Download Math 241 Midterm Exam 1, Spring 2014 - Prof. Stephen T. Avsec and more Exams Advanced Calculus in PDF only on Docsity!
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Section and TA name:
READ ALL INSTRUCTIONS CAREFULLY. Write legibly, and use the boxes and/or insert check marks ✓□ for
your final answer where provided.
Be sure to use correct notation; in particular, distinguish vectors from scalars by arrow notation, use explicit
clearly visible dots for dot products, etc.
An answer alone, without justification, will not earn full credit (with the exception of the two multiple
choice problems 3 and 4). If you make a mistake, cross out all of your incorrect work. We will take points
off for incorrect work that is not crossed out, even if the correct answer is given elsewhere on the page.
Problem Point Value Test Score
Total 50
- (4 + 2 points) Show your work.
(a) Find a scalar equation of the plane through the three points P(2; 4; − 3 ), Q(5; 6; − 3 ), and R(3; 3; − 2 ).
The two vectors →
PQ = ⟨3; 2; 0⟩
and →
PR = ⟨1; −1; 1⟩
lie in the plane through the three given points. (Answers may vary.) Thus their cross product is
a normal vector to that plane:
n =
PQ
PR = ⟨3; 2; 0⟩ ⟨1; −1; 1⟩ =
k
k = 2 ⃗ { − 3 ⃗ ȷ − 5
k
The scalar equation of a plane is
a(x − x 0 ) + b(y − y 0 ) + c(z − z 0 ) = 0;
where n = ⟨a; b; c⟩, and (x 0 ; y 0 ; z 0 ) is any point in the plane (for example, the point P).
x − (^2)
y − (^4)
z − − 3
(b) Is the line x = 4 + t, y = 1 − 6t, z = 4t parallel to the plane in part a?
If not, what is the point of intersection?
(If you could not solve part a, use the vector n = ⟨4; 3; 2⟩ for part b.)
Direction vector of the line: v = ⟨1; −6; 4⟩
Did get Part (a):
n v = ⟨2; −3; − 5 ⟩ ⟨1; −6; 4⟩ = 2 + 18 − 20 = 0
Thus n is perpendicular to v , and v is parallel to the plane.
Did not get Part (a):
n v = ⟨4; 3; 2⟩ ⟨1; −6; 4⟩ = 4 − 18 + 8 ̸= 0
Thus n is not perpendicular to v , and v is not parallel to the plane.
4 (x − 2 ) + 3 (y − 4 ) + 2 (z + 3 ) = 0
4 (( 4 + t) − 2 ) + 3 (( 1 − 6t) − 4 ) + 2 ((4t) + 3 ) = 0
4 ( 2 + t) + 3 (− 3 − 6t) + 2 (4t + 3 ) = 0
8 + 4t − 9 − 18t + 8t + 6 = 0
5 + −6t = 0
t = 5=
Point of intersection: ( 4 + 5=6; 1 − 6 (5=6); 4(5=6)) = (29=6; −4; 20=6)
Point of intersection: PARALLEL (enter PARALLEL if the line is parallel to the plane)
- (6 points) Match the functions with their graphs and with their level curves. Each level curves diagram
consists of level curves where the function is constant drawn for evenly spaced constants k.
(a) (1 + 1 point) f(x; y) = y
2 − x
2
(b) (1 + 1 point) g(x; y) = cos(x + y)
(c) (1 + 1 point) h(x; y) = x
2 e
−y
2
Each graph and level curves diagram below is labeled with the corresponding
line number in your
Scantron bubble sheet. Pencil in your answers using the following key:
A f(x; y) = y
2 − x
2 ,
B g(x; y) = cos(x + y),
C h(x; y) = x
2 e
−y
2
, or
D none of the above
(Option E is not a valid answer in this problem, and if marked, will automatically result in a score of
zero in that line.) Also label the graphs/diagrams in case the scanner is unable to read your answers.
D D D
B (^) B D
A (^) D C
C D^ A
- (4 points) Does
lim
(x,y)→(0,0)
(x + y)
2
x 2
exist? If yes, compute it. If not, show why it does not exist.
Enter the limit in the box below, or write DNE if the limit does not exist. You must show your work
to receive credit for this problem.
Along the line x = 0 :
lim
(x,y)→(0,0)
(x + y)
2
x
2
2
= lim
y→ 0
y
2
y
2
= lim
y→ 0
Along the line y = x:
lim
(x,y)→(0,0)
(x + y)
2
x
2
2
= lim x→ 0
(x + x)
2
x
2
2
= lim x→ 0
4x
2
2x
2
= lim x→ 0
(Answers may vary: you may choose limits along different curves to solve this problem.)
These two limits do not coincide, so the limit does not exist.
lim (x,y)→(0,0)
(x + y)
2
x
2
2
= DNE
- (6 points) Suppose that f(x; y) is a differentiable function of the two variables x and y, and has the
table of values shown below. Further suppose that x = s
2 − t and y = 2st in turn are differentiable
functions of the two variables s and t.
(x; y) f(x; y) f x (x; y) f y (x; y)
Let F(s; t) = f(x(s; t); y(s; t)). Use the table of values to calculate the partial derivative
@F
@t
You must show your work to receive credit for this problem.
x(2; 1) = 2
2 − 1 = 4 − 1 = 3
y(2; 1) = 2 ( 2 )( 1 ) = 4
@x
@t
@y
@t
= 2s
@F
@t
@f
@x
@x
@t
@f
@y
@y
@t
@F
@t
- (3 points) Find a scalar equation of the tangent plane to the hyperboloid
x
2
2 −
z
2
at the point (−4; 1; 3).
Pencil in your answer from the five choices below into line
17 of your Scantron bubble sheet:
A − 2 (x − 4 ) + 6 (y + 1 ) − 1 (z + 3 ) = 0
B − 4 (x + 4 ) + 1 (y + 3 ) + 3 (z −
3
2
C 4 (x + 4 ) + 3 (y − 1 ) − 6 (z − 3 ) = 0
D − 2 (x + 4 ) + 6 (y − 1 ) − 1 (z − 3 ) = 0
E 4 (x − 4 ) + 3 (y + 1 ) − 6 (z + 3 ) = 0
Also mark the correct answer ✓□ among the five choices above in case the scanner is unable to read
your answer.
Show your work.
You may use the space below for your computations.
f(x; y; z) =
x
2
2 −
z
2
∇f(x; y; z) =
2x
; 6y; −
2z
x
; 6y; −
z
∇f(−4; 1; 3) = ⟨−2; 6; − 1 ⟩ = ⟨a; b; c⟩
Then use the same format for the scalar equation of a plane as in problem 1.
- (5 points) Find the maximum value M and the minimum value m of the function f(x; y) = 3x + 4y
subject to the constraint x
2
2 = 13.
∇f(x; y) = ⟨3; 4⟩ and ∇g(x; y) = ⟨2x; 8y⟩ where g(x; y) = x
2
2
∇f = ∇g =⇒ ⟨3; 4⟩ = ⟨2x; 8y⟩
3 = 2x
4 = 8y
x
2
2 = 13
x; y; ̸= 0 by first two equations; solve for and then for x:
2x
8y
=⇒ 3y = x
(3y)
2
2 = 13
13 = 9y
2
2 = 13y
2
Thus: y = 1 and so x = 3 , or y = − 1 and so x = − 3.
Points: (3; 1) and (−3; − 1 )
Maximum: f(3; 1) = 3 ( 3 ) + 4 ( 1 ) = 9 + 4 = 13
Minimum: f(−3; − 1 ) = 3 (− 3 ) + 4 (− 1 ) = − 9 − 4 = − 13
Maximum value M = 13 at the point(s): (3; 1)
Minimum value m = − 13 at the point(s): (−3; − 1 )
