lim , along y = 0 we have lim 8 = 0, but along y = x2 we have 8 4 x→0 x x + 5y (x,y)→(0,0) 4 4 + 3x 4x8 + 3x9 = lim = 6= 0, so the limit diverges. lim x→0 x→0 6x8 6 6  3 3  x + 2y if (x, y) 6= (0, 0) 3. Let f (x, y) = x2 + y 2  0 if (x, y) = (0, 0). (a). Compute fx (0, 0) and fy (0, 0). (b). Compute D~u f (0, 0), where ~u is the unit vector in the direction of ~v = h1, 1i. (c). Is f differentiable at (0, 0)? Why or why not? h3 −0 2 f (h, 0) − f (0, 0) Answer. (a): fx (0, 0) = lim = lim h = lim 1 = 1, and h→0 h→0 h→0 h h 2h3 −0 2 f (0, h) − f (0, 0) fy (0, 0) = lim = lim h = lim 2 = 2. h→0 h→0 h→0 h h 1 (b): The unit vector ~u in the direction of ~v = h1, 1i is √ h1, 1i, and so √2 3 /(2 2) 3h √ √ −0 f (h/ 2, h/ 2) − f (0, 0) 3 3 2h2 /2 D~u f (0, 0) = lim = lim = lim √ = √ . h→0 h→0 h→0 2 2 h h 2 2 (c): No, f is not differentiable √ √at (0, 0).√If it were, D~u f (0, 0) would equal √ ∇f√(0, 0) · ~u = h1, 2i · h1/ 2, 1/ 2i = 3/ 2. However, we saw in part (b) that D~u f (0, 0) = 3/(2 2) 6= 3/ 2. So f cannot be differentiable at (0, 0). 4. State, but do not prove, the ε-δ definition of each of the following limit statements. p 2x2 − y = 4 lim Q(x, y, z) = R lim H(x, y) = −3 lim (x,y,z)→(0,−2,1) (x,y)→(4,−5) (x,y)→(3,2) p Answers. (a): ∀ε > 0, ∃δ > 0 such that ∀(x, y) such that 0 < (x − 4)2 + (y + 5)2 < δ, we have |H(x, y) + 3| < ε. p (x − 3)2 + (y − 2)2 < δ, we have (b): ∀ε > 0, ∃δ > 0 such that ∀(x, y) such that 0 < p 2 | 2x − y − 4| < ε. p (c): ∀ε > 0, ∃δ > 0 such that ∀(x, y, z) such that 0 < x2 + (y + 2)2 + (z − 1)2 < δ, we have |Q(x, y, z) − R| < ε. 20 + sin(x + 4y) . 10 + 3y (a). Find an equation for the tangent plane to the surface z = f (x, y) at the point where (x, y) = (0, 0). (b). Use your tangent plane to approximate the value of f (−0.3, 0.1). 5. Let f (x, y) = Answer. (a): We compute f (0, 0) = (20 + 0)/10 = 2. Also, fx (x, y) = cos(x + 4y) , 10 + 3y fy (x, y) = so fx (0, 0) = cos(0) = 0.1, 10 and 4 cos(x + 4y)(10 + 3y) − 3(20 + sin(x + 4y)) , (10 + 3y)2 so fy (0, 0) = 4(1)(10) − 3(20) 20 =− = −0.2. 102 100 Thus, the tangent plane has equation z = 2 + 0.1x − 0.2y. (b): Plugging x = −0.3 and y = 0.1 into the tangent plane equation yields f (−0.3, 0.1) ≈ 2.1 + (0.1)(−0.3) − (0.2)(0.1) = 2 − 0.03 − 0.02 = 1.95. [Side note: the actual value of f (−0.3, 0.1) is about 1.9514401. . . , so this is a good approximation!] 4y . x Is g differentiable at (0, 3)? Why or why not? Is g differentiable at (2, 3)? Why or why not? Compute D~u f (2, 3), where ~u is the unit vector in the direction of ~v = h3, −4i. At (2, 3), in what direction does g increase the fastest? What is the rate of increase of g in the direction from part (d)? 6. Let g(x, y) = x3 y − 2x2 + (a). (b). (c). (d). (e). Answer. (a): g is not differentiable at (0, 3), because g(0, 3) is not even defined. (b). g is differentiable at (2, 3), because g is a rational function which is defined at (2, 3). [Or, even more precisely: the partial derivatives gx (x, y) = 3x2 y − 4x − 4y/x2 and gy (x, y) = x3 + 4/x are defined and continuous everywhere near (2, 3), so g is differentiable at (2, 3).] (c). The unit vector in the direction of ~v = h3, −4i is ~u = h3/5, −4/5i. Meanwhile, ∇g = h3x2 y − 4x − 4y/x2 , x3 + 4/xi, so ∇g(2, 3) = h36 − 8 − 12/4, 8 + 2i = h25, 10i, and therefore D~u f (2, 3) = h25, 10i · h3/5, −4/5i = 15 − 8 = 7. (d). At (2, 3), g increases fastest in the direction of ∇g(2, 3) = h25, 10i. √ √ [FYI: that’s the same as the direction of h5, 2i or, if you prefer unit vectors, h5/ 29, 2/ 29i.] p p √ (e). The rate of increase of g in that direction is k∇g(2, 3)k = 252 + 102 = 52 (25 + 4) = 5 29 7. Suppose f (x, y), g(s, t), and h(s, t) are differentiable functions such that: fx (0, 0) = −4, fy (0, 0) = 7, fx (3, −2) = 5, f (0, 0) = 2, f (3, −2) = 9, gs (0, 0) = −1, hs (0, 0) = 2, ht (0, 0) = 3, g(0, 0) = 3, fy (3, −2) = −2, gt (0, 0) = 4, h(0, 0) = −2.
Let F (s, t) = f g(s, t), h(s, t) . Compute Fs (0, 0) and Ft (0, 0). ∂f ∂x ∂f ∂y ∂f ∂x ∂f ∂y + , and Ft = + . Answer. By the Chain Rule, Fs = ∂x ∂s ∂y ∂s ∂x ∂t ∂y ∂t So at (s, t) = (0, 0), which gives (x, y) = (g(0, 0), h(0, 0)) = (3, −2), we have Fs (0, 0) = fx (3, −2)gs (0, 0) + fy (3, −2)hs (0, 0) = 5 · (−1) + (−2) · 2 = −5 − 4 = −9, and Ft (0, 0) = fx (3, −2)gt (0, 0) + fy (3, −2)ht (0, 0) = 5 · 4 + (−2) · 3 = 20 − 6 = 14. 8. Suppose H(x, y, z) is a differentiable function such that ∇H(3, 2, 1) = h−4, 6, 5i. Let f (t) = H(t2 − 1, 8 − 3t, t − 1). (a). Compute f 0 (2). (b). At the point (3, 2, 1), in which direction does H(x, y, z) increase the fastest? How fast does H increase when we move in that direction? Answer. (a): Since H(x, y, z), and x = t2 − 1, y = 8 − 3t, and z = t − 1 are differentiable, the df ∂H dx ∂H dy ∂H dz Chain Rule says that = + + . Since x0 (t) = 2t, y 0 (t) = −3, and z 0 (t) = 1; and dt ∂x dt ∂y dt ∂z dt x(2) = 3, y(2) = 2, and z(2) = 1; and ∇H(3, 2, 1) = h−4, 6, 5i, we have f 0 (2) = −4 · 4 + 6 · (−3) + 5 · 1 = −16 − 18 + 5 = −29. (b): H increases fastest in the √ direction of the gradient vector, h−4, 6, 5i, and the rate of increase is √ the length 16 + 36 + 25 = 77. √ 9. Find an equation for the tangent plane to z = xy + 3y at the point (1, 1, 2). √ Answer: f (x, y) = xy + 3y has ∇f = h(1/2)(xy + 3y)−1/2 (y), (1/2)(xy + 3y)−1/2 (x + 3)i. So fx (1, 1) = (1/2)(4)−1/2 (1) = 1/4, and fy (1, 1) = (1/2)(4)−1/2 (4) = 1. Meanwhile, yes, indeed, f (1, 1) = 2. So the tangent plane is given by z = 2 + (x − 1)/4 + (y − 1), i.e., x + 4y − 4z = −3. 10. Find an equation for the tangent plane to x3 − cos(xz) + xy 3 − yz = 0 at the point (0, 1, −1). Answer: We are talking about the tangent plane to a level surface of the function g(x, y, z) = x3 − cos(xz) + xy 3 − yz, which has gradient ∇g = h3x2 + z sin(xz) + y 3 , 3xy 2 − z, x sin(xz) − yi. So ∇g(0, 1, −1) = h1, 1, −1i is orthogonal to the tangent plane. So the tangent plane has equation (x − 0) + (y − 1) − (z + 1) = 0, i.e., x + y − z = 2. 11. Find any and all point(s) on the ellipsoid 4x2 + y 2 + 4z 2 = 36 at which the tangent plane is parallel to the plane x + y − 2z = 3. Answer: Let f (x, y, z) = 4x2 + y 2 + 4z 2 . We need to find all points on the level surface f (x, y, z) = 36 at which the tangent plane is parallel to x + y − 2z = 3. That is, we want points on the surface at which the normal vector to the tangent plane is parallel to h1, 1, −2i. We compute ∇f (x, y, z) = h8x, 2y, 8zi. For this to be parallel to the desired vector, it must equal ch1, 1, −2i for some scalar c ∈ R. Thus, we need to solve the equations 8x = c, 2y = c, 8z = −2c, 4x2 + y 2 + 4z 2 = 36. The first two equations give 8x = 2y and hence y = 4x. The first and third give −4z = 8x, and hence z = −2x. Thus, the last equation gives 4x2 + 16x2 + 16x2 = 36, so that 36x2 = 36 and hence x = ±1. Plugging these values into y = 4x and z = −2x gives us the two points (1, 4, −2) and (−1, −4, 2). Thus, these are precisely the two points on the surface where the tangent plane is parallel to x + y − 2z = 3. 12. Find and classify (as local minimum, local maximum, or saddle point) all critical points of the function f (x, y) = 4xy − x4 − y 4 . Answer. We compute ∇f = h4y − 4x3 , 4x − 4y 3 i. Solving ∇f = ~0, we have y = x3 and x = y 3 , which means (substituting y = x3 into x = y 3 ) that x = x9 . Hence either x = 0 or we can divide by x to get x8 = 1, and so x = ±1. Using y = x3 therefore gives the three points (0, 0), (1, 1), and (−1, −1). Next, −12x2 4 D(x, y) = = 144x2 y 2 − 16. 4 −12y 2 Since D(0, 0) = −16 < 0, there is a saddle point at (0, 0). Meanwhile, D(1, 1) = 144 − 16 > 0 and fxx (1, 1) = −12 < 0, so there is a local maximum at (1, 1). Similarly, D(−1, −1) = 144 − 16 > 0 and fxx (−1, −1) = −12 < 0 again, meaning that there is also a local maximum at (−1, −1). 13. Find and classify (as local minimum, local maximum, or saddle point) all critical points of the function g(x, y) = 8x3 − 12xy + y 3 − 7. Answer. We compute ∇g = h24x2 − 12y, −12x + 3y 2 i. Solving ∇g = ~0, we have y = 2x2 and y 2 = 4x, which means (substituting y = 2x2 into y 2 = 4x) that 4x4 = 4x. That is, x4 = x, which means either x = 0 or x3 = 1. If x = 0, then y = 2 · 02 = 0; if x3 = 1, then x = 1, and hence y = 2 · 12 = 2. So there are two critical points: (0, 0) and (1, 2). Next, D(x, y) = 48x −12 −12 6y = 6 · 48xy − 122 = 144(2xy − 1). Since D(0, 0) = −144 < 0, there is a saddle point at (0, 0). Since D(1, 2) = 144 · 3 > 0 and fxx (1, 2) = 48 > 0, there is a local minimum at (1, 2). 14. Find the point(s) on the hyperboloid x2 + y 2 − 4z 2 = 10 closest to the point (4, 2, 0). Answer. We are asked to find the points that minimize the distance from (x, y, z) to (4, 2, 0) subject to the constraint g(x, y, z) = 10, where g(x, y, z) = x2 + y 2 − 4z 2 . To avoid square roots, we note that it’s equivalent to minimize the square of the distance, which is f (x, y, z) = (x − 4)2 + (y − 2)2 + z 2 . [And the minimum, although not a maximum, must actually be attained, as can be seen if we restrict ourselves to points (x, y, z) on the cube for which |x|, |y|, |z| ≤ 1000, where I just picked 1000 as a random large number. But you won’t need to say this comment on the exam. Intuitively, we expect to get a point of minimum distance.] Apply Lagrange multipliers: ∇f = h2(x − 4), 2(y − 2), 2zi and ∇g = h2x, 2y, −8zi. Then ∇f = λ∇g gives x − 4 = λx, y − 2 = λy, and z = −4λz, in addition to the original constraint. The third equation gives either z = 0 or λ = −1/4. If z = 0, then rewrite the first two equations as (1 − λ)x = 4 and (1 − λ)y = 2. Neither x nor y can be zero, so the first equation becomes (1 − λ) = 4/x, which in the second equation gives √ (4/x)y = 2, so 2 + y 2 = 10, i.e. y 2 = 2, so y = ± 2. Using x = 2y that x = 2y. In the original constraint, we get 4y √ √ √ √ and z = 0 means we have the two points (2 2, 2, 0) and (−2 2, − 2, 0). Next, consider λ = −1/4. Then the first two equations give x = 16/5 and y = 8/5, so that p the 2 = 10, i.e., 4z 2 = 14/5, which means z = ± 7/10. constraint equation gives 256/25 + 64/25 − 4z p That is, we have the two points (16/5, 8/5, ± 7/10). Testing all four points,√we get √ √ √ √ √ √ 2 + (2 − 2)2 = 16 − 16 2 + 8 + 4 − 4 2 + 2 = 30 − 20 2 f (2 √ 2, 2,√ 0) = (4 − 2 2)√ √ √ √ √ f (−2 2, − 2,p 0) = (4 + 2 2)2 + (2 + 2)2 = 16 + 16 2 + 8 + 4 + 4 2 + 2 = 30 + 20 2 2 2 + 7/10 = 20/25 + 7/10 = 15/10 = 1.5. f (16/5, √8/5, ± 7/10) = (4/5) + (2/5) √ Since 2 < 1.425, we have p 30 − 20 2 > 30 − 28.5 = 1.5. Thus, the minimum of f is attained at the two points (16/5, 8/5, ± 7/10), which are the points on the surface closest to (4, 2, 0). 15. Find the maximum and minimum values of the function g(x, y) = 4xy +y 2 on the ellipse 9x2 +y 2 = 45. Answer. Apply Lagrange multipliers; let h(x, y) = 9x2 + y 2 . So ∇g = h4y, 4x + 2yi and ∇h = h18x, 2yi, and we set ∇g = λ∇h. We get 4y = 18xλ and 4x + 2y = 2yλ, along with the original constraint 9x2 + y 2 = 45. The second equation gives x = (λ − 1)y/2, so that the first equation becomes 4y = 9yλ(λ − 1). Thus, either y = 0 or 9λ(λ − 1) = 4. If y = 0, then the original second equation gives x = 0, contradicting the constraint (0 = 45); so the case y = 0 does not occur. Thus, 9λ2 − 9λ − 4 = 0, which is to say, (3λ − 4)(3λ + 1) = 0, i.e., λ = 4/3 or λ = −1/3. If λ = 4/3, the first [or second] equation gives y = 6x. In the constraint, then, 9x2 + 36x2 = 45, and hence 45x2 = 45, so that x = ±1. That is, we get the two points (1, 6) and (−1, −6). If λ = −1/3, the first [or second] equation gives y = −(3/2)x. The constraint is then 9x2 +(9/4)x2 = 45, which is (45/4)x2 = 45, so that x2 = 4, and hence x = ±2. Using y = −(3/2)x gives the two points (2, −3) and (−2, 3). Testing these four points gives g(1, 6) = g(−1, −6) = 24 + 36 = 60, and g(−2, 3) = g(2, −3) = −24 + 9 = −15. Thus, the maximum value is 60, and the minimum is −15. 16. Find the maximum and minimum values of the function f (x, y, z) = xyz on the sphere x2 +y 2 +z 2 = 3. Answer. We use Lagrange multipliers. Setting g(x, y, z) = x2 + y 2 + z 2 , the equation ∇f = λ∇g gives yz = 2λx, xz = 2λy, and xy = 2λz. [FYI: this system can be solved a little more elegantly than what I’m about to do by multiplying the first of those equations by x, the second by y, and the third by z, to get xyz on the left in all three cases. That gives either λ = 0 or else x2 = y 2 = z 2 , which with a little more work gives the solution I’ll get below. But I won’t expect you to notice a sneaky trick like that on a timed exam.] If x = 0, then the first equation gives either y = 0 or z = 0. Similarly, if any of x, y, z is zero, √ then (at 2 2 2 least) two of those three variables are zero. If x = y = 0, then x + y + z = 0 gives z = ± 3, and similarly if x = z = 0 or y = z = 0. That’s six points so far. It remains to consider the case that none of x, y, z is zero. Then λ 6= 0 also. The first equation gives x = yz/(2λ), so the second equation gives yz 2 /(2λ) = 2λy, which implies z 2 = 4λ2 , and hence z = ±2λ. Similarly, the last equation gives y = ±2λ, and therefore also x = ±2λ. [Note: the ± signs are not in synch.] The equation x2 + y 2 + z 2 = 3 then gives 3(2λ)2 = 3, and therefore 2λ = ±1, giving us the eight points (±1, ±1, ±1). Plugging these fourteen points into f (x, y, z) = xyz gives 0 at each of the first six points, 1 at half of the remaining eight points (those with an even number of minus signs), and −1 at the remaining four points (those with an odd number of minus signs). Thus, the maximum value is 1, and the minimum value is −1. 17. Find the maximum and minimum values of the function g(x, y) = 3x2 − 4y + y 2 + 2 on the disk x2 + y 2 ≤ 9. Answer. First, let h(x, y) = x2 + y 2 and use Lagrange multipliers on the level curve h = 9. The equation ∇g = λ∇h gives 6x = 2λx and −4 + 2y = 2λy. If x = 0 (which satisfies the first equation), then h = 9 gives y = ±3. Otherwise, the first equation gives λ = 3, and √ so the second equation becomes −4 + 2y = 6y, and therefore y = −1; then h = 2 gives x = ±2 2. That’s a total of four points to check on the boundary. Inside, we set ∇g = ~0 to give 6x = 0 and −4 + 2y = 0, which means the single point (0, 2), which is indeed inside the disk (since 02 + 22 = 4 < 9). Finally,√we check the values of g at these five √ points: g(0, 3) = −12+9+2 = −1, g(0, −3) = 12+9+2 = 23, g(2 2, −1) = 24+4+1+2 = 31, g(−2 2, −1) = 24+4+1+2 = 31, and g(0, 2) = −8+4+2 = −2. Thus, the maximum value is 31, and the minimum is −2.