348 Higher Engineering Mathematics
10. In a thermodynamic system, k = Ae
T S−H
RT
,
where R, k and A are constants.
Find (a)
∂k
∂T
(b)
∂ A
∂T
(c)
∂((S)
∂T
(d)
∂((H )
∂T
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
(a)
∂k
∂T
=
AH
RT 2 e
T S−H
RT
(b)
∂ A
∂T
= −
kH
RT 2 e
H −T S
RT
(c)
∂((S)
∂T
=−
H
T 2
(d)
∂((H )
∂T
= S − R ln
k
A
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
34.3 Second order partial derivatives
As with ordinary differentiation, where a differential
coefficient may be differentiated again, a partial derivative may be differentiated partially again to give higher
order partial derivatives.
(i) Differentiating
∂V
∂r
of Section 34.2 with respect
to r, keeping h constant, gives
∂
∂r
∂V
∂r
which
is written as
∂ 2 V
∂r 2
Thus if V = πr
2 h,
then
∂ 2 V
∂r 2 =
∂
∂r
(2πrh) = 2π h.
(ii) Differentiating
∂V
∂h
with respect to h, keeping
r constant, gives
∂
∂h
∂V
∂h
which is written
as
∂ 2 V
∂h 2
Thus
∂ 2 V
∂h 2 =
∂
∂h
(πr 2 ) = 0.
(iii) Differentiating
∂V
∂h
with respect to r, keeping
h constant, gives
∂
∂r
∂V
∂h
which is written
as
∂ 2 V
∂r∂h
. Thus,
∂ 2 V
∂r∂h
=
∂
∂r
∂V
∂h
=
∂
∂r
(πr
2
) = 2π r.
(iv) Differentiating
∂V
∂r
with respect to h, keeping r
constant, gives
∂
∂h
∂V
∂r
, which is written as
∂
2 V
∂h∂r
. Thus,
∂ 2 V
∂h∂r
=
∂
∂h
∂V
∂r
=
∂
∂h
(2πrh) = 2π r.
(v)
∂ 2 V
∂r 2 ,
∂ 2 V
∂h 2 ,
∂ 2 V
∂r∂h
and
∂ 2 V
∂h∂r
are examples of
second order partial derivatives.
(vi) It is seen from (iii) and (iv) that
∂ 2 V
∂r∂h
=
∂ 2 V
∂h∂r
and such a result is always true for continuous
functions (i.e. a graph of the function which has
no sudden jumps or breaks).
Second order partial derivatives are used in the solution
of partial differential equations, in waveguide theory, in
such areas of thermodynamics covering entropy and the
continuity theorem, and when finding maxima, minima
and saddle points for functions of two variables (see
Chapter 36).
Problem 7. Given z =4x 2 y 3 − 2x 3 + 7y 2 find
(a)
∂ 2 z
∂x 2 (b)
∂ 2 z
∂ y 2 (c)
∂ 2 z
∂x∂ y
(d)
∂ 2 z
∂ y∂x
(a)
∂z
∂x
= 8x y 3 − 6x 2
∂ 2 z
∂x 2 =
∂
∂x
∂z
∂x
=
∂
∂x
(8x y 3 − 6x 2 )
= 8y 3 − 12 x
(b)
∂z
∂ y
= 12x 2 y 2 + 14y
∂ 2 z
∂ y 2 =
∂
∂ y
∂z
∂ y
=
∂
∂ y
(12x 2 y 2 + 14y)
= 24x 2 y + 14
10. In a thermodynamic system, k = Ae
T S−H
RT
,
where R, k and A are constants.
Find (a)
∂k
∂T
(b)
∂ A
∂T
(c)
∂((S)
∂T
(d)
∂((H )
∂T
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
(a)
∂k
∂T
=
AH
RT 2 e
T S−H
RT
(b)
∂ A
∂T
= −
kH
RT 2 e
H −T S
RT
(c)
∂((S)
∂T
=−
H
T 2
(d)
∂((H )
∂T
= S − R ln
k
A
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
34.3 Second order partial derivatives
As with ordinary differentiation, where a differential
coefficient may be differentiated again, a partial derivative may be differentiated partially again to give higher
order partial derivatives.
(i) Differentiating
∂V
∂r
of Section 34.2 with respect
to r, keeping h constant, gives
∂
∂r
∂V
∂r
which
is written as
∂ 2 V
∂r 2
Thus if V = πr
2 h,
then
∂ 2 V
∂r 2 =
∂
∂r
(2πrh) = 2π h.
(ii) Differentiating
∂V
∂h
with respect to h, keeping
r constant, gives
∂
∂h
∂V
∂h
which is written
as
∂ 2 V
∂h 2
Thus
∂ 2 V
∂h 2 =
∂
∂h
(πr 2 ) = 0.
(iii) Differentiating
∂V
∂h
with respect to r, keeping
h constant, gives
∂
∂r
∂V
∂h
which is written
as
∂ 2 V
∂r∂h
. Thus,
∂ 2 V
∂r∂h
=
∂
∂r
∂V
∂h
=
∂
∂r
(πr
2
) = 2π r.
(iv) Differentiating
∂V
∂r
with respect to h, keeping r
constant, gives
∂
∂h
∂V
∂r
, which is written as
∂
2 V
∂h∂r
. Thus,
∂ 2 V
∂h∂r
=
∂
∂h
∂V
∂r
=
∂
∂h
(2πrh) = 2π r.
(v)
∂ 2 V
∂r 2 ,
∂ 2 V
∂h 2 ,
∂ 2 V
∂r∂h
and
∂ 2 V
∂h∂r
are examples of
second order partial derivatives.
(vi) It is seen from (iii) and (iv) that
∂ 2 V
∂r∂h
=
∂ 2 V
∂h∂r
and such a result is always true for continuous
functions (i.e. a graph of the function which has
no sudden jumps or breaks).
Second order partial derivatives are used in the solution
of partial differential equations, in waveguide theory, in
such areas of thermodynamics covering entropy and the
continuity theorem, and when finding maxima, minima
and saddle points for functions of two variables (see
Chapter 36).
Problem 7. Given z =4x 2 y 3 − 2x 3 + 7y 2 find
(a)
∂ 2 z
∂x 2 (b)
∂ 2 z
∂ y 2 (c)
∂ 2 z
∂x∂ y
(d)
∂ 2 z
∂ y∂x
(a)
∂z
∂x
= 8x y 3 − 6x 2
∂ 2 z
∂x 2 =
∂
∂x
∂z
∂x
=
∂
∂x
(8x y 3 − 6x 2 )
= 8y 3 − 12 x
(b)
∂z
∂ y
= 12x 2 y 2 + 14y
∂ 2 z
∂ y 2 =
∂
∂ y
∂z
∂ y
=
∂
∂ y
(12x 2 y 2 + 14y)
= 24x 2 y + 14