Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 2
Due: at the end of the tutorial session Tuesday/Thursday, 2/4 February 2016
Name and student number:
1. Sketch the level curve z = k for the specified values of k
z = x2 − 2x + 4y 2 − 4y ,
k = −2, −1, 2 .
Solution
1.5
1.0
0.5
1
-1
2
3
-0.5
The level curve equation x2 − 2x + 4y 2 − 4y = k can be written as
1
(x − 1)2 + 4(y − )2 = k + 2 ,
2
and, therefore, the level curves are ellipses with the centre located at (1, 1/2). For k = −2
the level curve is just the point (1, 1/2).
2. Consider the function
z = 2ex sin y − 3ey cos x
(i) Find
a)
Solution:
∂z
(0, 0) ,
∂x
b)
∂z
(0, 0) ,
∂y
c)
∂ 2z
(0, 0) ,
∂x∂y
∂z
= 2ex sin y + 3ey sin x ⇒
∂x
∂z
b)
= 2ex cos y − 3ey cos x ⇒
∂y
a)
1
d)
∂ 2z
.
∂y∂x
∂z
(0, 0) = 0 .
∂x
∂z
(0, 0) = −1 .
∂y
c)
∂ ∂z
∂
∂ 2z
=
=
(2ex cos y − 3ey cos x) = 2ex cos y + 3ey sin x
∂x∂y
∂x ∂y
∂x
⇒
∂ 2z
(0, 0) = 2 .
∂x∂y
d)
∂ 2z
∂ ∂z
∂
=
=
(2ex sin y + 3ey sin x) = 2ex cos y + 3ey sin x
∂y∂x
∂y ∂x
∂y
⇒
∂ 2z
(0, 0) = 2 .
∂y∂x
(ii) Find the slope of the surface z = 2ex sin y − 3ey cos x
a) in the x-direction at the point (0, π2 );
Solution: The slope kx is equal to
kx =
∂z
π
(0, ) = 2 .
∂x
2
b) in the y-direction at the point ( π2 , π2 ).
Solution: The slope ky is equal to
ky =
∂z π π
( , ) = 0.
∂y 2 2
(iii) Show that the function satisfies Laplace’s equation
∂ 2z ∂ 2z
+
= 0.
∂x2 ∂y 2
Solution: To this end we compute the following derivatives
∂ ∂z
∂
∂ 2z
=
=
(2ex sin y + 3ey sin x) = 2ex sin y + 3ey cos x ,
2
∂x
∂x ∂x
∂x
∂ 2z
∂ ∂z
∂
=
=
(2ex cos y − 3ey cos x) = −2ex sin y − 3ey cos x .
2
∂y
∂y ∂y
∂y
The sum of these two expressions is obviously 0.
3. Show that the local linear approximation of the function
f (x, y) =
yα
+ xα − y β
xβ
at (1, 1) is
f (x, y) ≈ 1 + (α − β)(x − 1) + (α − β)(y − 1) .
Solution: The local linear approximation of the function at (1, 1) is given by the formula
L(x, y) = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1) .
We obviously have f (1, 1) = 1, and
fx (x, y) = −β
fy (x, y) = α
yα
+ α xα−1
xβ+1
y α−1
− β y β−1
β
x
2
⇒
⇒
fx (1, 1) = α − β ,
fy (1, 1) = α − β ,
which proves the formula.
4. Use appropriate forms of the chain rule to find
z = sin
x
cos 2y ;
2
∂z
∂u
and
∂z
∂v
where
x = 2u + 3v , y = u3 − 2v 2 .
Solution: We have
∂z
∂z ∂x ∂z ∂y
1
x
x
=
+
=
cos cos 2y × 2 − 2 sin sin 2y × 3u2
∂u
∂x ∂u ∂y ∂u
2
2
2
x
x
2
= cos cos 2y − 6u sin sin 2y
2
2
2u + 3v
2u + 3v
3
= cos
cos 2(u − 2v 2 ) − 6u2 sin
sin 2(u3 − 2v 2 ) ,
2
2
∂z
∂z ∂x ∂z ∂y
1
x
x
=
+
=
cos cos 2y × 3 − 2 sin sin 2y × (−4v)
∂v
∂x ∂v ∂y ∂v
2
2
2
3
x
x
=
cos cos 2y + 8v sin sin 2y
2
2
2
3
2u + 3v
2u + 3v
3
=
cos
cos 2(u − 2v 2 ) + 8v sin
sin 2(u3 − 2v 2 ) .
2
2
2
3