Math 211, Multivariable Calculus, Fall 2011 Final Exam Solutions

Math 211, Multivariable Calculus, Fall 2011 Final Exam Solutions 1. (10 points) Find the equation of the plane that contains both the point (−1, 1, 2) and the line given by x = 1 − t, y = 1 + 2t, z = 2 − 3t. Solutions: A point on the line is (1, 1, 2) and a vector parallel to the line is h−1, 2, −3i. Another vector parallel to the plane we want is h2, 0, 0i (the vector between the two points we have) so a vector perpendicular to the plane is given by n = h2, 0, 0i × h−1, 2, −3i = h0, 6, 4i . The equation of the plane is therefore 6(y − 1) + 4(z − 2) = 0 or 6y + 4z = 14 . 2. (10 points) Consider the function f (x, y) = x2 (y − 1). (a) What is the directional derivative of f at (1, 3) in the direction of the vector v = h3, −4i? Solution: ∇f = h2x(y − 1), x2 i, so ∇f (1, 3) = h4, 1i. The directional derivatives is therefore 8 h4, 1i · h3, −4i 8 =√ = . | h3, −4i | 5 25 (b) What is the maximum directional derivative of f at (1, 3), and in which direction does it occur? Solution: The maximum directional derivative is the length of the gradient vector √ 17 and it is in the direction of the gradient vector h4, 1i . 3. (10 points) Find the linear approximation to the function f (x, y) = 2 − sin(−x − 3y) at the point P = (0, π), and then use your answer to estimate f (0.001, π). Solution: The linear approximation is given by l(x, y) = f (0, π) + fx (0, π)(x − 0) + fy (0, π)(y − π). We have fx = cos(−x − 3y), so fx (0, π) = cos(−3π) = −1 and fy = 3 cos(−x − 3y), so fy (0, π) = 3 cos(−3π) = −3. Also f (0, π) = 2 − sin(−3π) = 2. Therefore l(x, y) = 2 − (x − 0) − 3(y − π) . So we have f (0.001, π) ≈ 1.999 . 4. (5 points) Prove that, for any curve described by a vector-valued function r(t), the unit tangent vector T(t) is always orthogonal to its derivative T0 (t). Solution: The unit tangent vector is a unit vector so T(t) · T(t) = 1. Differentiating both sides we get T0 (t) · T(t) + T(t) · T0 (t) = 0 so T0 (t) · T(t) = 0. Therefore T0 (t) is perpendicular to T(t). 5. (10 points) Let C be the curve given by r(t) = (cos t + t sin t)i + (sin t − t cos t)j, for t > 0. Find the unit tangent vector T(t), unit normal vector N(t), and curvature κ(t) for C. (Your answers should be functions of t.) Solution: We have r0 (t) = ht cos t, t sin ti so |r0 (t)| = p t2 cos2 t + t2 sin2 t = √ t2 = t since t > 0. Therefore the unit tangent vector is T(t) = r0 (t) = hcos t, sin ti . |r0 (t)| Differentiating this we get T0 (t) = h− sin t, cos ti . Then N(t) = T0 (t) h− sin t, cos ti = = h− sin t, cos ti . 0 |T (t)| 1 The curvature is then |T0 (t)| 1 κ(t) = 0 = . |r (t)| t 6. (5 points) Show that the function  2 2 x − y f (x, y) = x2 + y 2  0 is not continuous at (0, 0). if (x, y) 6= (0, 0); if (x, y) = (0, 0); Solution: We have to show that the limit of f (x, y) as (x, y) → (0, 0) is not equal to f (0, 0) = 0. But if (x, y) approaches (0, 0) along the line y = 0, then we have f (x, y) = f (x, 0) = x2 =1 x2 so the limit cannot be equal to 0. So f is not continuous at (0, 0). 7. (10 points) Let g be the function given by  2  xy g(x, y) = x2 + y 2  0 if (x, y) 6= (0, 0); if (x, y) = (0, 0). (a) Calculate the partial derivatives of g at (0, 0). Solution: We use the limit definition: g(h, 0) − g(0, 0) 0−0 = lim = 0 h→0 h→0 h h gx (0, 0) = lim and g(0, h) − g(0, 0) 0−0 = lim = 0 h→0 h→0 h h gy (0, 0) = lim (b) Show that g is not differentiable at (0, 0). Solution: If g were differentiable the linear approximation at (0, 0) would be l(x, y) = g(0, 0) + gx (0, 0)(x − 0) + gy (0, 0)(y − 0) = 0. So to show that g is not differentiable at (0, 0), we have to show that the limit g(x, y) − 0 p (x,y)→(0,0) x2 + y 2 lim is not equal to 0. The function in this limit is x2 y . (x2 + y 2 )3/2 Along the line y = x, this is equal to x3 x3 ±1 = = 3/2 . 2 3/2 3/2 3 (2x ) 2 |x | 2 (The plus or minus depends on whether x is positive or negative.) This certainly does not approach 0 as (x, y) approaches (0, 0) so the overall limit cannot be 0. Therefore g is not differentiable at (0, 0). 8. (10 points) Find the maximum and minimum values of the function f (x, y) = 4x + 4y − x2 − y 2 subject to the condition x2 + y 2 ≤ 2. Solution: We have to check the interior for critical points of f . We have ∇f = h4 − 2x, 4 − 2yi so the only critical point of f is (2, 2) which does not satisfy x2 + y 2 ≤ 2. The max and min (which must exist by the Extreme Value Theorem) therefore occur on the boundary x2 + y 2 = 2. We use the Lagrange multiplier method with g(x, y) = x2 + y 2 so ∇g = h2x, 2yi. This is only zero at (0, 0) which is not on the boundary, so the only cases we have to look at is then ∇f = λ∇g that is 4 − 2x = 2xλ, 4 − 2y = 2yλ. Therefore (4 − 2x)y = 2xyλ = (4 − 2y)x so 4y = 4x and so y = x. Substituting into the constraint, we get 2x2 = 2 and so x = ±1. The constrained critical points are therefore (1, 1) and (−1, −1). Evaluating f at each of these we get f (1, 1) = 6 and f (−1, −1) = −10, so the maximum value is 6 and the minimum value is −10 . p 9. (10 points) Find the volume of the solid region bounded by the cone z = 3x2 + 3y 2 and the sphere x2 + y 2 + z 2 = 9. √ Solution: In the xz-plane (y = 0), the cone is given by z =√ 3|x|. The angle between each of these lines and the z-axis is π6 since tan(π/6) = 1/ 3. The region is therefore given in spherical coordinates by 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π, 0≤φ≤ π . 6 The volume is given by integrating the function 1 which gives Z ρ=3 ρ=0 Z θ=2π θ=0 Z φ=π/6 ρ2 sin φ dφ dθ dρ φ=0 which equals φ=π/6 (9)(2π) [− cos φ]φ=0 = 18π(1 − √ 3/2) . 10. (10 points) Let R be the parallelogram with vertices (0, 0), (1, 1), (2, −1) and (3, 0). Use the change of variables x = u + 2v, y =u−v to evaluate the integral ZZ (x + 2y)2 e(x−y) dA. R Solution: The Jacobian is given by ∂(x, y) = (1)(−1) − (2)(1) = −3. ∂(u, v) The edges of the new region are found by y=x y = −x/2 y = −(x − 3)/2 y =x−3 u−v u−v u−v u−v = u + 2v v=0 = −u/2 − v u=0 = −u/2 − v + 3/2 u=1 = u + 2v − 3 v=1 so the new region is the square in the uv-plane given by 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. The new integral is therefore Z u=1 Z v=1 (3u)2 e3v | − 3| dv du u=0 which is Z v=0 u=1 9 u2 (e3 − 1) du = 3(e3 − 1) . u=0 11. (5 points for each part) Let C be the semicircle given by x2 + y 2 = 4 and y ≥ 0 oriented from left to right (with respect to the usual orientation of the x- and y-axes). Calculate each of the following integrals: Z (a) 1 ds C Solution: This integral is equal to the length of the curve which is half the length of a circle of radius 2, so is 2π . Z (b) F · dr where C F(x, y) = hx, 1 + yi Solution: We can parameterize the curve with r(t) = h2 cos t, 2 sin ti , 0 ≤ t ≤ π. This has the wrong orientation so the integral we want is Z t=π − F(r(t)) · r0 (t) dt t=0 which is Z t=π h2 cos t, 1 + 2 sin ti · h−2 sin t, 2 cos ti dt − t=0 which is Z t=π −4 cos t sin t + 2 cos t + 4 cos t sin t dt − t=0 which is Z t=π − 2 cos t dt t=0 which is [−2 sin t]t=π t=0 = 0 . Z (c) F · dr where F is the gradient vector field of the function f (x, y) = ln(x2 ) C Solution: The curve C goes from (−2, 0) to (2, 0) so by the Fundamental Theorem of Calculus, the integral is equal to f (2, 0) − f (−2, 0) = 0 . 12. (5 points for each part) Decide if each of the following vector fields F is conservative or not, justifying your answers. For each F that is conservative, find a function f such that F = ∇f . (a) F(x, y) = hxey , yex i Solution: We have ∂Q = yex , ∂x ∂P = xey . ∂y These are not equal so the vector field F is not conservative .   2x2 − 1 (b) F(x, y) = 4x ln(y), , for y > 0. y = ∂P and the domain of F is simply-connected so F is Solution: Here ∂Q ∂x ∂y conservative . We can find a suitable f by ‘partial integration. Since fx = 4x ln y we have f (x, y) = 2x2 ln y + g(y) for some function g(y). Since we want fy = (2x2 − 1)/y we must have 2x2 /y + g 0 (y) = (2x2 − 1)/y so g 0 (y) = −1/y so, for example, g(y) = − ln(y). Therefore a possible function f is given by f (x, y) = 2x2 ln y − ln y .   y −x (c) F(x, y) = , for (x, y) 6= (0, 0) , x2 + y 2 x2 + y 2 Solution: Here ∂Q = ∂P but the domain of F is not simply-connected so we cannot ∂x ∂y tell this way if F is conservative. (This was an example from class.) Instead look at the integral of F around the circle C given by x2 + y 2 = 1, say oriented clockwise. Sketching F you can see that it is always parallel to C pointing in the same direction. The integral of F around C is therefore positive. This tells you that F is not conservative . 13. (10 points) Let C be the boundary of the quadrilateral with vertices (1, 1), (1, 2), (2, 3) and (2, 1), oriented clockwise. Let F be the vector field given by D 4 E x 2 F(x, y) = e + y , xy + sin(ln(y)) . Calculate Z F · dr. C Solution: Use Green’s Theorem. This tells us that the integral is equal to ZZ ∂Q ∂P − − dA ∂y R ∂x (minus because C is oriented clockwise instead of counter-clockwise). Here R is the quadrilateral which can be described by 1 ≤ x ≤ 2, We have ∂Q ∂x − ∂P ∂y 1 ≤ y ≤ x + 1. = y − 2y = −y so the integral becomes Z x=2 Z y=x+1 − (−y) dy dx. x=1 y=1 This equals Z x=2  x=1 which is 1 2 Z x=2 (x + 1)2 − 1 dx x=1 which is 1 2 which is y=x+1 y 2 /2 y=1 dx Z x=2 x2 + 2x dx x=1  3 x=2 x /6 + x2 /2 x=1 which is 7/6 + 3/2 or 16 . 6