D-ARCH
Mathematics
Fall 2014
Problem set – Week 6
Chain rule for partial derivatives
change of bases and determinants
1. The lengths a, b, c of the edges of a rectangular box are changing with time. At
some fixed time t0 , we have the data a = 1, b = 2, c = 3, da/dt = db/dt = 1 and
dc/dt = −3.
At what rates are the volume and the surface area of the box changing at t0 ?
Are the interior diagonals of the box increasing or decreasing in length ?
Solutions : Volume V 0 (t0 ) = 3, surface area A0 (t0 ) = 0, diagonals are decreasing
in length at t0 .
Rb
Rb
2. Under mild continuity restrictions, if F (x) = a g(t, x)dt, then F 0 (x) = a gx (t, x)dt.
R f (x)
Together with the chain
rule,
this
allows
to
derivate
F
(x)
=
g(t, x)dt by
a
Ru
considering G(u, x) = a g(t, x)dt where u = f (x).
Find the derivative of
x2
Z
t2 + x3
F (x) =
3/2
dt.
0
√
√
√ Solution : 2x11/2 (x + 1)3/2 + 94 x9/2 x + 1 + 94 x5 ln x + 1 + x.
3. For the following sets of data, find the matrix B of the linear transformation
T (~x) = A~x with respect to basis B.
0 1
1
1
(a) A =
and B =
,
;
1 0
1
−1
1 0
Solution : B =
.
0 −1
(b) T (~x) = ~v2 × ~x and B = {~v1 , ~v2 , ~v3 } a basis of perpendicular vectors, such
that ~v3 = ~v1 × ~v2 .


0 0 −1
Solution : B = 0 0 0 .
0 0 0
4. Show that if the 3 × 3 matrix A represents the reflection about a plane, then A
is similar to


1 0 0
0 1 0  .
0 0 −1
Solution : To be discussed in class.
1
5. Find the derivative of

1
9

f (x) = det 
9
x
7
1
0
0
1
0
2
2
0
2
0
3
3
3
9
0

4
4

4
.
1
4
Solution : f 0 (x) = −3.


α
0  , where α ∈ R. How does the inverse
1
0 −1
2
6. Compute the inverse of −1
−2
4
depends on α ?
Solution :


−2 −1 − 4α 2α
−2α
α .
= −1
0
−2
1
A−1
7. Consider the following n × n matrix,

1 1

1 2


0 1
Tn :=  . .
 .. . .

 ..
.
0 ···

··· 0
.. 
0
.

. . . .. 
1
.
.
..
..
. 0
.

... ... 
1
0
1 2
···
0
1
2
..
.
...
···
(a) Using the Laplace expansion, prove that
det(Tn ) = 2 · det(Tn−1 ) − det(Tn−2 ) for n > 2.
(b) Prove by induction that
det(Tn ) = 1 for all n ∈ N.
Solution : To be discussed in class.
8. Verify
y
~v⊥ =
−c
a
w
~=
b
d
~v =
α
2
a
c
x
ad − bc = ~v⊥ · w
~ = ||~v⊥ || ||w||
~ cos
π
2
−α
= ||~v || ||w||
~ sin α.
What is the geometric interpretation of |det(~v |w)|
~ and how does det(~v |w)
~ depend
on α = ] (~v , w)
~ ?
Use your answer to find the area of the following region :
−7
7
5
5
3
−4
−5
−6
Solution : |det(~v w)|
~ is the area of the parallelogram spanned by ~v and w.
~ The
area of the region is 75.
3