McGill University April Calculus 3. Tuesday April 29, 2014 Solutions

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1 McGill University April 4 Faculty of Science Final Examination Calculus 3 Math Tuesday April 9, 4 Solutions Problem (6 points) Let r(t) = (t, cos t, sin t). i. Find the velocity r (t) and the acceleration r (t). ii. Find the tangential and normal components of the acceleration Hint: you can use the formulas sin(α) = sin α cos α and cos(α) = cos α sin α. Solution: We find that r (t) = (, cos t sin t, sin t, cos t) = (, sin(t), sin(t)). Next r (t) = (, cos(t), cos(t)). We note that r (t) = + sin (t). The tangential component of the acceleration is given by the formula a T = r (t) r (t) r (t) = 4 sin(t) cos(t) + sin (t) = sin(4t) + sin (t). The normal component of the acceleration is a N = r (t) r (t) r (t) = ( )(, cos(t), cos(t)) + sin (t) = cos(t) + sin (t) Problem (6 points) i. Write the Taylor series about x = of ln( + x). You do not need to justify your answer. ii. Use part i. to write the Taylor series about x = of ln( + x 3 ). iii. Write the Taylor series about x = of f(x) = x ln( + t3 )dt.

2 Math Final Exam Page April 9, 4 Solution: We have ln( + x) = n= ( )n x n /n = x x / + x 3 /3 ±.... It follows that ln( + x 3 ) = n= ( )n x 3n /n = x 3 x 6 / + x 9 /3 ±.... Integrating term by term, we get x ln( + t 3 )dt = Problem 3 (6 points) Let f(x, y, z) = x cos(y)z 3. n= ( ) n x 3n+ n(3n + ) = x 4 /4 x 7 /4 + x /3 ±... i. Find the gradient f(p) at the point p = (, π, ). In which direction does the function f increase the most? ii. Find the directional derivative u f(p) where u is the vector (,, ). iii. Let S be the level surface of f passing through (, π, ). Find an equation for the tangent plane to S at (, π, ). Solution: We have f = (f x, f y, f z ) = (x cos(y)z 3, x sin(y)z 3, 3x cos(y)z ). At the point p = (, π, ), f = (,, 3). The function f increases most in the direction f(p) = (,, 3). The directional derivative is equal to Uf(p) = f(p) u = (,, 3) (,, ) = 3 =. The equation of the tangent plane at the point p = (x, y, z ) is (x x, y y, z z ) f(p) =. In our case it becomes or x 3z = 5. = (x, y π, z + ) (,, 3) = x 3z 3 = x 3z 5, Problem 4 (6 points) Let x, y, z satisfy an equation: x y + z z = 4. Find z/ x and z/ y. Solution: ifferentiating with respect to x and using implicit differentiation, we find that x + z z x z x =. It follows that z x = x/( z). ifferentiating with respect to y and using implicit differentiation, we find that y + z z y z y =. It follows that z y = y/(z ). Problem 5 (6 points) Find all local maxima, local minima, and saddle points of the function in the region {(x, y) : y < π}. f(x, y) = x cos y x 3 /3

3 Math Final Exam Page 3 April 9, 4 Solution: At a critical point, f = (cos y x, x sin y) = (, ). Since x sin y =, either x = (case ); or sin y =, which for y < π only happens when y = (case ). In case, we cos y =, so cos y =, y = ±π/, so we get points (, π/) and (, π/) (we have used that y < π). In case, we have cos y = cos =, so x =, x = ±; we get points (, ) and (, ). The Hessian of f is equal to ( ) x sin y sin y x cos y At the points (, ±π/) the determinant is equal to sin (±π/) =, so we get saddle points by the Hessian criterion. At the point (, ), the Hessian is ( ), so that point is a local maximum. At the point (, ), the Hessian is ( so that point is a local minimum. Problem 6 (6 points) Use the method of Lagrange multipliers to find the minimum value of the function subject to the constraint g(x, y) = xy = 4. ), f(x, y) = (x 3 + y 3 )/3 Solution: The hyperbola g(x, y) = xy = 4 has two branches: on the first branch x >, y >, and so f(x, y) >, and also it is easy to see that f(x, y) as either x or y ; we shall restrict ourselves to that branch. On the nd branch, x <, y <, and it is easy to see that f(x, y) as either x or y ; so the minimum of f restricted to that branch is. On the first branch, we define F (x, y) := f λg = (x 3 + y 3 )/3 λ(xy 4). We have F x = x λy =, F y = y λx =, F λ = xy 4 =. If λ =, then from the first equation we have x =, and so the constraint equation xy = 4 is not satisfied, therefore λ. ividing the first two equations, we get x /y = y/x, or x 3 = y 3. This implies x = y. Substituting into the last equation, we find x = y =. The minimum is f(, ) = 6/3.

4 Math Final Exam Page 4 April 9, 4 Problem 7 (6 points) Use polar coordinates to find the volume of the region bounded by the paraboloids z = 3x + 3y and z = 4 x y. Solution: We use polar coordinates (r, θ) in the (x, y)-plane, or, equivalently, cylindrical coordinates in R 3. The equations of the paraboloids become z = 3r and z = 4 r. They intersect above the circle 4 r = 3r or r =. The paraboloid z = 3r lies below the paraboloid z = 4 r, so the limits of integration for z will be 3r r z 4 r. The volume is given by r= π 4 r θ= z=3r dzrdrdθ = π (4 4r )rdr = 8π[r / r 4 /4] = π. Problem 8 (6 points) Let be the bounded region of the plane which is enclosed by the curves y =, y = x and x =. Evaluate the following double integral: x sin y da. Solution: The limits of integration are x [, ], y [, x ]. The integral is x= x y= x sin ydydx = x[ cos(y)] x dx = x( cos(x ))dx = x cos(x )dx. To compute the last integral, we note that (d/dx)( sin(x )/) = x cos(x ), therefore the previous expression becomes (d/dx)(sin(x )/)dx = / [sin(x )/] = sin(). Problem 9 (6 points) Use spherical coordinates to evaluate the triple integral xyz dv, where is the region lying between the spheres of radius ρ = and ρ = 4, and above the cone φ = π/3. Solution: The limits of integration are φ [, π/3], ρ [, 4], θ [, π]. The function xyz = ρ 3 sin φ cos φ sin θ cos θ = ρ 3 sin φ sin(θ)/. The integral is 4 π 3 π ρ 3 sin φ cos φ sin(θ) ρ sin φdθdφdρ = 4 π ρ 5 3 π dρ sin 3 φ cos φdφ sin(θ) dθ. The last integral is equal to [ cos(θ)/4] π =, hence the product of the three integrals is also equal to. Answer:.

5 Math Final Exam Page 5 April 9, 4 Problem (6 points) Use the transformation x = u/v, y = v to evaluate the double integral xy da, where is the region in the first quadrant bounded by the lines y = x and y = 3x and the hyperbolas xy = and xy = 3. Solution: The integrand xy = u in the new coordinates. The limits of integration are xy = u [, 3]; straight lines y = kx become v = ku/v or u = v /k, so v /u [, 3] or equivalently v [ u, 3u]. The Jacobian is ( ) /v u/v det = /v. The integral is equal to u= 3u v= u u v dvdu = u[ln v] 3u u du = u (ln 3 + ln u ln u) du = ln 3 udu That expression is equal to ln 3(9 ) = ln 3.