Solution Midterm 2, Math 53, Summer (a) (10 points) Let f(x, y, z) be a differentiable function of three variables and define

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1 Solution Midterm, Math 5, Summer. (a) ( points) Let f(,, z) be a differentiable function of three variables and define F (s, t) = f(st, s + t, s t). Calculate the partial derivatives F s and F t in terms of the partial derivatives of f. (b) ( points) Compute the tangent plane to the surface z = + at the point (, 6, ). Solution: (a) Using the chain rule F s = f s + f s + f z z s = t f + f + sf z, and F t = f t + f t + f z z t = stf + f f z. Note that each f, f, f z is evaluated at (,, z) = (st, s + t, s t). (b) Letting f(, ) = + the tangent plane has equation Now z = f(, 6) + f (, 6)( ) + f (, 6)( 6). f (, ) = + and f (, ) = + so f(, 6) =, f (, 6) = 5, f (, 6) = 5. Then the equation of the tangent plane is or equivalentl z = + 5 ( ) + ( 6) 5 + 5z = 6.

2 . ( points) Let f(, ) = (a) ( points) Find all critical points of f. (b) ( points) Classif the critical points as local maimum, local minimum or saddle point using the second derivatives test. (a) Setting the partial derivatives equal to zero gives From () we obtain = or =. f = + = ( + ) = () f = = () If = : From () ( + 5) = so = or = 5. We obtain the critical points (, ) and (, 5). If = : From () = so = ± and we get the critical points (, ) and (, ). Critical points: (, ), (, 5 ), (, ), (, ). (b) The second order partial derivatives are f = +, f = +, f = f =. Then D(, ) = ( + )( + ). Evaluating at the critical points D(, ) = >, f (, ) = >. Then (, ) is a local minimum. D(, 5) = >, f (, 5) = <. Then (, 5 ) is a local maimum. D(, ) = 6 <. Then (, ) is a saddle point. D(, ) = 6. Then (, ) is a saddle point.

3 . (a) ( points) Let a be a constant. Evaluate the integral of the function f(, ) = ln(a + + ) over the region D in the plane described b D = {(, ) +, }. Hint: It ma (or ma not) be useful to know that ln d = ln + C. (b) ( points) Calculate ze + +z dv where E is the solid enclosed b the cone z = E + and the plane z =. (a) The integral is π r ln(a + r )drdθ = π r ln(a + r )dr. Substitute s = a + r, ds = rdr to obtain π Using the hint, the value of the integral is r ln(a + r )dr = π +a a ln s ds. π (( + a ) ln( + a ) a ln(a ) ). (b) Using clindrical coordinates, the cone becomes z = r and the integral is E ze + +z dv = = π z ze r +z rdrdzdθ = ze z ze z dz ( e z ) = π ez = π (e e + ) = π (e ). ze z er r=z r= dzdθ

4 . ( points) Let R be the region in the plane bounded b the lines =, = and the hperbola =. Calculate 6 da, using the change of variables u = +, v =. The inverse of the transformation is = u+v, = u v. The Jacobian is (, ) (u, v) = =. R We calculate the image of R in the uv-plane b mapping its boundar. The lines + = and + = map to the lines u = and u = respectivel. For the hperbola 6 = = u + v u v = u v, then u v = which is a hperbola. Soving for v gives v = ± u. The function equals u v. With this the integral is R da = = u (u v) dvdu = u u du u ( u ) / = 8 (5 5 ) = 8 (5 5 ).

5 5. ( points) Let I denote the integral z z ze dddz. (a) ( points) Rewrite the integral in the following orders ddzd, dzdd and dddz. (b) ( points) Evaluate I. (a) For the order ddzd switch the last two variables in the epression for I. This gives z ze ddzd. For the order dzdd switch the and z in the previous epression for I taking as a constant ze dzdd. For the order dddz we can go back to the original epression of I and switch and, (b) Using the last epression from (a) = z = e. z ze dddz = ze z + zdz = ( z z ze dddz. + e z ze ddz = ) = e z z e ddz 5

6 6. ( points) Let f(,, z) = + + z. Find all solutions (,, z) of the sstem of equations coming from minimizing f(,, z) subject to the constraint z = using the method of Lagrange multipliers. Then find the point (or points) where the minimum happens and write what that minimum value is. Let g(,, z) = z, so that the restriction is g(,, z) =. Then f =,, z and g =, z,. The sstem of eqautions for the method of Lagrange multipliers is f = λ g, = λ () = λz () z = λ () z = () Using () and () we obtain = λ, that is ( λ ) = from where = or λ = or λ =. We stud each case. Case : =, then from (), z =. From (), =, so = /. We obtain the point ( /,, ). Case : λ =, then from (), = and from (), = z. For = we have that either = or =. Subcase : = and = z. From (), =, so = ± = z and we obtain the points (,, ), (,, ). Subcase : = and = z. From (), =, so = ± points (,, ), (,, ). = z and we obtain the Case : λ =. Then from (), = and from (), = z. For = we have that either = or =. In either case, from () we obtain = for the case = and = for the case =, none of which has a solution. There are five solution to the sstem of equations: ( /,, ), (,, ), (,, ), (,, ), (,, ). Evaluating the function f(,, ) =, f(,, ) =, f(,, f( /,, ) = /. ) = 7, f(,, The minimum is attained at (,, ) and (,, ) and the minimum value is. ) = 7, 6

7 7. ( points) If ou take the circle ( ) + z = in the z-plane and rotate it about the z-ais, the resulting surface is called torus. Its equation in spherical coordinates is ρ = sin φ. The surface of equation ρ = cos φ is a sphere. (a) ( points) Convert the equation of the sphere ρ = cos φ to cartesian coordinates and identif its radius and center. (b) (6 points) Calculate the mass of the solid E that is inside the sphere ρ = cos φ and outside the torus ρ = sin φ if the densit equals σ(,, z) = + + z. (a) Using that ρ cos φ = z we get ρ = z that is ρ ρ = z. Then + + z = z. Completing square gives + + ( z ) =, a sphere of radius centered at (,, ). (b) We see that the angle θ moves from to π. To find the range of φ we find the intersection of ρ = sin φ and ρ = cos φ, that is we set sin φ = cos φ giving φ = π. The description of E in spherical coordinates is E = {(ρ, φ, θ) θ π, φ π, sin φ ρ cos φ}. The densit in spherical coordinates is σ(ρ, φ, θ) =. The total mass m is ρ m = = π = π σdv = E cos φ sin φ cos φ sin φ ρ sin φdθdρdφ = π (cos φ sin φ) sin φdφ = π ( = π ) π cos φ + cos φ = π ( ). ρ ρ sin φdθdρdφ ρ cos φ sin φdφ sin φ ( cos φ ) sin φdφ 7