Study Guides (248,234)
Canada (121,419)
Mathematics (143)
MATH209 (14)
Midterm

midterm-solutions.pdf

4 Pages
384 Views
Unlock Document

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 )
More Less

Related notes for MATH209

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit