Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 5
To be solved during the tutorial session Thursday/Friday, 18/19 February 2016
1. Use appropriate forms of the chain rule to find
z = sin
x
cos 2y ;
2
∂z
∂u
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
= cos cos 2y − 6u2 sin sin 2y
2
2
2u + 3v
2u + 3v
= cos
cos 2(u3 − 2v 2 ) − 6u2 sin
sin 2(u3 − 2v 2 ) ,
2
2
2. A function f (x1 , . . . , xn ) is said to be homogeneous of degree k if f (tx1 , . . . txn ) = tk f (x1 , . . . , xn )
for t > 0. Show that it satisfies
n
X
∂f
= kf .
xi
∂x
i
i=1
Solution: Consider the function
F (t, x1 , . . . , xn ) = f (tx1 , . . . , txn ) − tk f (x1 , . . . , xn ) .
(1)
Compute
∂
∂
∂
F (t, x1 , . . . , xn ) = f (tx1 , . . . , txn ) − tk f (x1 , . . . , xn )
∂t
∂t
∂t
n
X
∂f (tx1 , . . . , txi−1 , yi , txi+1 , . . . , txn )
=
xi
|yi =txi − ktk−1 f (x1 , . . . xn ) .
∂yi
i=1
(2)
If f is homogeneous of degree k, then F = 0 for any t > 0, and setting t = 1 in the
formula, one gets
n
X
∂f (x1 , . . . , xn )
0=
xi
− kf (x1 , . . . , xn ) .
(3)
∂xi
i=1
3. Let r =
Pn
i=1
xi ei and r = |r|, where ei form an orthonormal basis of vectors in Rn . Find
∂r
,
∂xi
i = 1, 2, · · · , n ,
and ∇r ,
|∇r| .
Solution: We have
∂r
2xi
xi
= p 2
= ,
∂xi
r
2 x1 + · · · + x2n
1
∇r =
n
X
∂r
r
ei = ,
∂xi
r
i=1
|∇r| = 1 .
4. Let ur be a unit vector whose counterclockwise angle from the positive x-axis is θ, and let
uθ be a unit vector 90o counterclockwise from ur . Show that if z = f (x, y), x = r cos θ,
y = r sin θ, then
1 ∂z
∂z
ur +
uθ .
∇z =
∂r
r ∂θ
Solution: We have
ur =
and
r
= cos θ e1 + sin θ e2 ,
r
∂z
∂z
∇z =
e1 +
e2 =
∂x
∂y
Then
r=
p
x2 + y 2
y
θ = arctan
x
⇒
⇒
uθ = − sin θ e1 + cos θ e2 .
∂z ∂r ∂z ∂θ
+
∂r ∂x ∂θ ∂x
e1 +
∂z ∂r ∂z ∂θ
+
∂r ∂y ∂θ ∂y
(4)
e2 .
∂r
x
∂r
y
= = cos θ ,
= = sin θ ,
∂x
r
∂y
r
∂θ
y
sin θ
∂θ
x
cos θ
=− 2 =−
,
= 2 =
.
∂x
r
r
∂y
r
r
(5)
(6)
Collecting the terms one gets the formula.
5. Consider the surface
p
3
2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18
z = f (x, y) = ln
2
(a) Find an equation for the tangent plane to the surface at the point P (3, −2, z0 ) where
z0 = f (3, −2).
(b) Sketch the tangent plane.
(c) Find parametric equations for the normal line to the surface at the point P (3, −2, z0 ).
(d) Sketch the normal line to the surface at the point P (3, −2, z0 ).
Show the details of your work.
Solution:
(a) We first simplify
p
3
2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18
1
z = ln
= ln 2x2 −3xy 2 +3 cos(2x+3y)−3y 3 +18 −ln 2 ,
2
3
and compute z0
z0 = z|x=3,y=−2 =
1
3
ln 3 cos(0) + 18 + 24 − 36 + 18 − ln 2 = ln ≈ 0.405465
3
2
Then, we compute the partial derivatives at P (3, −2, z0 )
∂z
−6 sin(2x + 3y) + 4x − 3y 2
=
∂x
3 (2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18)
2
⇒
∂z
|x=3,y=−2 = 0 .
∂x
∂z
−6xy − 9 sin(2x + 3y) − 9y 2
=
∂y
3 (2x2 − 3xy 2 + 3 cos(2x + 3y) − 3y 3 + 18)
⇒
∂z
|x=3,y=−2 = 0 .
∂y
The tangent plane equation is given by
z = z0 + 0(x − 3) + 0(y + 2) = ln
3
.
2
(b) It is a plane through the point (3, −2, ln 32 ) parallel to the xy-plane.
(c) The normal line to the surface (and the tangent plane) is given by
r = 3i − 2j + ln
3
3
+ t 0i + 0j + k = 3i − 2j + (t + ln )k .
2
2
Note that the same line is given by
r = 3i − 2j + tk .
(d) It is parallel to the z-axis.
3