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MTH2010-MTH2015 - Multivariable Calculus - S2 2021
Started on Saturday, 23 October 2021, 9:07 PM
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Completed on Saturday, 23 October 2021, 9:08 PM
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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
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Question 2
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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
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(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
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Question 5
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(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
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Question 7
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(a) For each of the following equations, determine whether it is right or wrong.

  
  
  


(b) The tangent plane to the graph of at the point can be described by the equation
   .
(c) Using the linear approximation of near tells us that is approximately  . Please
enter your answer as a decimal.
(d) The most positive rate of change of at the point is in the direction of the vector
(2,  ).
(e) If and , then at the point , is equal to  .
f(x,y) = x + y .
y
3
e
x−2
  (x,y) = + yf
x
y
3
e
x−2
  (x,y) = 3x +f
y
y
2
e
x−2
  (x,y) = yf
xx
e
x−2
  (x,y) = 3 + yf
xy
y
2
e
x−2
  (x,y) = 3 +f
yx
y
2
e
x−2
  (x,y) = 6xf
yy
z= f(x,y) (2,1,3)
x  +   y  −   z  =  
f(x,y) (x,y) = (2,1) f(1.99,1.01)
f(x,y) (x,y) = (2,1)
x= −2s
2
t
2
y= s+ t (s,t) = (2,−1)
∂f
∂s
Consider the function
defined on the set .
(a) There is one critical point and it occurs at (  ,  ).
(b) Using the second derivatives test, we find that the critical point is  .
(c) The global minimum of the
occurs at an interior critical point.
occurs at .
occurs on the boundary of .
is not achieved.
f(x,y) = y+9 − y,x
3
x
2
D= {(x,y) ∣ + ≤ 100}x
2
y
2
f(x,y)
(0,0)
D
Consider the double integral
y dA,∬
x
2

Question 8
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where is the semicircular region given by .
(a) After converting the double integral above to an iterated integral, one obtains
.
.
.
.
.
.
.
(b) After taking the iterated integral from part (a) and changing the order of integration, one obtains
.
.
.
.
.
.
.
.
(c) After converting the double integral above to an iterated integral in polar coordinates, one obtains
.
.
.
.
.
.
.
.
(d) The value of the double integral above can be expressed as the fraction in lowest terms, where  and
 .
y ,∬
D
D {(x,y) ∣ + ≤ 4 and y≥ 0}x
2
y
2
y dy dx∫
2
0
∫
4−x
2
√
0
x
2
y dy dx∫
2
−2
∫
4−y
2
√
− 4−y
2
√
x
2
y dy dx
∫
2
−2
∫
4−y
2
√
0
x
2
y dy dx
∫
2
−2
∫
4−x
2
√
− 4−x
2
√
x
2
y dy dx∫
2
0
∫
4−y
2
√
0
x
2
y dy dx∫
2
−2
∫
4−x
2
√
0
x
2
y dy dx∫
2
0
∫
4−y
2
√
− 4−
y
2
√
x
2
y dy dx
∫
2
0
∫
4−x
2
√
− 4−x
2
√
x
2
y dxdy∫
2
−2
∫
4−y
2
√
0
x
2
y dxdy∫
2
0
∫
4−
y
2
√
− 4−y
2
√
x
2
y dxdy∫
2
0
∫
4−x
2
√
0
x
2
y dxdy∫
2
0
∫
4−
x
2
√
− 4−x
2
√
x
2
y dxdy∫
2
−2
∫
4−x
2
√
0
x
2
y dxdy∫
2
0
∫
4−
y
2
√
0
x
2
y dxdy∫
2
−2
∫
4−x
2
√
− 4−x
2
√
x
2
y dxdy∫
2
−2
∫
4−
y
2
√
− 4−y
2
√
x
2
θsinθ dr dθ∫
2π
0
∫
2
0
r
4
cos
2
θcosθ dr dθ∫
2π
0
∫
2
0
r
4
sin
2
θsinθ dr dθ∫
π
0
∫
2
0
r
3
cos
2
θsinθ dr dθ∫
π
0
∫
2
0
r
4
cos
2
θcosθ dr dθ∫
π
0
∫
2
0
r
4
sin
2
θcosθ dr dθ∫
π
0
∫
2
0
r
3
sin
2
θcosθ dr dθ∫
2π
0
∫
2
0
r
3
sin
2
θsinθ dr dθ∫
2π
0
∫
2
0
r
3
cos
2
a
b
a= b=

Question 9
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Question 10
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(a) To change from cylindrical coordinates to Cartesian coordinates, we use the equations
\(x=r \cos \theta, y = r \sin \theta, z = z\).
\(x=r \cos \theta, y = r \sin \theta, z = rz\).
\(x=r \sin \theta, y = r \cos \theta, z = rz\).
\(x=r \sin \theta, y = r \cos \theta, z = z\).
(b) To convert a triple integral from Cartesian coordinates to cylindrical coordinates, we use the fact that \(dV = dx \, dy \, dz\) is
\(rz \, dr \, d\theta \, dz\).
\(dr \, d\theta \, dz\).
\(\theta \, dr \, d\theta \, dz\).
\(r \, dr \, d\theta \, dz\).
(c) Consider the triple integral \(\iiint_E z \, dV\), where \(E\) is the region enclosed by the paraboloid \(z = x^2 + y^2\) and the plane \(z =
4\). After converting this triple integral to an iterated integral in cylindrical coordinates, one obtains an expression of the form \[
\int_0^{a\pi} \int_0^b \int_c^d g(r, \theta, z) \, dz \, dr \, d\theta, \]
where \(a\) is equal to  ,
\(b\) is equal to  ,
\(c\) is the expression
\(0\) \(r\) \(r^2\) \(4\)
and
\(g(r, \theta, z)\) is the expression
\(z\) \(z\theta\) \(zr\) \(z r \theta\).
(d) The value of the triple integral from part (c) can be expressed as \(\dfrac{a}{b} \pi\) with \(\dfrac{a}{b}\) in lowest terms, where \(a =
\)  and \(b = \)  .  
Consider the lamina that occupies the triangular region with vertices \((0,0)\), \((2,2)\) and \((2,-2)\), with density \(\rho(x,y) =x^2\).
Recall that the mass of the lamina is given by
\[ m = \iint_D \rho(x,y) \, \mathrm{d}A, \]
where \(D\) is the region occupied by the lamina.
(a) The mass of the lamina is  .
(b) The centre of mass of the lamina is (  ,  ). Please enter your answer as decimals.
Consider the vector field \[ \mathbf{F}(x,y) = (x-2y) \, \mathbf{i} + xy \, \mathbf{j}. \]
Let \(C\) denote the curve parametrised by the vector function \[ \mathbf{r}(t) = (t^2+2t, t), \] for \(0 \leq t \leq 1\).
(a) Using the above parametrisation for \(C\), the line integral \(\int_C \mathbf{F} \cdot \mathrm{d}\mathbf{r}\) becomes \[ \int_a^b
(Pt^3 + Qt^2 + Rt + S) \, dt, \] where \(a =\)  , \(b =\)  , \(P =\)  , \(Q =\)  , \(R =\)  , \ 
Question 11
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Question 12
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Question 13
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Question 14
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(S =\)  .
(b) The value of the line integral can be expressed as the fraction \(\dfrac{a}{b}\) in lowest terms, where \(a =\)  and \(b =\)
 .
(a) Which one of the following three-dimensional vector fields is conservative?
\((y^2, 2y + \sin z, y \cos z)\)
\((4xz^2, z e^y, 4x^2z + e^y)\)
\((xz, -z^2 \sin y, x + 2z \cos y)\)
(b) Let \(\mathbf{F}(x,y,z)\) denote the vector field from part (a) that is conservative. Let \(C\) be the curve parametrised by the vector
function \[ \mathbf{r}(t) = (t e^{t^2-t}, \sin \pi t, t^{100}), \] for \(0 \leq t \leq 1\). The value of the line integral \(\int_C \mathbf{F} \cdot
d\mathbf{r}\) is  .
Consider the surface integral \[ \iint_S xy \, dS, \] where \(S\) is the surface \(z = x^2 + \sqrt{8}\,y\) for \(0 \leq x \leq 2\) and \(0 \leq y
\leq 6\).
(a) The surface integral can be expressed as \[ \int_0^6 \int_0^2 xy \sqrt{g(x,y)} \, dx \, dy, \] where \(g(x,y)\) is the function
\(2x+\sqrt{8}\)
\(2x+\sqrt{8}+1\)
\(2x+\sqrt{8}\,y\)
\(2x+\sqrt{8}\,y+1\)
\(4x^2+8\)
\(4x^2+9\)
\(4x^2+8y^2\)
\(4x^2+8y^2+1\)
(b) The value of the surface integral is  .  
Let \(C\) be the space curve \(r(t) = (3\sin(t)-1,4\sin(t)+1)\), \(0\leq t\leq \frac{\pi}{2}\).
The length of \(C\) is  .

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Question 15
Not answered
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Question 16
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Consider the surface integral \[ \iint_S \mathbf{F} \cdot d\mathbf{S}, \] where \(\mathbf{F}(x,y,z) = (xye^z, xy^3, -ye^z)\) and \(S\) is the
surface of the box bounded by the coordinate planes and the planes \(x = 1\), \(y = 2\) and \(z = 3\).
(a) The divergence theorem allows us to express the surface integral as a triple iterated integral of the form \[ \int_0^a \int_0^b \int_0^c
g(x, y, z) \, dx \, dy \, dz, \]
where \(a\) is the constant  ,
\(b\) is the constant  ,
\(c\) is the constant  ,
\(g(x,y,z)\) is a function that satisfies \(g(2,3,4) =\)  .
(b) The value of the surface integral is  .
Determine the minimum distance from the point \((5,-1)\) to the parabola given by\(y = x^2\), using Lagrange multipliers in the following
way.
(a) Let \(f(x,y)\) denote the square of the distance between the point \((5,-1)\) and a point \((x,y)\). Write down an expression for \
(f(x,y)\).
(b) Using \(g(x,y) = x^2 - y\), write down the three simultaneous equations that \(x\), \(y\) and the Lagrange multiplier \(\lambda\) must
satisfy at an extreme value of \(f(x,y)\) subject to the constraint \(g(x,y)=0\).
(c) Use the simultaneous equations from part (b) to express\(x\) and \(y\) in terms of \(\lambda\). Then substitute these expressions
into the constraint \(g(x,y) = 0\) to obtain a cubic equation for \(\lambda\).
(d) Show that \(\lambda = -4\) is the only solution to the cubic equation from part (c). Hence, find the point on the parabola that is
closest to \((5,-1)\) and calculate the corresponding minimum distance.
(a) Determine the Jacobian \(\dfrac{\partial(x,y)}{\partial(u,v)}\) for the change of variables \(x = \frac{1}{2}(u+v)\) and \(y = \frac{1}{2}
(v-u)\). 
Question 17
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(b) Use the change of variables from part (a) to evaluate the integral \[ \iint_R \frac{\sin(x+y)}{4+x-y} \, \mathrm{d}A, \] where \(R\) is the
region in \(\mathbb{R}^2\) bounded by the lines\(y = x-2\), \(y = x+2\), \(y = -x + 2\) and \(y = -x+6\).
Let \(S\) be the surface in \(\mathbb{R}^3\) that is the part of the ellipsoid \(2x^2 + y^2 + z^2 = 2\) satisfying \(y \geq 1\). Consider the
vector field \[ \mathbf{F}(x,y,z) = (z, z^2 \log_e(y) + \cos(xyz), -x). \] Compute \(\iint_S \mathrm{curl}~\mathbf{F} \cdot d\mathbf{S}\).
(You should assume that \(S\) has the same orientation as the positive orientation of the ellipsoid.)
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