Review for the Final Test

Transcription

1 Math 7 Review for the Final Test () Decide if the limit exists and if it exists, evaluate it. lim (x,y,z) (0,0,0) xz. x +y +z () Use implicit differentiation to find z if x + y z = 9 () Find the unit tangent vector T for r (t) = (e t sin t) i (e t cos t) j + e t k at t = 0. = 0, using the indicated change of variables: u = x + y, v = x + y. (0) (5) Let f(x, y) = tan y, x (a). find df(x, y); (b). find f(x, y) at (, ); (c). find D a f at p, where p = (, ), and a = i + j ; (6) Use Lagrange multipliers to find the maximum and minimum values of f(x, y, z) = xy + yz + xz subject to constraint xyz = where x, y, and z are positive real numbers. (7) Evaluate the integral by changing it to the polar coordinates: () Solve the partial differential equation: f x f dxdy, where D = {(x, y) : y x + y }. D (+x +y) (8) Without changing the integrand express the integral as an equivalent one in which the order of integral is reversed (i.e. dxdydz): x x y f(x, y, z)dzdydx (9) Evaluate the integral C y dx x dy, where C is the boundary of the region R = {(x, y) : x + y 9, and x 0} with a counterclockwise orientation. (0) Use the Divergence theorem to evaluate σ F n ds, where n is the outer unit normal vector to σ, F (x, y, z) = (x y) i + (xz) j + (xyz) k, and σ is the surface of the solid bounded above by z = 9 and below by z = x + y. () Use Stokes theorem to estimate the integral C F d r, where F (x, y, z) = (y) i + z j + (x) k, C is the triangle with vertices (, 0, 0), (0,, ), (0, 0, 0) with a counterclockwise orientation looking down the positive z-axis.

2 SOLUTIONS Problem. On the one hand the limit of this expression along the y-axis is lim xz = 0. x + y + z (0, y,0) 0 On the other hand the limit of the same expression along the line x= y = zis obviously ¼. Therefore the limit in this problem does not exist. Problem. We can rewrite this relation as 9 9 z = x + y 9 whence z z = get 9 y and z z = 9 y. Differentiating both parts by y once more we z z + z = 9, whence z 9 z 9 8y 9(z 9 y ) = = =. z z z 6z 6z Problem. t t t t t r ( t) = ( e cost+ e sin t) i+ ( e cost+ e sin t) j+ ek; t t t t t t r ( t) = ( e cost+ e sin t) + ( e cost+ e sin t) + e = e

3 whence T( t) = [(sint+ cos t) i+ (sint cos t) j+ k ]. Problem. Therefore f f = f x x u f f u f v f f = + = + ; x u x v x u v f f u f v f f = + = + ; u v u v f = u. Our equation is equivalent to 0which means that f depends only on v. The general solution of this equation is where f is an arbitrary differentiable function of one variable. f( x+ y) Problem 5. (a) f y y =, y = x x x + y + x f x =, whence x y = x x + y + x df ( x, y) = ( ydx + xdy). x + y

4 (b) f( x, y) = ( yi+ xj), in particular, x + y f (,) = ( i j). 5 (c) f (,) = ( ij, ), Df a = f(,) a =. Problem 6. Let Fxyz (,,) = xy+ yz+ xz λxyz. The equations F F F = = = 0provide the system x z y+ z = λ yz x+ z = λxz y+ x= λxy If we multiply both parts of each equation by x, y, and z, respectively we will getxy + xz = xy + yz = yz + xz. Because none of the variables can be 0 we immediately obtain x= z = y. Plugging these equalities into the relation xyz = we get x =, y =, z =. The minimum value of the function xy + yz + xz is attained at the point (,,) and equals. It is easy to see that the function does not attain the greatest value. Indeed if we take a positive number A and put xy + yz + xz = A +. A A x= y= A, z = then xyz = A but

5 Problem 7. The inequality r sin r y x y θ or equivalentlysinθ r D + in polar coordinates becomes π ( + u). Therefore dxdy r = drdθ = ( + x + y ) ( + r ) 0 sinθ = π 0 sin θ du, where u = r. Next notice that ( + u) + u sin + sin θ θ sin θ du = =. whence D π dxdy π = dθ = ( + x + y ) + sin θ 0 π/ π/ π π = dθ = sec θdθ = + sin θ sec θ + tan θ 0 0 π = du (where u = tan θ ) = + u 0 π π π π( ) = dv (where v = u) = = + v 0

6 Problem 8. The region of integration in 0 x, 0 y x, 0 z x y. R is defined by the inequalities From the last inequality we see that the total range of the variable z is 0 z. From the same inequality we see that z ywhence y z. Finally, because our answer is x z y we get x z y and x x y z z y f( x, y, z) dzdydx= f( x, y, z) dxdydz

7 Problem 9. The contour of integration is shown on the picture below. The contour integral can be represented as sum of four terms: integrals along the semicircles and integrals along the vertical segments. To integrate along the semicircles it is convenient to use polar coordinates: x= rcos t, y= rsint Next we compute: π π π I = 8 ( sin t + cos t) dt = 8 =, π π π I = 0, π I = ( sin t + cos t) dt =, I = 0. I + I + I + I = 60 π.

8 Another way to solve the problem is to use the Green s theorem. We have Q x P = + ( x y ) and D Q P da r drd x y = = π / π / θ 60 π. Problem 0. Let E be the solid bounded by the surfaceσ. Then by the Divergence theorem σ = ( xy) ( xz) ( xyz) F nds = divfdv = + + dv = x z E E E xydv. To evaluate the triple integral it is convenient to use the cylindrical coordinates: x= rcos θ, y= rsin θ, z= z. Then we compute E 9π z xydv = r cosθsinθdrdθdz = 0, π because cosθsinθdθ = 0.

9 Problem. Let us first computecurlf. By the Stokes Theorem i j k curlf = = i j k. x z y z x F dr= curlf ds= curlf n ds. C S S Let us compute the unit normal vectorn. The region S is a part of a plane defined by the vectors a= j and = + cross product b j k. The vector n is proportional to the i j k a b= 0 0 = j+ k 0 This vector is already oriented upward so we have n= j+ kand curlfn =. Therefore curlfn ds = Awhere Ais the area of the S triangle S. This area can be computed as a b = 9. Finally, the answer is 9.