Week 18 - Directional Derivatives and the Gradient; the Chain Rule
Hughes-Hallett Section 14.6
SUGGESTED PROBLEMS
For Exercises 1-6, find dz
dt using the chain rule. Assume the variables are restricted to domains
on which the functions are defined.
1. z = xy 2 , x = e−t , y = sin t
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
2
−t
= (y )(−e ) + (2xy)(cos(t))
Now put everything in terms of t:
= −(sin2 t)e−t + 2e−t sin(t) cos(t)
We can also solve this by writing everything in terms of t to begin with, and then doing
d
taking
:
dt
d −t 2
dz
=
e sin t
dt
dt
= −e−t sin2 t + e−t (2 sin t cos t)
This is the same answer as we achieved through using the multivariate chain rule. In
many cases, however, using the multi-variate chain rule will result in easier steps along
the way than writing everything in terms of t before you begin.
3. z = xey , x = 2t,y = 1 − t2
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= (ey )(2) + (xey )(−2t)
Put in terms of t:
∂z
For Exercises 7-14, find ∂u
and
the functions are defined.
9. z = xey , x = u2 + v 2 ,y = u2 − v 2
∂z
∂v
2
2
= 2e1−t − 4t2 e1−t
. The variables are restricted to domains on which
2
MATH 122 - Section H-H 14.6 Solutions
Put in terms of u, v:
Put in terms of u, v:
∂z
∂z ∂x
∂z ∂y
=
+
∂u
∂x ∂u
∂y ∂u
y
= (e )(2u) + (xey )(2u)
∂z
2
2
2
2
= (eu −v )(2u) + 2u(u2 + v 2 )eu −v
∂u
∂z ∂x
∂z ∂y
∂z
=
+
∂v
∂x ∂v
∂y ∂v
y
= (e )(2v) + (xey )(−2v)
∂z
2
2
2
2
= (eu −v )(2v) − 2v(u2 + v 2 )eu −v
∂v
11. z = sin(x/y), x = ln u,y = v
Put in terms of u, v:
Put in terms of u, v:
∂z
∂u
∂z
∂u
∂z
∂v
∂z
∂v
11
−x
= cos(x/y)
+ cos(x/y) 2 (0)
yu
y
ln u
1
=
cos
uv
v
1
−x
= cos(x/y) (0) + cos(x/y) 2 (1)
y
y
ln u ln u
= − cos
v
v2
17. Corn production, C, is a function of rainfall, R, and temperature, T . Figures 14.49 and
14.50 show how rainfall and temperature are predicted to vary with time because of global
warming. Suppose we know that ∆C ≈ 3.3.∆R − 5∆T . Use this to estimate the change
dC
in corn production between the year 2020 and the year 2021. Hence, estimate
when
dt
t = 2020.
Figure 14.49:
Rainfall as a function of time
3
MATH 122 - Section H-H 14.6 Solutions
Figure 14.50:
Temperature as a function of time
We will need to estimate the rates
dR
dT
and
from the graphs.
dt
dt
−2
∆T
3
∆R
≈
and
≈ .
∆t
30
∆t
40
Now both R and T are functions of time t (in years), and we want to find the effect of a
small change in time, ∆t, on R and T . For a change of ∆t = 1, we can estimate
From these, we can say that
−2
−2
∆t =
30
30
3
3
∆T = ∆t =
40
40
∆R =
From those, we can estimate ∆C:
∆C = 3.3∆R − 5∆T
3
−2
−5
= 3.3
30
40
= −0.595
From these calculations we would estimate that corn production will drop by 0.595 over
the next year 2020 to 2021.
4
MATH 122 - Section H-H 14.6 Solutions
19. The voltage, V , (in volts) across a circuit is given by Ohm’s law: V = IR, where I is the
current (in amps) flowing through the circuit and R is the resistance (in ohms). If we
place two circuits, with resistance R1 and R2 , in parallel, then their combined resistance,
R, is given by
1
1
1
=
+
R
R1 R2
Suppose the current is 2 amps and increasing at 10−2 amp/sec and R1 is 3 ohms and
increasing at 0.5 ohm/sec, while R2 is 5 ohms and decreasing at 0.1 ohm/sec. Calculate
the rate at which the voltage is changing.
V =I R
dI
dR
dV
=
R+I
dt
dt
dt
dI
∂R dR1
∂R dR2
=
R+I
+
dt
∂R1 dt
∂R2 dt
∂R
To determine
, it is easier to express
∂R1
R1 R2
R=
R1 + R2
R2 (R1 + R2 ) − R1 R2
∂R
=
From the quotient rule,
∂R1
(R1 + R2 )2
∂R
R1 (R1 + R2 ) − R1 R2
and
=
∂R2
(R1 + R2 )2
Subbing in R1 = 3, R2 = 5
5(3 + 5) − (3)(5)
25
∂R
=
=
2
∂R1
(3 + 5)
64
∂R
3(3 + 5) − (3)(5)
9
=
=
2
∂R2
(3 + 5)
64
dR2
dI
(3)(5)
15
dR1
= 0.5,
= −0.1, I = 2,
= 0.01, R =
=
Using the other values,
dt
dt
dt
3+5 8
15
dV
25
9
= (0.01)
(0.5) + (−0.1)
+ (2)
dt
8
64
64
≈ 0.3812
20. Air pressure decreases at a rate of 2 pascals per kilometer in the eastward direction. In
addition, the air pressure is dropping at a constant rate with respect to time everywhere.
A ship sailing eastward at 10 km/hour past an island takes barometer readings and records
a pressure drop of 50 pascals in 2 hours. Estimate the time rate of change of air pressure
on the island. (A pascal is a unit of air pressure.)
Pressure, P is a function of x, the distance east of the island, in kilometers, and t, the
time in hours since the ship passes the island.
5
MATH 122 - Section H-H 14.6 Solutions
The S(t) be the air pressure on the ship at time t, so that S(t) = p(10t, t). By the chain
rule we have
∂P
dS
=
dt
∂x
∂P
=
∂x
dx
∂P dt
+
dt
∂t dt
dx
∂P
+
dt
∂t
Of these terms, we know
dS
−50 Pa
=
dt
2 hours
∂P
Pa
= −2
∂x
km
dx
km
= 10
dt
hour
Using these, we can solve for the rate of temperature change,
dP
:
dt
∂P
−50
= (−2)(10) +
2
∂t
∂P
= −5
∂t
From this, the pressure is dropping everywhere at 5 Pa per hour.