MAST10008 Accelerated Maths 1 Version 2024
Tutorial Week 12
Calculus of functions in two variables
Q1. Consider the function
f (x, y) = 2x3 + 3xy − 6y.
Let S be the surface defined by the equation z = f (x, y).
(a) Find the gradient vector ∇f at (x, y).
(b) Show that the point (0, 1, −6) lies on S.
(c) Find a Cartesian equation of the tangent plane to S at the point (0, 1, −6).
(d) Write down the linear approximation to f (x, y) at (1, −2), and use this to estimate the value
16
7
,−
of f
.
10 10
(e) Suppose that x = − cos(2t) and y = sin(3t). Using the chain rule for functions of two
df
variables, calculate
at t = 0.
dt
(f) Find the directional derivative of f at (1, −2) in the direction of the unit vector from the
point (1, −2) towards the point (−2, 2).
" ∂f #
∂x
∂f
∂y
A1. (a) ∇f =
"
=
#
6x2 + 3y
.
3x − 6
(b) Substituting x = 0 and y = 1 into f (x, y) gives
f (0, 1) = 2 × 03 + 3 × 0 × 1 − 6 × 1 = −6 as required.
"
(c) (∇f )(0, 1) =
#
3
. Thus the equation of the tangent plane at (0, 1, −6) is
−6
z = −6 + 3(x − 0) + (−6)(y − 1) ⇒ 3x − 6y − z = 0.
"
(d) We first find (∇f )(1, −2) =
0
−3
#
and f (1, −2) = 8.
So near (x, y) = (1, −2), we have
z = f (x, y) ≈ f (1, −2) + fx (1, −2)(x − 1) + fy (1, −2)(y + 2) = 8 − 3(y + 2).
Thus f
7
, − 16
10
10
16
≈ 8 − 3 × − 10
+ 2 = 8 − 12
= 68
= 6.8
10
10
dx
dy
= 2 sin(2t) = 0 and
= 3 cos(3t) = 3.
dt
dt
Also when t = 0, (x, y) = (− cos(0), sin(0)) = (−1, 0).
(e) We first note that when t = 0,
"
Thus when t = 0, ∇f =
fx (−1, 0)
fy (−1, 0)
#
"
=
#
6
.
−9
Now when t = 0,
∂f
∂f dx ∂f dy
=
+
= 6 × 0 − 9 × 3 = −27.
∂t
∂x dt
∂y dt
"
(f) A vector in the direction from (1, −2) to (−2, 2) is u =
"
Thus a unit vector in this direction is û = 51
Then Du f
"
(1, −2) = û · (∇f )(1, −2) = 15
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−2 − 1
2 − (−2)
#
"
=
#
−3
.
4
#
−3
.
4
−3
4
# "
·
0
−3
#
= − 12
.
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MAST10008 Accelerated Maths 1 Version 2024
Tutorial Week 12
Q2. Consider the function
f (x, y) = 2y 3 + 3xy 2 + 12xy + 9x + 6y 2 .
(a) Find the first and second order partial derivatives.
(b) Locate and classify the stationary points of the function.
A2. (a)
fx (x, y) = 3y 2 + 12y + 9
fy (x, y) = 6y 2 + 6xy + 12x + 12y
fxx (x, y) = 0
fyy (x, y) = 12y + 6x + 12
fxy (x, y) = fyx (x, y) = 6y + 12.
(b) fx (x, y) = 3(y 2 + 4y + 3) = 3(y + 3)(y + 1) = 0 when y = −3, −1.
fy (x, y) = 6y 2 + 6xy + 12x + 12y = 6(y 2 + xy + 2x + 2y) = 0 when x =
−y 2 − 2y
.
y+2
−9 + 6
= 3.
−3 + 2
−1 + 2
Substitute y = −1 to obtain x =
= 1.
−1 + 2
The stationary points are (3, −3) and (1, −1).
Substitute y = −3 to obtain x =
Now fxx fyy − (fxy )2 = −(6y + 12)2 < 0 regardless of the value of x and y.
Thus both stationary points are saddle points.
Q3. Evaluate the double integral
Z 3Z 1
0
e2x+y dx dy.
0
A3.
Z 3Z 1
0
0
2x+y
e
1 2x+y x=1
dy
dx dy =
e
2
0
x=0
Z 3 1 2+y
=
e
− ey dy
0 2
y=3
1 2+y
y
=
e
−e
2
y=0
1 5
=
e − e3 − (e2 − 1)
2
1 5
=
e − e3 − e2 + 1 .
2
Z 3
Q4. Use a double integral to find the volume of the region lying below the graph z = x2 + y 2 and above
the rectangle 1 ⩽ x ⩽ 2, 0 ⩽ y ⩽ 1 in the plane z = 0.
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MAST10008 Accelerated Maths 1 Version 2024
Tutorial Week 12
A4. The volume is
Z 2Z 1
1
2
2
(x + y ) dy dx =
Z 2h
iy=1
1
y=0
yx2 + y 3 /3
0
=
Z 2
dx
x2 + 1/3 dx
1
ix=2
h
= x3 /3 + x/3
x=1
8 2
1 1
= + −
+
3 3
3 3
8
= cubic units.
3
Q5. Let A be a symmetric 2 × 2 matrix with real entries and let λ1 , λ2 be its eigenvalues. Show that
(a) The determinant of A is equal to λ1 λ2 and the trace of A is equal to λ1 + λ2 .
(b) det(A) < 0 if and only if A has one positive eigenvalue and one negative eigenvalue;
(c) A has two positive eigenvalues if and only if det(A) > 0 and a11 > 0;
(d) A has two negative eigenvalues if and only if det(A) > 0 and a11 < 0.
A5. (a) The characteristic polynomial of A is
det(A − xI) = (a11 − x)(a22 − x) − a212 = x2 − (a11 + a22 )x + (a11 a22 − a212 )
At the same time, in terms of the eigenvalues, it is
det(A − xI) = (x − λ1 )(x − λ2 ) = x2 − (λ1 + λ2 )x + λ1 λ2
By comparing the two right hand sides we see
λ1 + λ2 = a11 + a22 = trace of A
λ1 λ2 = a11 a22 − a212 = determinant of A
(b) Clearly det(A) = λ1 λ2 < 0 if and only if one of the factors is positive and the other one is
negative.
(c) Let t = a11 + a22 denote the trace of A. The quadratic formula tells us that the eigenvalues
are
q
t ± t2 − 4 det(A)
λ=
2
Since det(A) > 0, the square root is < |t|, so the eigenvalues are both positive if and only
if t = a11 + a22 > 0. This forces a11 > 0, since otherwise a11 < 0 and a22 > 0, which makes
det(A) = a11 a22 − a212 < 0, contradiction.
(d) Follows from the previous part as it’s the only possibility left.
Q6. Let f : {x2 + y 2 ⩽ 1} −→ R be given by
f (x, y) =
q
1 − x2 − y 2 .
(a) Draw a sketch of the surface S that is the graph of f .
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MAST10008 Accelerated Maths 1 Version 2024
Tutorial Week 12
(b) Locate the point P on S with x = y = 21 .
(c) Based only on the geometry of the surface S, draw the curve of steepest descent originating
at the point P ; in other words, if a mountain spring comes to the surface at the point P ,
what path does the water follow down the mountain?
(d) Challenge: Without relying on the geometry of S, use properties of the gradient to derive a
formula for the curve of steepest descent passing through P . [Note: at some point you will
have to solve a differential equation.]
A6. (a) S is the upper unit hemisphere in R3 .
√
(b) The point is of course P (1/2, 1/2, 1/ 2).
sphere and P ; the curve we
(c) Find the great circle passing through the topmost point of
√ the √
want is the arc of this circle that joins P to the point (1/ 2, 1/ 2, 0) on the equator.
(d) The gradient of f is given by
1
∇f = − √
1 − x2 − y 2
"
x
y
#
,
so the direction of steepest descent at any point is given by the unit vector in the direction of
−∇f , that is
"
#
1
x
u= √ 2
.
x + y2 y
The projection onto the xy plane of the curve of steepest descent must have tangent direction
u, in other words it should satisfy
√
dy
y/ x2 + y 2
y
= √ 2
=
.
dx
x
x/ x + y 2
This is a separable differential equation with general solution given by
log |y| = log |x| + C ⇒ y = Ax,
A ∈ R.
It must also pass through the projection of the point P , that is (1/2, 1/2), forcing A = 1.
The curve of steepest descent is then the space curve with projection y = x, in other words is
given by
(
"
#)
√
1
1
x, x, 1 − 2x2 : x ∈ , √
.
2 2
This of course is the same as the curve sketched in part (c).
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