MTH2010-MTH2015 - Multivariable Calculus - S2 2021
Started on Saturday, 23 October 2021, 9:07 PM
State Finished
Completed on Saturday, 23 October 2021, 9:08 PM
Time taken 1 min 41 secs
Grade 0.00 out of 60.00 (0%)
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PLEASE READ ALL INSTRUCTIONS CAREFULLY.
There are 60 marks available and the exam is worth 60% of your final unit mark.
For each problem, you will be required to select a response from a drop-down menu, to select responses from various options, or to
input a number into a box. Numerical answers are integers, possibly with a negative sign in front, unless otherwise specified. Please
enter as "5" or "-5" as required.
You are permitted and encouraged to use pen or pencil and blank sheets of paper for your working out.
Symmetry of second derivatives. If has continuous second partial derivatives, then
Second derivatives test. At a critical point of where , let . If and , then has a
local minimum; if and , then has a local maximum; and if , then has a saddle point.
Lagrange multipliers. To maximise/minimise subject to the constraint , solve
Polar coordinates. Cartesian and polar coordinates are related by
Line integrals of functions. For a parametrisation for of a curve in and a function ,
Line integrals of vector fields. For a parametrisation for of a curve in and a vector field
,
f(x,y)
= .
f∂
2
∂x∂y
f∂
2
∂y∂x
f(x,y) ∇f = 0 D= −f
xx
f
yy
f
2
xy
D> 0 > 0f
xx
f
D> 0 < 0f
xx
f D< 0 f
f(x) g(x) = 0
∇f(x) = λ∇g(x).
x= rcosθ and y= rsinθ.
r(t) = (x(t),y(t)) a≤ t≤ b C R
2
f(x,y)
f(x,y)ds= f(x(t),y(t)) dt.∫
C
∫
b
a
(t + (tx
′
)
2
y
′
)
2
− −−−−−−−−−−
√
r(t) = (x(t),y(t)) a≤ t≤ b C R
2
F(x,y)
d ( )d∫ ∫
b
′
Question 1
Not answered
Marked out of 2.00
Question 2
Not answered
Marked out of 2.00
Fundamental theorem for line integrals. For a parametrisation for of a curve and a function ,
Surface integrals of functions. For a parametrisation for of a surface in and
a function ,
Surface integrals of vector fields. For a parametrisation for of a surface in
and a vector field ,
Green's theorem. For a region in the plane and functions and ,
Stokes' theorem. For a surface in space and a vector field ,
Divergence theorem. For a solid region in space and a vector field ,
F ⋅ dr= F ⋅ (t)dt.∫
C
∫
a
r
′
r(t) = (x(t),y(t)) a≤ t≤ b C f
∇f ⋅ dr= f(r(b))− f(r(a)).∫
C
r(u,v) = (x(u,v),y(u,v),z(u,v)) (u,v) ∈D S R
3
f(x,y,z)
f(x,y,z)dS = f(x(u,v),y(u,v),z(u,v)) | × |dA.∬
S
∬
D
r
u
r
v
r(u,v) = (x(u,v),y(u,v),z(u,v)) (u,v) ∈D S R
3
F(x,y,z)
F(x,y,z) ⋅ dS= F(x(u,v),y(u,v),z(u,v)) ⋅ ( × )dA.∬
S
∬
D
r
u
r
v
D P(x,y) Q(x,y)
( − )dA= P dx+Qdy.∬
D
∂Q
∂x
∂P
∂y
∫
∂D
S F(x,y,z)
curl F ⋅ dS= F ⋅ dr.∬
S
∫
∂S
E F(x,y,z)
div FdV = F ⋅ dS.
∭
E
∬
∂E
Consider the following two lines in :
and
(a) The two lines intersect at
( , , ).
(b) If is the angle between these two lines, then . You may enter your answer as a decimal.
R
3
{(t,1+2t,2−2t)}
{(1+4s,3− s,s)}.
θ cosθ=
Consider the vector function given by
and let be the corresponding space curve.
r :R→R
3
r(t) = (2 + t,t+1, −2),t
2
t
2
C
Question 3
Not answered
Marked out of 3.00
(a) The plane given by intersects at the points
(1, 0, -1) and ( , , ).
(b) At the point (3, 2, -1), the vector
( , 1 , )
is tangent to C.
2x+ y−3z= 5 C
Match the following functions with their contour maps below.
Contour map A
Contour map B
Contour map C
xy+ +1y
2
8 −8x
3
y
3
+xyx
2
y
2
x− y
+ +2x
2
y
2
cosx+siny
+ sin
x
2
y
2
pContour map D
Contour map E
Contour map F
Question 4
Not answered
Marked out of 4.00
Question 5
Not answered
Marked out of 4.00
(a) Consider the limit . Which one of the following statements is correct?
The limit does not exist.
The limit is infinite.
The limit exists and is equal to .
The limit exists and is equal to .
None of the previous statements is correct.
(b) Consider the limit . Suppose that you want to test the limit along all lines passing through . Which one of the
following statements is correct?
One obtains the same limit along all lines, so the limit must exist.
One obtains the same limit along all lines, but the limit might not exist.
One obtains two different limits along two different lines, so the limit does not exist.
One obtains two different limits along two different lines, but the limit might still exist.
None of the previous statements is correct.
(c) Consider the function defined by y for and .
Which one of the following statements is correct?
The function is continuous at all points.
The function is continuous at all points except for .
The function is continuous at all points except for those of the form. for .
The function is continuous at all points except for those of the form. for .
The function is never continuous, because it is not well-defined.
None of the previous statements is correct.
(d) Consider the function n . Which one of the following statements is correct?
The function is differentiable at (0,0).
The function is not differentiable at (0,0), because the partial derivative with respect to x does not exist.
The function is not differentiable at (0,0), because the partial derivative with respect to y does not exist.
The function is not differentiable, because the partial derivatives are not defined for some points near (0,0).
The function is not differentiable, because the partial derivatives are not continuous near (0,0).
None of the previous statements is correct.
lim
(x,y)→(1,0)
2xy
+x
2
y
2
0
1
lim
(x,y)→(0,0)
x+ y
2
+ yx
2
(0,0)
f(x,y) =
−x
4
y
4
+x
2
y
2
(x,y) ≠ (0,0) f(0,0) =
1
2
(0,0)
(x,x) x ∈R
(x,x) x ∈R ∖ {1,−1}
f(x,y) = |x| ⋅ y
1/3
Consider the function given by
f : →RR
2
Question 6
Not answered
Marked out of 4.00