Download The Chain Rule

Calculus 3
Lia Vas
The Chain Rule
Recall the chain rule for functions of single variable:
If y = f (x) and x = g(t), then y 0 (t) = y 0 (x) · x0 (t) ( or
dy dx
dy
=
).
dt
dx dt
The chain rule for function z = f (x, y) with x = g(t) and y = h(t) is
z 0 (t) = zx · x0 (t) + zy · y 0 (t) or
dz
∂z dx ∂z dy
=
+
.
dt
∂x dt
∂y dt
The chain rule for function z = f (x, y) with x = g(s, t) and y = h(s, t) is
zs = zx · xs + zy · ys ,
or
zt = zx · xt + zy · yt
∂z
∂z ∂x ∂z ∂y
=
+
∂s
∂x ∂s ∂y ∂s
∂z
∂z ∂x ∂z ∂y
=
+
∂t
∂x ∂t
∂y ∂t
Implicit functions. In some cases, an implicit function F (x, y, z) = 0 cannot be solved for z. To
find the derivatives zx and zy in these cases, use implicit differentiation. Differentiating the equation
with respect to x we obtain Fx + Fz zx = 0. Solving for zx gives us zx = − FFxz . Differentiating the
equation with respect to y we obtain Fy + Fz zy = 0. Solving for zy gives us zy = − FFyz . Thus, the
partial derivatives zx and zy are given by:
zx = −
Fx
Fz
and
zy = −
Fy
Fz
Practice problems.
1. Find the indicated derivatives.
(a) z = 3x2 + 2xy − 5y 2 , x = 2 + t2 , y = 1 − t3 ; z 0 (t) when t = 0.
(b) z = x ln(x + 2y), x = cos t, y = sin t; z 0 (t) when t = 0.
(c) z = ex y + xy 2 , x = st, y = s2 + t2 ; zs and zt at (1, 1).
(d) z = x2 + xy, x = et cos s, y = et sin s; zs and zt at (π, 0).
(e) xy 2 + yz 2 + zx2 = 3; zx and zy at (1, 1, 1).
(f) x − yz = cos(x + y + z); zx and zy at (0, 1, −1).
2. The pressure of 1 mole of an ideal gas is increasing at a rate of 0.05 kPa/s and the temperature
is rising at a rate of 0.15 K/s. The pressure P , volume V and temperature T are related by
the equation P V = 8.31T. Find the rate of change of the volume when the pressure is 20 kPa
and temperature 320 K.
3. The number N of bacteria in a culture depends on temperature T and pressure P which
depend on time t in minutes. Assume that 3 minutes after the experiment started, the pressure
is increasing at a rate of 0.1 kPa/min and the temperature at a rate of 0.5 K/min. The number
of bacteria changes at the rates of 3 bacteria per kPa and 5 bacteria per Kelvin. Find the rate
at which the number of bacteria is increasing 3 minutes after the experiment started.
4. The number of flowers N in a closed environment depends on the amount of sunlight S that
the flowers receive and the temperature T of the environment. Assume that the number of
flowers changes at the rates NS = 2 and NT = 4. If the temperature depends on time as
1
8
T (t) = 85 − 1+t
2 and the amount of sunlight decreases on time as S(t) = t , find the rate of
change of the flower population at time t = 2 days.
5. The temperature at a point (x, y) is T (x, y),√measured in degrees Celsius. A bug crawls so that
its position after t seconds is given by x = 1 + t, y = 2 + 31 t where x and y are measured in
centimeters. The temperature function satisfies Tx (2, 3) = 4 and Ty (2, 3) = 3. Determine how
fast the temperature is rising on the bug’s path after 3 seconds.
6. If the trajectory of an object is given by a parametric curve, its speed at a point can be found
as the length of the tangent vector at that point. Assume that an object moves in
√ space
away from its initial position so that after t hours it is at x = 2t and y = 2t2 −1 and z = 3t + 1
miles from its initial position. Find the speed of that object 5 hours after it started moving.
7. An object moves in xy-plane so that after t hours it is at x = x(t) and y = y(t) miles from its
initial position. Suppose that after 2 hours, the object has the velocity so that x0 (2) = 3 and
y 0 (2) = 5. Find the speed of the object after 2 hours.
Solutions.
∂z dx
∂z dy
= ∂x
+ ∂y
= (6x + 2y)(2t) + (2x − 10y)(−3t2 ). When t = 0, x = 2 + 02 = 2 and
1. (a) dz
dt
dt
dt
y = 1 − t3 = 1 = 0 = 1. Thus, z 0 (0) = (12 + 2)(0) + (4 − 10)0 = 0.
∂z dx
∂z dy
x
2x
= ∂x
+ ∂y
= (ln(x + 2y) + x+2y
)(− sin t) + x+2y
(cos t). When t = 0, x = cos 0 = 1
(b) dz
dt
dt
dt
1
2
0
and y = sin 0 = 0. Thus, z (0) = (ln(1 + 0) + 1+0 )(0) + 1+0 (1) = 0 + 2 = 2.
(c) zs = zx xs +zy ys = (ex y+y 2 )t+(ex +2xy)2s, zt = zx xt +zy yt = (ex y+y 2 )s+(ex +2xy)2t. When
s = 1, t = 1, x = 1(1) = 1 and y = 12 + 12 = 2. Thus zs (1, 1) = (e1 2 + 22 )1 + (e1 + 2(1)2)2 =
4e + 12 and zt (1, 1) = (e1 2 + 22 )1 + (e1 + 2(1)2)2 = 4e + 12.
(d) zs = zx xs + zy ys = (2x + y)(−et sin s) + x(et cos s), zt = zx xt + zy yt = (2x + y)(et cos s) +
x(et sin s). When s = π and t = 0, x = e0 cos π = −1 and y = e0 sin π = 0. Thus, zs (π, 0) =
(−2)(0) + (−1)(−1) = 1 and zt (π, 0) = (−2)(−1) + (−1)(0) = 2.
(e) Use the formulas for derivatives of implicit function. Let F = xy 2 + yz 2 + zx2 − 3. Then
2 +2xz
Fy
Fx = y 2 + 2xz, Fy = 2xy + z 2 , Fz = 2yz + x2 and so zx = − FFxz = − y2yz+x
=
2 and zy = − F
z
2
2xy+z
1+2
2+1
− 2yz+x
2 . At (1, 1, 1), zx = − 2+1 = −1 and zy = − 2+1 = −1.
(f) Use the formulas for derivatives of implicit function. Let F = x − yz − cos(x + y + z).
Then Fx = 1 + sin(x + y + z), Fy = −z + sin(x + y + z), Fz = −y + sin(x + y + z) and so
1+sin(x+y+z)
−z+sin(x+y+z)
1+0
zx = − FFxz = − −y+sin(x+y+z)
and zy = − FFyz = − −y+sin(x+y+z)
. At (0, 1, −1), zx = − −1+0
= 1 and
1+0
zy = − −1+0 = 1.
2. We are given the rates dP
= 0.05 kPa/s,
dt
dV
need to find the rate dt at (P0 , T0 ).
dT
dt
= 0.15 K/s at P0 = 20 kPa and T0 = 320 K. We
Since the derivatives are with respect to P and T , treat V as the dependent variable and
. By the chain
solve for V in terms of P and T to obtain the dependence formula V = 8.31T
P
dV
∂V dP
∂V dT
8.31T
∂V
−8.31T
rule, dt = ∂P dt + ∂T dt . From V = P we find that ∂P = P 2 and that ∂V
= 8.31
.
∂T
P
−8.31(320)
∂V
At P0 = 20 and T0 = 320, we calculate that ∂P =
= −6.648 liters/kPa and that
400
∂V
8.31
dV
= 20 = 0.4155 liters/K. Thus, dt = (−6.648)(0.05) + (0.4155)(0.15) = −.27 liter per
∂T
second.
3. We are given the rates P 0 (3) = 0.1, T 0 (3) = 0.5, NP = 3, and NT = 5. By the chain rule
formula, N 0 (t) = NT T 0 + NP P 0 . Thus N 0 (3) = 5 · 0.5 + 3 · 0.1 = 2.8 bacteria/minute.
4. We are given the rates NS = 2 and NT = 4. The remaining two rates S 0 and T 0 needed for the
chain rule formula N 0 = NS S 0 + NT T 0 can be found from the formulas for S and T.
S=
1
t
⇒ S0 =
T = 85 −
8
1+t2
−1
.
t2
At t = 2, S 0 =
⇒ T0 =
16t
.
(1+t2 )2
−1
.
4
At t = 2, T 0 =
32
.
25
32
Thus, N 0 (2) = NS S 0 + NT T 0 = 2 −1
+ 4 25
= 4.62 flowers/day.
4
5. We are given the rates Tx (2, 3) = 4 and Ty (2, 3) = 3 and the problem is asking for the rate
at t = 3.
dT
dt
dy
dx
Using chain rule, dT
= ∂T
+ ∂T
. x0 = 2√11+t and y 0 = 13 . At t = 3, x0 = 2√1 4 = 14 and y 0 = 13 .
dt
∂x dt
∂y dt
√
When t = 3, x = 1 + 3 = 2 and y = 2 + 1 = 3 so the point (2, 3) at which the derivatives Tx
= 4 14 + 3 31 = 1 + 1 = 2 degrees Celsius
and Ty are given corresponds exactly to t = 3. Thus, dT
dt
per second.
6. x0 = 2, y 0 = 4t and z 0 =
√3
.
2 3t+1
When t = 5, x0 = 2, y 0 = 20 and z 0 = 38 . Thus, the tangent
vector is h2, 20, 38 i. The speed is the length |h2, 20, 83 i| =
q
22 + 202 + ( 83 )2 = 20.1 miles/hour.
√
7. At t = 2, hx0 , y 0 i = h3, 5i. The speed = length of h3, 5i = 32 + 52 = 5.83 miles/ hour.