2004-2005 spring mt2

# 2004-2005 spring mt2 - METU Department of Mathematics Group...

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Unformatted text preview: METU Department of Mathematics Group Code Acad. Year Semester Instructor : Math 120 : 2004-2005 : Spring : Date Time Duration Calculus II MidTerm 2 Last Name : : Name Department : : Signature : May 7th,2005 : 09:30 : 120 minutes 1 2 3 4 List No. Student No. : : Section 6 QUESTIONS ON 5 PAGES TOTAL 60 POINTS 5 6 SOLUTION KEY Question 1 (10 points) Show that for any tangent plane to the surface √ √ √ x+ √ y+ z = a , (a > 0) the sum of x, y, and z axis intercepts is constant. ( x0 is called x intercept if the graph intersects x axis at the point (x0 , 0, 0). ) √ √ √ √ Solution: Let f (x, y, z) = x+ y+ z− a then the surface , S, is given by f (x, y, z) = 0. Let the point P0 (x0 , y0 , z0 ) ∈ S be the point of tangency. Then 1 1 1 f (x, y, z) = √ i + √ j + √ k 2 y 2 x 2 z and hence n= 1 1 1 f (x0 , y0 , z0 ) = √ i + √ j + √ k 2 x0 2 y0 2 z0 is a normal vector to the tangent plane. Therefore, the equation of the tangent plane is obtained as 1 1 1 √ (x − x0 ) + √ (y − y0 ) + √ (z − z0 ) = 0. 2 x0 2 y0 2 z0 Now, √ √ √ √ x−intercept is obtained, for y = z = 0, as: ( x0 + y0 + z0 ) x0 √ √ √ √ y−intercept is obtained, for x = z = 0, as: ( x0 + y0 + z0 ) y0 √ √ √ √ z−intercept is obtained, for x = y = 0, as: ( x0 + y0 + z0 ) z0 . Therefore the sum of intercepts is: √ √ √ √ ( x0 + y0 + z0 )2 = ( a)2 = a which is a constant. Question 2 ( 12 points) Consider the function f (x, y) = x3 if (x, y) = (0, 0), f (0, 0) = 0. Show that; x2 + y 2 a)f is continuous everywhere Solution: We need to check only continuity at the point (0, 0). 0≤ x2 x2 .x x3 = 2 ≤ |x| 2 +y x + y2 so by the squeeze theorem, we have lim f (x, y) = (x,y)→(0,0) lim (x,y)→(0,0) x2 x3 = 0 = f (0, 0) + y2 meaning that f is continuous at (0, 0). b)all directional derivatives of f exist at (0, 0), using limit deﬁnition of directional derivative Solution: Let u = ai + bj with a2 + b2 = 1. Then a3 h3 /[h2 (a2 + b2 )] − 0 f (ah, bh) − f (0, 0) = lim = a3 h→0 h→0 h h Du f (0, 0) = lim which completes the proof. c)f is not diﬀerentiable (0, 0) f (h, k) − f (0, 0) − hfx (0, 0) − kfy (0, 0) √ (h,k)→(0,0) h2 + k 2 = f (h, k) − h √ (h,k)→(0,0) h2 + k 2 = lim h3 h2 +k2 − h √ lim (h,k)→(0,0) h2 + k 2 lim along the line k = h we have −h/2 lim √ 2|h| h→0 which does not exist. Hence, f is not diﬀerentiable at (0, 0). d)ﬁnd the set of all vectors v for which the equation Dv f (0, 0) = f (0, 0) · v is satisﬁed. 3 Solution: Let v = v1 i + v2 j. Then, by (b) we have Dv f (0, 0) = v1 and f (0, 0) = i. which gives us v1 = Therefore, f (0, 0) · v = v1 . Now, the required condition is v1 = 0, v1 = ±1. For, v1 = 0 we have v2 = ±1 and for v1 = ±1 we have v2 = 0. Thus, the directions are ±i and ±j. 3 v1 Question 3 (8 points) Let z = f (x, y) = 20 − x2 − 7y 2 . Use Tangent Plane Ap- proximation to approximate f (1.95, 1.08). Solution: We have −x fx (x, y) = 20 − x2 − 7y 2 , fy (x, y) = −7y 20 − x2 − 7y 2 . So fx (2, 1) = −2/3 and fy (2, 1) = −7/3. Therefore f (1.95, 1.08) ≈ f (2, 1) + fx (2, 1)(1.95 − 2) + fy (2, 1)(1.08 − 1) = = 3 + (−2/3)(−0.05) + (−7/3)(0.08) 0.1 − 0.56 3+ . 3 x y y x derivatives in the neighborhood of the point (x, y) = (1, −1) with f1 (−1, −1) = 2 = f2 (1, −1), f1 (1, −1) = −2 = f2 (−1, −1), f11 (−1, −1) = 6 = f12 (1, −1), f11 (1, −1) = 8 = f21 (1, −1), f12 (−1, −1) = −5 = f21 (−1, −1), f22 (1, −1) = 1 = f22 (−1, −1). ∂2z Find z12 = at the point (x, y) = (1, −1). ∂y∂x Question 4 (10 points) Given that the function z = f ( , ) has continuous partial Solution: Let s = x/y, t = y/x then (x, y) = (1, −1) gives (s, t) = (−1, −1) and z = f (x/y, y/x) = f (s, t). Now, z1 = ∂z ∂z ∂s ∂z ∂t 1 y = + = f1 (s, t) − 2 f2 (s, t). ∂x ∂s ∂x ∂t ∂x y x Similarly, z12 ∂ 1 y f1 (s, t) − 2 f2 (s, t) ∂y y x −1 1 ∂s ∂t = f (s, t) + f11 (s, t) + f12 (s, t) 2 1 y y ∂y ∂y 1 ∂t y ∂s − 2 f2 (s, t) − 2 f21 (s, t) + f22 (s, t) x x ∂y ∂y −1 1 −x 1 = f1 (s, t) + f11 (s, t) 2 + f12 (s, t) y2 y y x 1 y −x 1 − 2 f2 (s, t) − 2 f21 (s, t) 2 + f22 (s, t) x x y x = Put x = 1, y = −1 to get z12 (1, −1) = −f1 (−1, −1) − [−f11 (−1, −1) + f12 (−1, −1)] −f2 (−1, −1) + [−f21 (−1, −1) + f22 (−1, −1)] = 17. Question 5 (10 points) Use the Method of Lagrange Multipliers to ﬁnd the least and greatest distance between the origin and the ellipse 17x2 + 12xy + 8y 2 − 100 = 0. Solution: Let the point be (x, y) then the distance is d = x2 + y 2 . Now, we deﬁne L(x, y, λ) = x2 + y 2 + λ(17x2 + 12xy + 8y 2 − 100) ∂L ∂x ∂L ∂y ∂L ∂λ = 2x + λ(34x + 12y) = 0 (1) = 2y + λ(12x + 16y) = 0 (2) = 17x2 + 12xy + 8y 2 − 100 = 0 (3) −y −x and λ = , respectively. 17x + 6y 6x + 8y Equating them and performing cross multiplication we get From the equations (1) and (2) we write λ = 2x2 − 3xy − 2y 2 = 0. (4) Multipying, (4) by 4, and adding with (3) we obtain, 25x2 − 100 = 0 or x = ±2. For x = 2, (4) becomes 8 − 6y − 2y 2 = 0 which has the roots y = 1 and y = −4. For x = −2, (4) becomes 8 + 6y − 2y 2 = 0 which has the roots y = −1 and y = 4. Therefore we have P1 (2, 1), P2 (2, −4), P3 (−2, −1), P4 (−2, 4). Therefore, the least distance is √ √ 5 and the greatest distance is 2 5. Surname, Name: ............................, ............................ Student Id:....................... Question 6 (10 points) Find and classify all the critical points of the function 1 f (x, y) = x3 + xy 2 − x. 3 Solution: We ﬁrst ﬁnd the critical points: f1 (x, y) = x2 + y 2 − 1 = 0 (5) f2 (x, y) = 2xy = 0 (6) The equation in (6) gives us x = 0 or y = 0. If x = 0, (5) implies that y = ±1 and if y = 0, (5) implies that x = ±1. Therefore, we have P1 (0, 1), P2 (0, −1), P3 (1, 0), P4 (−1, 0) as critical points. f11 (x, y) = 2x, f12 (x, y) = f21 (x, y) = 2y, f22 (x, y) = 2x ∆= f11 f21 f12 f22 = 2x 2y 2y 2x = 4(x2 − y 2 ) and ∆(P1 ) = −4 < 0 so P1 is a SADDLE point, ∆(P2 ) = −4 < 0 so P2 is a SADDLE point, ∆(P3 ) = 4 > 0 and f11 (P3 ) = 2 > 0 so P3 is a LOCAL MIN point, ∆(P4 ) = 4 > 0 and f11 (P4 ) = −2 < 0 so P4 is a LOCAL MAX point. ...
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