therefore
Df · Dx
1 1
= y cos xy x cos xy
2s 2t
= y cos xy + 2sx cos xy y cos xy + 2tx cos xy
Section 2.5 selected answers
Math 131 Multivariate Calculus
D Joyce, Spring 2014
Thus,
Exercises 1, 2, 5, 8, 11, 15, 19, 22, 23.
∂f
= y cos xy + 2sx cos xy
∂s
2. If f (x, y) = sin xy, and x = s + t, y = s2 + t2 ,
and
find ∂f /∂s and ∂f /∂t in two ways:
∂f
= y cos xy + 2tx cos xy
∂t
a. By substitution.
Since
which agrees with the part a after x is replaced by
s + t and y is replaced by s2 + t2 .
f (x, y) = sin xy = sin(s + t)(s2 + t2 )
= sin(s3 + st2 + s2 t + t3 )
5. This is a word problem where we have a box
of dimensions L × W × H. At the initial time
the dimensions are L = 7, W = 5, and H = 4,
and they’re changing at the rates dL/dt = 0.75,
dW/dt = 0.5, and dH/dt = −15. You have to find
dV /dt at that instant, where V is the volume of the
box, V = LW H.
There’s only one independent variable, t, so these
are all ordinary derivatives. In fact, you may wonder why this problem is here in a section on the
chain rule for multivariate functions when this looks
like ordinary calculus. It’s because you can treat
the dimensions of the box as a vector (L, W, H) ∈
R3 , or actually, since they depend on time, they’re
a function R → R3 . Then the volume is a scalar
valued function R3 → R. The composition of these
two functions is a function R → R.
Well,
therefore
∂f
= (3s2 + t2 + 2st) cos(s3 + st2 + s2 t + t3 )
∂s
and
∂f
= (2st + s2 + 3t2 ) cos(s3 + st2 + s2 t + t3 ).
∂t
b. by means of the chain rule.
For this exercise, the chain rule,
D(f ◦ x) = Df Dx
looks like
∂f
∂s
∂f
∂t
=
h
∂f
∂x
∂f
∂y
i ∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t
dV
dt
Since
Df =
h
∂f
∂x
∂f
∂y
i
= y cos xy x cos xy
and
∂x
Dx =
∂s
∂y
∂s
∂x
∂t
∂y
∂t
1 1
=
,
2s 2t
d
(LW H)
dt
= L0 W H + LW 0 H + LW H 0
= 0.75 · 5 · 4 + 7 · (0.5) · 4 + 7 · 5 · (−1)
= 15 + 14 − 36 = −6
=
Therefore, the volume is decreasing.
1
8. The Centers for Disease Control provide infor- that
2 2
mation on the body mass index (BMI) which de∂z
∂z
∂z
∂z
−2r
pends on both a person’s weight and height. It’s
+
.
+
=e
∂x
∂y
∂r
∂θ
given by the formula
First note that although r and θ appear in this
exercise, it’s not actually polar coordinates we’ve
got. However, you can do this exercise in the same
where w is an individual’s mass in kilograms and h way that polar coordinates are done in the text.
is the person’s height in centimeters.
We can treat the vector x = (x, y) as a function
While monitoring a child’s growth, you estimate R2 → R2 as
that at the time he turned 10 years old, his height
x(r, θ) = (x(r, θ), y(r, θ)) = (er cos θ, er sin θ).
showed a growth rate of 0.6 cm per month. At the
same time, his mass showed a growth rate of 0.4 kg
Then we can apply the general chain rule D(f ◦x) =
per month. Suppose that he was 140 cm tall and
Df Dx. We’ll get the matrix equation
weighed 33 kg on his tenth birthday.
∂z ∂z ∂z ∂z ∂x ∂x
a. At what rate is his BMI changing on his tenth
∂r
∂θ
= ∂x ∂y ∂y
∂y
∂r
∂θ
birthday?
∂r ∂θ
∂z ∂z er cos θ −er sin θ
Using the product formula and the chain formula
= ∂x ∂y
we find
er sin θ er cos θ
BM I =
10000w
h2
10000 ∂w
h2 − 10000w(2h) ∂h
d BMI
∂t
∂t
=
dt
h4
10000h ∂w
− 20000w ∂h
∂t
∂t
=
h3
Therefore,
∂z
∂z
∂z
= er cos θ
+ er sin θ
∂r
∂x
∂y
and
At 10 years we get
10000 · 140 · 0.4 − 20000 · 33 · 0.6
d BMI
=
= 0.060
dt
1403
∂z
∂z
∂z
= −er sin θ
+ er cos θ
∂θ
∂x
∂y
We need to eliminate θ form these two equations. You can do that if you square each equation and add the resulting equations. The middle terms cancel out, and the Pythagorean identity
sin2 θ + cos2 θ = 1 simplifies the result to give
" 2 2
2 #
2
∂z
∂z
∂z
∂z
+
= e2r
+
.
∂r
∂θ
∂x
∂y
Note that all measurements are in months, cm, and
kg, so the answer is in BMI/month.
b. The BMI of a typical 10-year-old male increases at an average rate of 0.04 BMI/month.
Should you be concerned about the child’s weight
gain?
The rate 0.06 is 50% greater than the rate 0.04,
2r
so his BMI is growing significantly faster than aver- Divide by e to get the required equation.
age, but, given the lack of information, that could
xy
be good or bad. Maybe his BMI was exception15. If w = f
is a differentiable funcx2 + y 2
ally low to begin with. Maybe the BMI was a bad
xy
measurement to begin with.
tion of u = 2
, show that
x + y2
11. Suppose z = f (x, y) has continuous partial
derivatives. Let x = er cos θ, and y = er sin θ. Show
x
2
∂w
∂w
+y
= 0.
∂x
∂y
Here we have a composition of u : R2 → R and Therefore,
∂w ∂w f : R → R. The chain rule looks like the matrix
∂s
∂t
D(f ◦ g) =
equation
∂z
∂z
∂s ∂t 4
15(s − 7t) −105(s − 7t)4
∂w ∂w
∂u
∂u
dw
=
=
2e2s−14t
−14e2s−14t
∂x ∂y
∂x ∂y
du
Part b. For this exercise, the chain rule,
As two equations, this says
D(f ◦ g) = Df Dg
dw ∂u
∂w
=
∂x
du ∂x
looks like
∂w
∂s
∂z
∂s
and
∂w
dw ∂u
=
.
∂y
du ∂y
∂w
∂t
∂z
∂t
D(f ◦ g) =
∂w
∂w
+y
∂x
∂y
dw ∂u
dw ∂u
x
+y
du ∂x
du ∂y
dw
∂u
∂u
+y
x
du
∂x
∂y
3
2
y −x y
x3 − xy 2
dw
x 2
+y 2
du
(x + y 2 )2
(x + y 2 )2
dw
0
= 0.
2
du (x + y 2 )2
x
∂s
∂z
∂s
∂w
∂t
∂z
∂t
Once x is replaced by s − 7t, this is the same as the
answer in part a.
22. Calculate D(f ◦ g) in two ways: (a) by first
evaluating f ◦ g and (b) by using the chain rule and
the derivative matrices Df and Dg.
f (x, y) = x2 − 3y 2 ,
19. Calculate D(f ◦ g) in two ways: (a) by first
evaluating f ◦ g and (b) by using the chain rule and
the derivative matrices Df and g.
f (x) = (3x5 , e2x ),
= Df Dg
∂w ∂x ∂x ∂x
=
∂z
∂s
∂t
∂x 4 15x 1 −7
=
2x
2e
15x4 −105x4
=
2e2x −14e2x
Let’s see if that’s enough to finish the exercise.
=
∂x
∂t
∂w
2
∂u
x − xy
= 2
.
∂y
(x + y 2 )2
=
∂s
Therefore,
and
=
∂x
4
15x
Df = ∂x
=
∂z
2e2x
∂x
∂x
Dg = ∂x
= 1 −7
∂s
∂t
y 3 − x2 y
∂u
= 2
∂x
(x + y 2 )2
=
∂x
∂z
∂x
=
∂w Now,
3
∂w g(s, t) = (st, s + t2 ).
Part a.
z =
=
=
=
=
g(s, t) = s − 7t
Part a.
(w, z) = f (x) = (3x5 , e2x ) = (3(s − 7t)5 , e2s−14t )
3
(f ◦ g)(s, t)
f (st, s + t2 )
s2 t2 − 3(s + t2 )2
s2 t2 − 3(s2 + 2st2 + t4 )
s2 t2 − 3s2 − 6st2 − 3t4
So
Therefore,
∂z
= 2st2 − 6s − 6t2
∂s
and
D(f ◦ g)
∂w ∂w ∂z
= 2s2 t − 12st − 12t3
∂t
=
Part b. In this exercise, the chain rule, D(f ◦
g) = Df Dg, looks like
∂f ∂f h ∂f ∂f i ∂x ∂x
∂s
∂t
= ∂x ∂y ∂y ∂y
∂s
∂t
=
∂s
Df =
h
∂f
∂x
∂x
Dg =
∂s
∂y
∂s
∂f
∂y
i
∂x
∂t
∂y
∂t
∂t
∂z
∂t
2
3s − t2
−2st
−2 −2
−s t + 6s5 t3 −2s−1 t−3 + 3s6 t2
Part b. Here’s the setup for this part.
∂w ∂w ∂x
∂y
Df =
∂z
∂z
∂x
∂y
y + x−2 y
x − x−1
=
y −1
−xy −2 + 3y 2
∂t
= 2x −6y
∂s
∂z
∂s
t s
=
1 2t
∂x
Therefore,
Dg =
∂s
∂y
∂s
−1
∂x
∂t
∂y
∂t
−2
D(f ◦ g) = Df Dg
t
−st
=
t s
2st
s2
= 2x −6y
1 2t
Therefore,
= 2xt − 6y 2xs − 12yt
= 2st2 − 6s − 6t2 2s2 t − 12st − 12t3
D(f ◦ g)
∂w ∂w ∂s
∂t
=
which agrees with the results in part a.
∂z
∂z
∂s
∂t
∂w ∂w ∂x ∂x ∂x
∂y
∂s
∂t
=
23. Calculate D(f ◦ g) in two ways: (a) by first
∂y
∂y
∂z
∂z
∂x
∂y
∂s
∂t
evaluating f ◦ g and (b) by using the chain rule and
−1
−2
y
+
x
y
x
− x−1
t
−st−2
the derivative matrices Df and Dg.
=
y −1
−xy −2 + 3y 2 2st
s2
y x
f (x, y) = xy − , + y 3
When these matrices are multiplied and the varix y
ables x and y replaced by expressions in s and t,
s
the result should be the same as in part a.
g(s, t) =
, s2 t
t
Math 131 Home Page at
Part a.
http://math.clarku.edu/~djoyce/ma131/
(w, z) = (f ◦ g)(s, t)
s
= f
, s2 t
t
s 2
s2 t s/t
2 3
=
s t−
,
+ (s t)
t
s/t s2 t
= s3 − st2 , s−1 t−2 + s6 t3
4