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A set of problems and solutions related to multivariable calculus, including topics such as parallelogram area, matrix transformation, vector products, volume of parallelepiped, projections, level curves, cylindrical and spherical coordinates, particle motion, removable discontinuities, and planes perpendicular to lines. The exam is divided into five problems, each with a different score.
Typology: Exams
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EXAM I - SEPTEMBER 29, 2006
Instruction: Read each question carefully. Explain ALL your work and give reasons to support your answers. Advice: DON’T spend too much time on a single problem.
Problems Maximum Score Your Score
1
2 EXAM I - SEPTEMBER 29, 2006
(5 pts) (i) Find the area of P. Let a = (− 1 , 4) − (2, 1) = (− 3 , 3) and b = (6, 3) − (2, 1) = (4, 2). Then the area of P is equal to ||a × b||=18. (3 pts) (ii) Suppose T (x 1 , x 2 ) = (5x 1 + 4x 2 , 5 x 1 + 3x 2 ). Find the associated matrix A such that T (x) = Ax where x = (x 1 , x 2 ). The associated matrix is
A =
(4 pts) (iii) What are the vertices of the image T (P )? The vertices of T (P ) are (11, 7), (14, 13), (42, 39), and (39, 33). (4 pts) (iv) What is the area of T (P )? Area of T (P ) is given by | det A| · area of P = |(5)(3) − (5)(4)| · 18 = 90.
(4 pts) (v) What is the angle of the parallelogram P at the vertex (6, 3)? (You may express it in terms of inverse trig function.) Since ||a × b|| = 18 = (||a||)(||b||) sin θ. It follows that θ = arcsin
√ 10
4 EXAM I - SEPTEMBER 29, 2006
level curve of f(x,y)=−
level curve of f(x,y)= 3
level curve of f(x,y)= −5 is empty
(5 pts) (ii)Describe or sketch the set of points in R^3 that satisfy the equation f (x, y) = x^2 + 2y^2 − 1 (or the graph of z = f (x, y))
elliptic paraboloid
(5 pts) (iii)What are the cylindrical coordinates of the point (1, 2 , 8)? In cylindrical coordinates, we have r =
x^2 + y^2 =
5 and tan θ = y/x so that θ = arctan 2. The point (1, 2 , 8) becomes (
5 , arctan 2, 8) in cylindrical coordinates. (5 pts) (iv) Write the equation z = x^2 + 2y^2 − 1 in spherical coordinates. In spherical coordinates, x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. Thus the equation in spherical coordinates becomes ρ cos φ = (ρ sin φ)^2 cos^2 θ+ 2 ρ^2 sin^2 φ sin^2 θ − 1.
MATH206A MULTIVARIABLE CALCULUS - PROF. P. WONG 5
v(t) = i + (1 + t)j + cos tk.
(10 pts) (i) Find the position r(t) of the particle with the initial point r(0) = i + k. Note that r(t) =
v(t) dt
= (t + C 1 )i + (t + t
2 2 +^ C^2 )j^ + (sin^ t^ +^ C^3 )k where C 1 , C 2 , and C 3 are constants. Since r(0) = i + k, it follows that C 1 = 1, C 2 = 0, and C 3 = 1. Hence,
r(t) = (t + 1)i + (t + t
2 2 )j^ + (sin^ t^ + 1)k.
(10 pts) (ii) Give an equation (in vector form) of the line tangent to the path of the particle at the point r(π). The vector v(π) = i + (1 + π)j − k is in the same direction as the desired tangent line. Hence, the tangent line to the path at r(π) is given by
x(t) = r(π) + tv(π)
= (π + 1)i + (π + π
2 2 )j^ +^ k^ +^ t[i^ + (1 +^ π)j^ −^ k]
