Mathematical Analysis II, 208/9 First semester Yoh Tanimoto Dipartimento di Matematica, Università di Roma Tor Vergata Via della Ricerca Scientifica, I-0033 Roma, Italy email: hoyt@mat.uniroma2.it We basically follow the textbook Calculus Vol. I,II by Tom M. Apostol, Wiley. Nov 26. Implicit functions and partial derivatives. The equation x+z +y +z 2 6 defines implicitly a function fx, y z. Compute, in terms of x, y, z. Check that,, satisfies the equation, and compute,,,. Solution. Put F x, y, z x + z + y + z 2 6. The partial derivatives are, 2y + z, + 2y + z. For the partial derivaties of f, using the formula of the lecture, x, y x, y, fx, y x, y, fx, y + 2y + fx, y, x, y x, y, fx, y 2y + fx, y x, y, fx, y + 2y + fx, y. By putting x, y, fx, y,,,, 5,, x, y, fx, y x, y, fx, y 4 5. 2. Consider two surfaces 2x 2 + 3y 2 z 2 25 0, x 2 + y 2 z 2 0. The intersection C can be parametrized as Xz, Y z, z. a Check that C passes the point P 7, 3, 4. b Find a tangent vector of C at P. Solution. a By substituting x, y, z 7, 3, 4. b i. By implicit computations. Put F x, y, z 2x 2 + 3y 2 z 2 25, Gx, y, z X x 2 + y 2 z 2. We use the formula Y G G g. These partial derivatives can be computed: 4x, 6y, 2z, G 2x, G 2y, G 2z.
By putting the value P 7, 3, 4, we obtain X 4 Y 4 4 7 8 2 8 7 6 8 2 6 8 7 2 7 4 8 7 8 8 7 4 3. ii. By direct computations. We have, from 3G F 0, x 2 + 25 2z 2, which is equivalent to x Xz ± 2z 2 25. Similarly, from F 2G 0, it follows that y 2 25 z 2, which is equivalent to y Y z ± 25 z 2. Since we are interested in the point 7, 3, 4, we take the + solutions. By differentiating them, X 2z z, Y 2z z z X. At P, 4 7 8 2 25 25 z 2 Y 4 4. 3 Hence a tangent vector is Nov 26. Stationary points. 8 7 4 3.. Locate and classify the stationary points. a fx, y x 2 + y 2 b fx, y 2x 2 xy 3y 2 3x + 7y c fx, y sin x cosh y Solution. 2 0 a fx, y 2x, 2y. fx, y 0 x, y 0,. Hx, y, 0 2 2 0 hence H0,, and this has positive eigenvalues 2, 2. Therefore, 0, is 0 2 a local minumum. b fx, y 4x y 3, x 6y + 7. fx, y 0 x, y,. Hx, y 4 4, hence H,, and this has positive and negative 6 6 eigenvalues, because its determinant is 26. Therefore,, is a saddle point. c fx, y cos x cosh y, sin x sinh y. fx, y 0 cos x 0 and sinh y 0 x, y m + sin x cosh y cos x sinh y 2 π, 0. Hx, y. Note that cos x sinh y sin x cosh y sin m + 2 0. hence H m + 2, 0 has the determinant, and hence has positive and negative eigenvalues. Therefore, m + 2, 0 is a saddle point. 2. Let x,, x n be distinct numbers, y,, y n R. Let a, b R, fx ax + b. With Ea, b n j fx j y j 2. Find a, b which minimize Ea, b. Solution. We can write Ea, b ax j + b y j 2. Therefore, j Ea, b 2x j ax j + b y j, j 2 2ax j + b y j j
From Ea, b 0, we obtain a x 2 j + b x j x j y j, j j j a x j + b j j j y j Put x n n j x j, y n n j y j, then the second equation is x a + b y, or x 2 + x b x y. Set u j x j x, then the first equation is a n x 2 j + x b n j By subtracting x 2 + x b x y, we have a n x j u j n j x j y j. j u j y j, hence by noting that j u j 0, a n j u jy j / n j x ju j n j u jy j / n j u2 j, b y x a. Nov. 9. Lagrange s multiplier method. Find the maximum and minimum distances from the origin to the curve 5x 2 +6xy+5y 2 8. 2. Assume a, b R, a, b > 0. a Find the extreme values of fx, y x a + y b on x2 + y 2. b Find the extreme values of fx, y x 2 + y 2 on x a + y b. 3. Find the nearest point from the origin to the curve of intersection of x 2 xy+y 2 z 2 0 and x 2 + y 2. Nov. 9. Line integrals. Compute the line integrals f dα j a fx, y x 2 2xy, y 2 2xy, αt t, t 2, t [, ]. b fx, y, z y 2 z 2, 2yz, x 2, αt t, t 2, t 3, t [, ]. c fx, y y, x, αt cos t, sin t, t [0, π] and βt t, t 2, t [, ]. 2. A wire has a shape x 2 + y 2 a 2, a > 0 with density ϕx, y x + y. Compute the mass. Nov. 9. Gradients and line integrals. Show that the following vector fields f are not gradient. Find a closed path α such that f dα 0. a fx, y, z y, x, x b fx, y, z xy, x 2 +, z 2 2. Show that, for a continuous function f, the vector field fx, y xf x 2 + y 2, yf x 2 + y 2 is a gradient. 3
3. Let S {x, y R 2, x, y 0, 0}, fx, y y, x 2 +y 2 a Show that 2 f f 2. x x 2 +y 2. b For αt cos t, sin t, t [0, 2π], show that f dα 2π, therefore, f is not a gradient on S. Nov. 6. Potentials. Determine whether the following vector fields f are a gradient. If so, find a potential. a fx, y e x 2y on R 2. e x +y 2 e x +y 2 b fx, y, z 2xyz + z 2 2y 2 +, x 2 z 4xy, x 2 y + 2xy 2 on R 3. c fx, y, z 2xz 3, x 2 z 3, 3x 2 yz 2. 2. Solve the following differential equations. a dy dx 3x2 +6xy 2 6x 2 y+4y 3. b y + 2xy 0. Nov. 6. Double integrals. Show that the function fx, y xy 3 on Q [0, ] [0, ] is integrable. 2. The following functions are integrable. Compute Q fx, ydxdy. a fx, y xyx + y, Q [0, ] [0, ]. b fx, y sinx + y, Q [0, π 2 ] [0, π 2 ]. c fx, y y 3 e x/y, Q [0, ] [, 2]. Nov. 23. Double integrals. Compute the following integrals. a Q x sin y yex dxdy, Q [, ] [0, π 2 ]. b Q y x 2 dxdy, Q [, ] [0, 2]. 2. Compute the integral S fdxdy. a fx, y x cosx + y, S {x, y : 0 x π, 0 y x}. b fx, y x 2 y 2, S {x, y : 0 x π, 0 y sin x}. c fx, y 3x + y, S {x, y : 4x 2 + 9y 2 36, x 0, y 0}. d fx, y y + 2x + 20, S {x, y : x 2 + y 2 6}. 3. Write S as a type II region. a S {x, y : 0 x, x 3 y x 2 }. b S {x, y : x e, 0 y log x}. 4. Find the centroid of S. a S {x, y : 0 x π 4, sin x y cos x}. b S {x, y : x e, 0 y log x}. 4
Nov. 30. Green s theorem. Compute the following line integrals. a C f dα, fx, y y 2, x and C is the boundary of [0, 2] [0, 2]. b C f dα, fx, y 3x 3y, 4y + x and αt cos t, sin t, t [0, 2π]. 2. With S {x, y R 2, x, y 0, 0}, fx, y y + y, 2x +, x show that x 2 +y 2 x 2 +y 2 C f dα, where αt a cos t, b sin t, t [0, 2π] does not depend on a, b > 0. Nov. 30. Change of coordinates. Find the corresponding region in the new coordinates. a S {x, y : 0 x, 0 y, x + y 2}, x 2 v u, y 2 v + u. b S {x, y : 0 x, x 2 + y 2 }, x r cos θ, y r sin θ. c S {x, y : x a 2 + y 2 a 2 }, x r cos θ, y r sin θ. 2. Computer the integrals in the new coordinages. a S ey x/y+x dxdy, S {x, y : 0 x, 0 y, x+y 2}, x 2 v u, y 2 v+u. b S x2 + y 2 dxdy, S {x, y : x a 2 + y 2 a 2 }, x r cos θ, y r sin θ. 3. Compute the volume of the sphere V {x, y, z : x 2 + y 2 + z 2 a 2 }. Dec. 4. Surface. Find a parametrization of the cylinder {x, y, z : x 2 + y 2 a 2, 0 z }. 2. Compute r u r v. a ru, v u + v, u v, 4v 2. b ru, v a sin u cosh v, b cos u cosh v, c sinh v. 3. Compute the area. a the intersection of x + y + z a, x 2 + y 2 a 2. Dec. 4. Surface integrals. Let S : x 2 + y 2 + z 2, z 0 and F x, y, z x, y, 0. Compute S F nds with the parametrization z x 2 y 2. 2. Let S be a triangle with vertices, 0, 0, 0,, 0, 0, 0, and F x, y, z x, y, z. Compute S F nds, where n has positive z-component. 3. Compute curl and div. a F x, y, z 2z 3y, 3x z, y 2x b F x, y, z e xy, cos xy, cos xz 2 5
Dec. 2. Stokes theorem. Let C be the curve of the intersection x 2 + y 2 + z 2 a 2, x + y + z 0. Compute F dα, where F x, y, z y, z, x. 2. Let F x, y, z e zy2, e yx2, e xz2, C be the boundary of the square which has the vertices 0, 0, 0,, 0, 0,,, 0, 0,, 0. Compute C F dα, where α is a parametrization of C going counterclockwise. Dec. 2. Gauss theorem. Let S be the surface of the unit cube V {x, y, z : 0 x, y, z }, n be the outgoing unit vector on S, F x, y, z x 2, y 2, z 2. Compute S F nds and V div F dxdydz. 2. Let F x, y, z x 3, y 3, z 3, S : {x, y, z : x 2 + y 2 + z 2 a 2 }, and n the outgoing normal unit vector on S at each point of S. Compute the surface integral F n ds. S 6