Home Work Solution Set #6
Math 14: Gupta, Limon
1. Determine D(f ◦ g) in terms of s and t for the following functions:
(a) f (x, y) = x4 − 5y 3 and g(s, t) = (st2 , s + t2 )
Solution (5 points):
By the Chain Rule, we know that D(f ◦ g) = Df · Dg, so let’s
compute
i
h
∂f
,
= [4x3 , −15y 2 ] and
Df = ∂f
∂x ∂y
∂g1 ∂g1 2
t 2st
∂s
∂t
Dg = ∂g2 ∂g2 =
.
1 2t
∂s
∂t
Thus, D(f ) · D(g) = [4x3 t2 − 15y 2 , 8x3 st − 30y 2 t], where x = st2
and y = s + t2 .
2
(b) f (x, y, z) = (x2 y + yz 2 , x5 − eyz ) and g(s, t) = (st2 , st, s + t)
Solution (5 points):
Again by the Chain Rule, we know that D(f ◦ g) = Df · Dg, so
let’s compute
# "
∂f1
∂f1
∂f1
2xy x2 + z 2
2yz
∂x
∂y
∂z
Df = ∂f2 ∂f2 ∂f2 =
and
2
2
5x4 −z 2 eyz −2yzeyz
∂x
∂y
∂z

 ∂g1 ∂g1   2
t
2st
∂s
∂t
∂g2 
2
Dg =  ∂g
=  tst−1 st ln s .
∂s
∂t
∂g3
∂g3
1
1
∂s
∂t
Thus, D(f ) · D(g) =
2xyt2 + (x2 + z 2 )tst−1 + 2yz 4xyst + (x2 + z 2 )st ln s + 2yz
,
2
2
2
2
5x4 t2 − z 2 eyz tst−1 − 2yzeyz 10x4 st − z 2 eyz st ln s − 2yzeyz
where x = st2 , y = st and z=s + t.
2. Suppose that f : R → R is a differentiable function with non-vanishing
2 2
−y
derivative. That is, the derivative of f is never zero. Let w = f xx2 +y
2 .
Show that x ∂w
+ y ∂w
= 0.
∂x
∂y
1
Solution (5 points):
Let’s start
the derivative with
to x,
by computing
respect
2x(x2 +y 2 )−2x(x2 −y 2 )
4xy 2
∂w
0 x2 −y 2
0 x2 −y 2
= f x2 +y2 ·
= f x2 +y2 · (x2 +y2 )2 ,
∂x
(x2 +y 2 )2
and with respect
to y, 2 2
2y
−2y(x +y )−2y(x2 −y 2 )
∂w
0 x2 −y 2
0 x2 −y 2
=
f
·
=
f
· (x−4x
2 +y 2 )2 .
∂y
x2 +y 2
(x2 +y 2 )2
x2 +y 2
And now, let’s compute
and make sure it is zero.
the2 sum
4x y 2 −4x2 y 2
∂w
∂w
0 x2 −y 2
x ∂x + y ∂y = f x2 +y2 · (x2 +y2 )2 = 0
3. Section 2.5, Problem 20
Let g : R3 → R2be a differentiable
function s.t. g(1, −1, 3) = (2, 5) and
1 −1 0
Dg(1, −1, 3) =
. Suppose that f : R2 → R2 is defined by
4 0 7
f (x, y) = (2xy, 3x − y + 5). What is D(f ◦ g)(1, −1, 3)?
Solution (5 points):
We know D(f ◦ g)(1, −1, 3) = Df (g(1, −1, 3)) · Dg(1, −1, 3) via the
Chain-Rule, and from the problem statement, g(1, −1, 3) = (2, 5), so
D(f ◦ g)(1, −1, 3) = Df (2, 5) · Dg(1, −1, 3). Moreover, we also know
Dg, so we only
work out Df at (2, 5), which is
" need to#
∂f1
∂f1
2y 2x 10 4
∂x
∂y
=
=
.
Df (2, 5) = ∂f2 ∂f2 3 −1 (2,5)
3 −1
∂x
∂y
(x,y)=(2,5)
Therefore,
D(f ◦ g)(1, −1, 3) =
10 4
3 −1
1 −1 0
26 −10 28
·
=
.
4 0 7
−1 −3 −7
4. Section 2.6, Problem 4
Calculate the directional derivative of the function f at the point ~a in
1
a = (3, −2) and ~u = î − ĵ.
the direction ~u, where f (x, y) = (x2 +y
2) , ~
Solution (5 points):
We want to compute the directional
derivative, so
let’s first compute
−2y
−2x
−6
4
the gradient at ~a, ∇f (3, −2) = (x2 +y2 )2 (x2 +y2 )2 = 169
, 169
,
(3,−2)
−1
then the unit vector û = √12 , √
, such that
2
D~u f = ∇f (3, −2) · û =
−6
√
169 2
−
4√
169 2
=
−10
√
169 2
5. Section 2.6, Problem 12
A ladybug (who is very sensitive to temperature) is crawling on a
graph paper. She is at point (3,7) and notices that if she moves in
2
the î-direction, the temperature increases at a rate of 3 deg/cm. If she
moves in the ĵ-direction, she finds that her temperature decreases at a
rate of 2 deg/cm. In what direction should the ladybug move if
(a) she wants to warm up the most rapidly?
Solution (5 points):
The gradient, ∇f (3, 7) = (3, −2), points in the direction of great∇f
√
= (3,−2)
est increase, so ~u = k∇f
.
k
13
(b) she wants to cool off the most rapidly?
Solution (5 points):
The negative gradient points in the direction of greatest decrease,
∇f
√
= (−3,2)
.
so ~u = − k∇f
k
13
(c) she desires her temperature not to change?
Solution (5 points):
The temperature should not change, so D~u f = 0, which implies
√
or
that ∇f · ~u = 0, or ∇f ⊥~u. Thus, any path along ~u = (2,3)
13
(−2,−3)
√
13
will not change the temperature.
3