MA1132 (Advanced Calculus) Solutions to tutorial sheet 9
1. Evaluate
Z
0
3
Z
√
y= 3x
y=0
1
p
dy dx
x2 + y 2
by making a change of variables to polar coordinates. [Hint: Sketch the region first. Then
do
R the dr integral first, before dθ; for fixed θ find the limits for r in terms of θ. Recall that
sec θ dθ = ln | sec θ + tan θ| + C].
Solution: This iterated integral is the same (by Fubini’s theorem) as the double integral
ZZ
1
p
dx dy
x2 + y 2
R
where
√ R is the triangle in the plane bounded by the x-axis, the line x = 3 and the line
y = 3x.
3
The
√ limit x = 3 is r cos θ = 3 or r = 3/ cos θ and the angle is π/3 (because tan(π/3) =
3).
Changing this integral to polar coordinates, and remembering that dx dy = r dr dθ, we get
!
Z θ=π/3 Z r=3/ cos θ
Z θ=π/3
1
cos θ
[r]r=3/
dθ
r dr dθ =
r=0
r
θ=0
r=0
θ=0
Z θ=π/3
3
=
dθ
cos θ
θ=0
Z θ=π/3
=
3 sec θ dθ
θ=0
= [3 ln | sec θ + tan θ|]π/3
0
√
= 3 ln(2 + 3)
2. Consider the function
f (x, y, z) = 1 +
x2 ex y + xey z
x+y+z
(a) Find ∇f (evaluated at an unspecified point (x, y, z)).
Solution:
(2xex y + x2 ex y + ey z)(x + y + z) − (x2 ex y + xey z)
∂f
=
∂x
(x + y + z)2
∂f
(x2 ex + xey z)(x + y + z) − (x2 ex y + xey z)
=
∂y
(x + y + z)2
xey (x + y + z) − (x2 ex y + xey z)
∂f
=
∂z
(x + y + z)2
∂f ∂f ∂f
∇f =
,
,
∂x ∂y ∂z
(2xex y + x2 ex y + ey z)(x + y + z) − (x2 ex y + xey z)
,
=
(x + y + z)2
(x2 ex + xey z)(x + y + z) − (x2 ex y + xey z)
,
(x + y + z)2
xey (x + y + z) − (x2 ex y + xey z)
(x + y + z)2
(b) Find the equation of the tangent plane to the surface f (x, y, z) = 1 +
2e
at the point
3
(1, 1, 1).
Solution: We are getting the tangent plane to a level surface f (x, y, z) = 2e/3. So the
normal vector to the tangent plane is the gradient of f evaluated at the point (1, 1, 1).
10e 4e e
(4e)(3) − 2e (2e)(3) − 2e 3e − 2e
∇f |(1,1,1) =
,
,
=
, ,
.
32
32
32
9 9 9
2
For the tangent plane we get
10e
4e
e
(x − 1) + (y − 1) + (z − 1) = 0
9
9
9
and we could simplify this by dividing by e/9 to get the equivalent equation
10(x − 1) + 4(y − 1) + (z − 1) = 0.
(c) Find the direction u = (u1 , u2 , u3 ) for which the directional derivative Du f (1, 1, −1)
is as small as possible. Also, what is that smallest possible value of Du f (1, 1, −1)?
Solution: What we need is u to be the unit vector in the direction opposite to ∇f |(1,1,−1)
and the smallest possible value is −k∇f |(1,1,−1) k.
e(1) − 0
(2e + e − e)(1) − 0
, 0 − 0,
= (2e, 0, e)
∇f |(1,1,−1) =
12
12
So
u=−
1
k∇f |(1,1,−1)
−1
∇f |(1,1,−1) = √ (2e, 0, e) =
k
e 5
2
1
− √ , 0, − √
5
5
and the corresponing smallest value of the directional derivative is
√
Du f (1, 1, −1) = −k∇f |(1,1,−1) k = − 5e
3