(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)


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1 eview Exam Math 43 Name Id ead each question carefully. Avoid simple mistakes. Put a box around the final answer to a question (use the back of the page if necessary). For full credit you must show your work. Incomplete solutions may receive partial credit if you have written down a reasonable partial solution. (1) Let u = (1, 3, ) and v = ( 4i k). Find: (a) the vector projection of u onto v. (b) the angle between u and v. (c) the area of the triangle spanned by u and v. (d) the equation of the plane through P (1, 1, 1) with normal perpendicular to both u and v. () Let r(t) = ( sin t, 5t, cos t). Find (a) the length of the curve (helix) over t 1. (b) the equation of the tangent line to the curve when t = π/4 (c) the curvature at any given time t (hint: κ(t) = r (t) r (t) / r (t) 3 ) (3) Find a vector orthogonal to the plane containing the points: P (1,, ), Q(, 3, 4), and (1, 1, 1). Find the equation of the plane. Find the area of the triangle P Q. (4) Let h(x, y) = 9 x y (a) Find and sketch the domain of the function h y x (b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (, ) f(x, y) = { 3x y x +y if (x, y) (, ) if (x, y) = (, ) (6) (a) Find all critical points, extrema and saddle points of g(x, y) = x 4 + y 4 4xy + 1. (b) Find the absolute maximum and absolute minimum of values of f(x, y) = x + y + x y + 4 on the region D = {(x, y) x 1, y 1}
2 (7) (a) Use the limit definition to find the partial derivative f(x,y) y where f(x, y) = x y. (b) Find the indicated partial derivatives: f xy, f yx, where f(x, y) = ye x y (c) Find the directional derivative of f(x, y, z) = x x y 3 + 1z + 3 at (1, 1, 1) in the direction v = (,, 1). (8) Let w = cos(x + 3y), x = r + s and y = rs (a) Use the hain rule to express w r and w s as functions of r and s. (b) Evaluate w r and w s at (r, s) = (, π). (9) Let g(x, y) = ln(x + y ). (a) Find the partial derivatives g x and g y. (b) Find the equation of the tangential plane to the surface z = g(x, y) at (, 1, ). (c) Find the linearization L(x, y) of the function g(x, y) at the point (, 1). (d) Give an estimate for the value g(.1,.9). (e) alculate g := ( x + y)g. (1) (a) Find the differential of u = x + y + z. (b) Find the tangent plane to the surface z = y ln x at (1, 4, ). (c) Use the hain rule (plot a tree diagram) to find the indicated partial derivatives u = x + y + z, x = sin(r) cos(s), y = sin(r) sin(s), z = 3. Find u u r and s. Your final answer should be given in terms of r and s only. (11) (a) *(optional) Find the area of the surface for the part of the hyperbolic paraboloid z = y x that lies between the cylinders x + y = 1 and x + y = 4. (b) Find the volume of the tetrahedron enclosed by the planes: x+y+z =, x =, z =, and y = x. (1) Quick answer: (a) The curvature of x + y = 49 is (b) How is the gradient at a point related to the level curve through that point? (c) If z depends upon u and v, and u and v depend upon t, then (d) The graph of r(t) is a (e) The graph of z = f(x, y) is a. (f) The graph of r(t) = t + 1, t 1, 3t is (g) Given a vector valued function r(t), where does the acceleration lie? (13) (a) Evaluate the line integral xydx + (x + y)dy along: (i) That part of the graph of y = x from (, 1) to (, 4) (ii) The line segment from (, 1) to (, 4) (iii) The broken line from (, 1) to (, 1) to (, 4). (b) Evaluate the line integral: c xy3 ds, is the left half of the circle x + y = 4. (14) (a) Determine whether or not F(x, y) = (x cos y y cos x)i + ( x sin y sin x)j is a conservative vector field. If it is, find a function f such that F = f.
3 (b) Show that F(x, y) = (e x (sin(x + 3y) + cos(x + 3y)), 3e x cos(x + 3y)) is conservative and find a potential function. (hint: verify that M y = N x ) (c) Show that F(x, y) = (M, N, P ) = (xyz + 3y, x z + 6xy z 3, x y 6yz ) is conservative and find a potential function. (hint: show that curl F = (P y N z, M z P x, N x M y ) = (15) Use the Fundamental Theorem for line integrals to evaluate F dr along c a very complicated curve from (1,, ) to (,, 3), and F(x, y, z) = yzi + xzj + (xy + z)k. (16) Let T be the unit tangent and N the (outward) unit normal for a curve : t r(t) on [a,b]. Let be a simply connected bounded open domain in, where F = Mi + Nj is continuously differentiable. (a) Scalar line integral on is: b f(x, y)ds = f(x(t), y(t)) r (t) dt = a (b) Vector line integral b W = F T ds = F (r(t)) r (t)dt = a (c) (optional)* Green s Theorem (circulation): F T ds = Mdx + Ndy = ( F )dxdy = (d) (optional)* Green s Theorem (flux): F Nds = Mdy Ndx = ( F )dxdy =. (e) (optional)* Use Green s Theorem to evaluate the line integral along the given positively oriented curve: (y + e x )dx + (x + cos(y ))dy where is the boundary of the region enclosed by the parabolas y = x and x = y. (f) (optional)* Use Green s Theorem to evaluate F dr where c F(x, y) =< 8 x + y, x y >, consists of the arc of the 8 curve y = sin x from (, ) to (π, ) and the line segment from (π, ) to (, ). (17) (optional*) Let F(x, y, z) = e x sin y i + e x cos y j + z k. (a) find curl F and div F (b) find div curl F, i.e. the divergence of the vector field curl F (c) What is the physical meaning of curl F? (18) (optional)* Find the area of the surface for the part of the sphere x + y + z = 9 that lies within the cylinder (x ) + y = 1 and above the xyplane. (19) (optional*) (a) Evaluate the integral x da where is the region bounded by y = x and y = x x. (b) Find the area of one loop of the polar curve r = sin 3θ
4 (Hint: You can think of the area formula by Green s Theorem Area = 1 xdy ydx = 1 xdx + ydy, a closed curve with counterclockwise orientation). () Evaluate the triple integrals: (a) 1 1 x x +y x +y z 3 dzdydx. (b) 1 z y ze y dxdydz. (1) Set up a triple integral but do not evaluate it. (a) (x3 + xy)dv in cylindrical coordinates where is the solid in the first octant that lies beneath the paraboloid z = 1 x y. (b) (optional*) using spherical coordinates for the mass of the icecream cone inside the cone z = x + y and topped off by x + y + z = 1, with density δ(x, y, z) = 4z. Solutions: 1. (a) 4,, (b) θ = cos ( ) = rad = 113 (c) Area = 1 u v = (d) 3x 7y 1z = 6. (a) 9 (b) r (t) = ( cos t, 5, sin t), T (t) = ( cos t, 5, sin t)/ 9, l(t) = (, 5π/4, ) + t(, 5, ) (c) /9. Indeed, κ = dt ds = 1 v dt dt = 1 9 ( sin t,, cos t) / 9 = /9 where v = r (t) = 4 cos t sin t = Using cross product we find N = P Q P is orthogonal to the plane. The equation of the plane is given by the standard form: N x x, y y, z z = where (x, y, z ) can be taken to be any of the three point P, Q, or. Now, P Q = Q P = (, 3, 4) (1,, ) = 3, 4, 4, P = P = (1, 1, 1) (1,, ) =,, 1 i j k N = = 1i + 3j 6k 1 Hence the equation of the plane is 1(x 1) + 3(y 1) 6(z 1) = or 4x + y z = 3 Second method. Let the plane equation be A(x)+B(y)+(z) =. Substituting the coordinates of P and Q into this equation to obtain a relation between A, B, which will tell the three components of N.
5 4. It is easy to check xy x + y x + y Hence 3x y x + y 3 x as (x, y), which agrees with f(, ), and we know that f(x, y) is continuous at (, ). 5 (a) Saddle at (, ), local at (1, 1), (, ). Indeed, Solve { g x = 4x 3 4y = g y = 4y 3 4x = { x 3 y = y 3 x = Sub y = x 3 into x = y 3 to get x = x 9 x =, x = ±1. 6(a) By definition 6 (b) f(x, y) y = lim k f(x, y + k) f(x, y) k x (y + k) x y = lim k k = lim k x = x. = lim k x (k) k f xy = f yx = x y ex/y provided y. u (,,1) [6 (c)] D u f = f (1,1,1) u = (, 3, 1) 5 = 5 7 (a) By definition du = u x dx + u y dy + u z dz x = x + y + z dx + y x + y + z dy + z x + y + z dz (b) If z = f(x, y), then the graph of this function is a surface. At any point (x, y, z ) on the surface, there is a tangent plane whose normal vector is given by N = f x, f y,. Then the equation is given by (c) The hain rule reads: N x x, y y, z z =. u r = u x x r + u y y r + u z z r u s = u x x s + u y y s + u z z s
6 8 (a) Let z = f(x, y) = y x and the region on the xyplane {(x, y) : 1 x + y }. Then the Area of the surface = S = θ=π r= θ= 1 + f x + f y dxdy r= r rdrdθ where f x = x, f y = y. (b) The tetrahedron is bounded on the bottom by y = x, x = and x + y =. The volume V can be evaluated over this triangle OB by a double integral. Here O = (, ), B = (/5, 4/5), = (, 1). Let z = x y, which is the equation of the plane on the top. V = zdxdy = x=/5 x= y=1 x y=x ( x y)dydx. 9. (a) 1/7 (b) perpendicular to the tangent vector at the point, that is F T =, given that F (x, y) = is a level curve or alternatively, orthogonal to the level curve. (c) z is a function of t (d) curve (e) surface (f) line (g) a lies in the plane spanned by the tangential and principle normal of r = r(t) 1 (a) (i) (t) = (t, t ), (t) = (1, t), (t 3 + t(t + t ))dt = =( 3 t t4 ) = 69/4 (t + 3t 3 )dt (ii) (t) = (, 1) + (3, 3)t = ( + 3t, 1 + 3t), (t) = (3, 3), (iii) =3 w = 3( + 3t)(1 + 3t) + (3t + 3t)dt (9t 1 + 6t)dt = 99 w = xdx ( + y)dy = x + (y + y ) 4 1 = 15.
7 b) The parametric equation for the left semicircle is r(t) = ( cos t, sin t), t [ π, 3π ]. So s (t) = v(t) = dr/dt = ( sin t, cos t) =. We have = xy 3 ds = 3π/ π/ 3π/ π/ cos t(sin t) 3 dt =. cos t(sin t) 3 v(t) dt 11 a) check that M y = N x F is conservative. To find f, notice that f x = M, f(x, y) = Mdx = (x cos y y cos x)dx =x cos y y sin x + h(y) To find h(y), taking derivative in y we get y f(x, y) = N(x, y) = x sin y sin x + h (y) = x sin y sin x h (y) = and so h(y) = b) f(x, y) = e x sin(x + 3y) c) It is conservative because: M y = xz + 6y = N x = xz + 6y N z = P y = x 6z P x = M z = xy. In order to find the potential function f such that f = F, that is, f x = M, f y = N, f z = P, we integrate f = Mdx = (xyz + 3y )dx = x (yz) + (3y )x + g(y, z). Then, to recover g(y, z), taking derivative in y both sides we get f y = N or x z + (6y)x + g y(y, z) = x z + 6xy z 3 simplify g y(y, z) = z 3 g = ( z 3 )dy = z 3 y + h(z). It remains to recover h(z). To do that we take derivative in z both sides of f = x (yz) + (3y )x z 3 y + h(z) to obtain f z = P or x y 6z y + h (z) = x y 6yz simplify h (z) = h(z) = constant. Hence, f(x, y, z) = x zy + 3xy z 3 y +.
8 1. The function F has a potential function (antiderivative) f = xyz + z, because f = F. F.T. tells that F dr = f(,, 3) f(1,, ) = a) b a f(r(t)) v(t) dt b) b a F (r(t)) dr = Mdx + Ndy c) Let F = M, N, then F = ( N d) F = x, y M, N = M x x M y )k. ( N x M y )dxdy + N y ( M x + N y )dxdy 14. (a) 64/5 (b) π/1 15*. Find the area of the surface for the part of the sphere x + y + z = 9 that lies within the cylinder (x ) + y = 1 and above the xyplane. [Solution] The surface z = f(x, y) = 9 x y is defined over : (x ) + y = 1. x f x = 9 x y, f y y = 9 x y 1 + fx + fy 9 = 9 x y Use polar coordinates x = r cos θ, y = r sin θ to get Surface Area = 1 + fx + fy da = π/6 r=g(θ) π/6 r=h(θ) 3 9 r rdrdθ where g(θ) = cos θ 4 cos θ 3, h(θ) = cos θ + 4 cos θ 3. Note that 4 cos θ 3 if π/6 θ π/6. 19 (a) Area= 1 xdy ydx (b) The curve has three leaves or loops. Because of symmetry, we look at the one in the first quadrant, bounded by θ = and θ = π/3. Hence π/3 sin 3θ Area of one loop = da = rdrdθ = 1 π/3 (sin 3θ) dθ = 1 4 π/3 (1 cos 6θ)dθ = π (a) Hint: sketch a picture. Use cylindrical coordinates to write the integral as = 1 4 π 1 r π 1 r z 3 dzrdrdθ = π 1 (r 4 r 8 )rdrdθ = π/6. z 4 4 r r rdrdθ
9 (b) 1/(4e). Indeed, 1 z y ze y dxdydz = = (use sub u = y to finish it) 1 1 r 1 z 18. (a) π (r 3 cos 3 θ + r cos θ sin θ)dzrdrdθ (b) Spherical coordinates (x, y, z) = (ρ, φ, θ) x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ yze y dydz where ρ, φ [, π], θ [, π). The side of the cone z = x + y and the zaxis make an angle φ = π/4, which is the upper limit for φ. The sphere has an equation ρ = x + y + z = 1. Hence the mass of the icecream cone π π/4 1 M = δdv = 4zρ sin φdρdφdθ = θ= π π/4 1 φ= 4ρ cos φ ρ sin φ dρdφdθ.
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