e x3 dx dy. 0 y x 2, 0 x 1.
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1 Problem 1. Evaluate by changing the order of integration y e x3 dx dy. Solution:We change the order of integration over the region y x 1. We find and x e x3 dy dx = y x, x 1. x e x3 dx = 1 x=1 3 ex3 x= = 1 (e 1). 3 1
2 Problem. Find the critical points of the function f(x, y) = xy 4xy + 1 x and determine their nature. Solution:We calculate f x = y 4y + x, f y = xy 4x = x(y ). Since f y = = x = or y =. When x = we obtain from f x = = y 4y = = y = or y = 4. When y = we find f x = = x = 4. There are three critical points (, ), (, 4) and (4, ). We find the second derivatives f xx = 1, f xy = y 4, f yy = x. At the points where x =, the determinant of the hessian is D = f xx f yy fxy = 1 (y 4) < hence both (, ) and (, 4) are saddle points. For (4, ), the determinant of the Hessian is f xx f yy fxy = 1 8 > hence (4, ) is a local minimum.
3 Problem 3. Consider the function (i) Show that the limit f(x, y) = x y x 4 + y. lim f(x, y) (x,y) (,) along any line of fixed slope m through the origin equals. (ii) Evaluate the limit lim (x,y) (,) f(x, y) along the parabola y = x. (iii) What is the value of the limit lim (x,y) (,) f(x, y)? Solution: (i) We have y = mx and f(x, y) = x (mx) x 4 + (mx) = mx x + m m provided x. Thus f(x, y) along the line y = mx provided m. When m =, we have y = and f(x, y) =. (ii) When y = x, we have f(x, y) = x y x 4 +y = 1, so the limit equals 1. (iii) Since the answers in (i) and (ii) are different, the limit does not exist.
4 Problem 4. Consider the function f(x, y) = 6 x y. (i) Find the direction of steepest increase of f at the point P (1, ). (ii) Draw the graph of the function f. (iii) Find the directional derivative D v f(1, ) in the direction v = 3 5i 4 5j. (iv) Find the linear approximation f((1, ) +.1 v). (v) Find the tangent plane to the surface z x + zf(x, y) = at (1,, 1). Solution: (i) We have Similarly, f x = 1 x 6 x y = x 6 x y = f x(1, ) = 1. y f y = 6 x y = f y(1, ) =. Thus f = ( 1, ) and the direction of steepest increase is ( 1, ). (ii) The graph is z = f(x, y) = 6 x y = x + y + z = 6, z which is the half sphere of radius 6. (iii) We have D v f = f v = ( 1, ) (3/5, 4/5) = 1. (iv) We evaluate f((1, ) +.1 v) f(1, ) +.1 D v f = = 1.1. (v) We have g(x, y, z) = z x + zf(x, y) hence g x = z + zxf x (x, y) = g x (1,, 1) = 1 + ( 1) = 1 g y = zf y (x, y) = g y (1,, 1) = 1( ) = g z = zx + f(x, y) = g z (1,, 1) = 3. The normal vector is ( 1,, 3) hence the plane is x y + 3z =.
5 Problem 5. Find the minimum and the maxim value of the function along the sphere of center (1,, 1) and radius 3. Solution:We have We need which gives Using f(x, y, z) = x y + z g(x, y, z) = (x 1) + y + (z + 1) = 9. f = λ g (,, 1) = λ(x 1, y, z + 1) = x 1 = λ, y = λ, z + 1 = 1 λ. (x 1) + y + (z + 1) = 9 = When λ = 1 we find the point When λ = 1 we find the point ( ) ( + ) ( ) 1 + = 9 = λ = ±1. λ λ λ (3,, ) with f(3,, ) = 1. ( 1,, ) with f( 1,, ) = 8.
6 Problem 6. Find the area of the region bounded in the first quadrant by the hyperbola xy = 1 and the parabolas x = y and x = 8y. Express the answer in the simplest possible form. Solution:The intersection of the parabola x = y and the hyperbola xy = 1 occurs at (1, 1). The intersection of the parabola x = 8y and the hyperbola xy = 1 is the point (, 1 ). The integral is to be split into two regions. For the region x 1, we have x y min = 8, y max = x. For the region 1 x, we have We calculate Area = = y min = x 8, y max = 1 x. x x/8 dy dx + x x 8 dx + /x = 3 x3/ (1 1 8 ) x=1 x= + ln x x= x=1 3 1 dy dx 1 x/8 1 x x 8 dx x 3/ 8 x= x=1 = ln.
7 Problem 7. Find the average value of the function over the first octant of the ball x + y + z 1. f(x, y, z) = xyz x, y, z Solution:We use spherical coordinates. We have 1 f = xyz dv. volume Thus 1 1 π π f = (ρ sin φ cos θ)(ρ sin φ sin θ)(ρ cos φ)(ρ sin φ) dρ dφ dθ 4π/3 1/8 = 6 π = 6 π = 6 π ρ6 6 ρ=1 ρ= 1 π ρ 5 dρ π π 4 sin4 φ φ= π φ= 1 ρ 5 sin 3 φ cos φ cos θ sin θ dρ dφ dθ π sin 3 φ cos φ dφ sin θ cos θ dθ sin θ θ= π θ= = 6 π = 1 8π.
8 Problem 8. Using cylindrical coordinates, find the mass of the solid with density ρ = z bounded by the sphere x + y + z = and the cone z = x + y. Solution:We calculate The intersection is and therefore the cylindrical coordinates satisfy We have π r = π r ρ dv. x + y = z = 1 r 1, r z r. 1 z dz(r dr) dθ = π r z z=r z= r dr = (1 r )r dr = π( r r4 4 ) 1 r= = π 1 4 = π.
9 Problem 9. Find the volume of the solid bounded by the cylinders and the planes x = and x + z = 1. Solution:We calculate y z 1 y 1 = dx dz dy = 1 z = 1 y, z = y 1 y 1 y 1 (1 z) dz dy = 1 (1 y ) dy = y 3 y3 y=1 y= 1 = = 8 3. (z 1 z ) z=1 y z=y 1 dy
10 Problem 1. Consider the parametric curve given by ( t 4 r(t) = 4, t6 6 Find the arclength parametrization of the curve. Solution:We have The arclength function is s(t) = Solving s(t) = s we find Thus t ), t 1. r (t) = (t 3, t 5 ) = r (t) = (t 3 ) + (t 5 ) = t t 4. r (u) du = t u u 4 du = 1 6 (1 + u4 ) 3 u=t u= = (1 + t4 ) (1 + t 4 ) 3 1 = s = t 4 = (6s + 1) ( ) 1 r(t) = 4 ((6s + 1) 1 3 1), 6 ((6s + 1) 3 3 1). Finally, t = corresponds to s = and t = 1 corresponds to s = 1 6 ( 3 1).
11 Problem 11. Assume that where Using the chain rule, calculate the derivative Express your answer in the simplest possible form. Solution:We have We compute Similarly Next, Therefore w x = w s w = ln(x y + z ) x = s + t, y = s t, z = st. w s. w s = w x x s + w y y s + w z z s. x x y + z = (s + t) (s + t) (s + t) (s t) = + 4st 1st = (s + t) 6st w t) = (s y 6st w z = 4 st 1st = 1 3 st. x s = 1, y s = 1, z s = 1 s t t = s. + (s t) 6st = s + t. 6st st t s = 4t 6st + 1 3s = 1 s.
12 Problem 1. The ellipsoid x + y + z = 4 and the plane x + y + 3z = 6 intersect in an ellipse passing through the point (1, 1, 1). Find the parametrization of the tangent line to the ellipse at (1, 1, 1). Solution:The normal vector to the ellipsoid is n 1 = (x, 4y, z) which gives at (1, 1, 1) the vector n 1 = (, 4, ). The normal vector to the plane is n = (, 1, 3). The vector tangent to the ellipse is perpendicular to both n 1 and n hence t = n 1 n = (, 4, ) (, 1, 3) = (1,, 6). The tangent line is (1, 1, 1) + t(1,, 6).
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