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Midterm

# midterm-solutions.pdf

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School
Department
Mathematics
Course
MATH209
Professor
Dragos Hrimiuc
Semester
Fall

Description
Math 209 Midterm | Solutions 1. Either compute the following limit or show that it does not exist: p x y + 1 ▯ 1 lim : (x;y)!(0;0) x + y 2 Solution We have (by passing to polar coordinates x = r cos▯ and y = r sin▯) p x y + 1 ▯ 1 x y 2 1 lim 2 2 = lim 2 2 lim p (x;y)!(0;0) x + y x;y)!(0;0x + y (x;y)!(0;0) x y + 1 + 1 2 2 x y 1 = x;y)!(0;0) + y 2 ▯2 1 r cos ▯ sin ▯ 1 = lim = lim r cos ▯ sin ▯: 2r!0 r 2 2 r!0 By applying the squeeze property 2 2 2 2 0 ▯ r cos ▯ sin ▯ ▯ r ▯! 0 as r ! 0; we obtain p x y + 1 ▯ 1 1 lim 2 2 = ▯ 0 = 0: (x;y)!(0;0) x + y 2 ax+4y 2. (a) Let u(x;y;z) = e cos(5z). Find all values of the constant \a" which make the following true: 2 2 2 @ u + @ u + @ u = 0: @x 2 @y2 @z 2 @z @z y z (b) Find and at the point P : (1;ln2;ln3) ifxe + ye + 2lnx ▯ 2 ▯ 3ln 2 = 0. @x @y Solution (a) Taking derivatives we have: 2 @u = aeax+4ycos(5z) = au; @ u = a @u = a u; @x @x 2 @x 2 @u ax+4y @ u @u @y = 4e cos(5z) = 4u; @y 2= 4 @y = 16u; @u ax+4y @ u 2 ax+4y = ▯5e sin(5z); 2= ▯5 e cos(5z) = ▯25u: @z @z Therfore @ u @ u @ u + + = 0 when (a2+ 16 ▯ 25)u = 0 i.e. when a = ▯3: @x 2 @y2 @z 2 1 Math 209 Midterm | Solutions 2 y z (b) Let F(x;y;z) = xe + ye + 2lnx ▯ 2 ▯ 3ln2. Taking derivatives we have: y 2 Fx(x;y;z) = e + ; Fx(1;ln2;ln3) = 4; x F (x;y;z) = xe + e ; z F (1;ln2;ln3) = 5; y y F zx;y;z) = ye ;z Fz(1;ln2;ln3) = 3ln2: Therefore ▯ ▯ @z ▯ F x1;ln2;ln3) 4 @z ▯ F y1;ln2;ln 3) 5 ▯ = ▯ = ▯ ; ▯ = ▯ = ▯ : @x P F z1;ln2;ln3) 3ln2 @y P F z1;ln2;ln3) 3ln2 3. (a) Find the equation of the tangent plane to the surface x + 2y + 3z = 6 at the point (1;▯1;1). (b) Find the di▯erential du for the function u = xy + xz + yz . Solution (a) Let F(x;y;z) = x + 2y + 3z : 2 Then rF = (F ;Fx;Fy) =z(2x;4y;6z); which implies rF(1;▯1;1) = (2;▯4;6): Thus, the equation of the tangent plane to the surface F(x;y;z) = 6 at the point (1;▯1;1) is (2;▯4;6) ▯ (x ▯ 1;y + 1;z ▯ 1) = 0; that is 2(x ▯ 1) ▯ 4(y + 1) + 6(z ▯ 1) = 0: Simplifying this last equation gives x ▯ 2y + 3z = 6: (b) Note that u = y + z; u = x + z; u = x + y: x y z Thus du = u dx + u dy + u dz x y z = (y + z)dx + (x + z)dy + (x + y)dz: 4. The Temperature at a point (x;y;z) is given by 2 3 T(x;y;z) = 100e ▯x ▯xy▯z +2 (a) Find the rate of change of the temperature at the point (1;0;1) in direction of v = h1;1;1i. (b) In which direction does the temperature increase fastest at P? (c) Find the maximum rate of change of the temperature at P. Solution Math 209 Midterm | Solutions 3 (a) 2 ▯x ▯xy▯z +2 rT(x;y;z) = (▯2x ▯ y;▯x;▯3z )100e rT(1;0;1) = 100(▯2;▯1;▯3) 1 1 1 1 kvk = p u := (p ;p ;p )
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