348 Higher Engineering Mathematics
- In a thermodynamic system,k=Ae
TS−H
RT ,
whereR,kandAare constants.Find(a)∂k
∂T(b)∂A
∂T(c)∂(S)
∂T(d)∂(H)
∂T
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
(a)∂k
∂T=AH
RT^2eTS−H
RT(b)∂A
∂T=−kH
RT^2e
H−RTTS(c)∂(S)
∂T=−H
T^2(d)∂(H)
∂T=S−Rln(
k
A)⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦34.3 Second order partial derivatives
As with ordinary differentiation, where a differential
coefficient may be differentiated again, a partial deriva-
tive may be differentiated partially again to give higher
order partial derivatives.(i) Differentiating∂V
∂rof Section 34.2 with respecttor, keepinghconstant, gives∂
∂r(
∂V
∂r)
whichis written as∂^2 V
∂r^2Thus if V=πr^2 h,then∂^2 V
∂r^2=∂
∂r( 2 πrh)= 2 πh.(ii) Differentiating∂V
∂hwith respect toh, keepingr constant, gives∂
∂h(
∂V
∂h)
which is writtenas∂^2 V
∂h^2Thus∂^2 V
∂h^2=∂
∂h(πr^2 )= 0.(iii) Differentiating∂V
∂hwith respect tor, keepingh constant, gives∂
∂r(
∂V
∂h)
which is writtenas∂^2 V
∂r∂h. Thus,
∂^2 V
∂r∂h=∂
∂r(
∂V
∂h)
=∂
∂r(πr^2 )= 2 πr.(iv) Differentiating∂V
∂rwith respect toh, keepingrconstant, gives∂
∂h(
∂V
∂r)
, which is written as∂^2 V
∂h∂r. Thus,
∂^2 V
∂h∂r=∂
∂h(
∂V
∂r)
=∂
∂h( 2 πrh)= 2 πr.(v)∂^2 V
∂r^2,∂^2 V
∂h^2,∂^2 V
∂r∂hand∂^2 V
∂h∂rare examples ofsecond 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. Givenz= 4 x^2 y^3 − 2 x^3 + 7 y^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= 8 xy^3 − 6 x^2∂^2 z
∂x^2=∂
∂x(
∂z
∂x)
=∂
∂x( 8 xy^3 − 6 x^2 )= 8 y^3 − 12 x(b)∂z
∂y= 12 x^2 y^2 + 14 y∂^2 z
∂y^2=∂
∂y(
∂z
∂y)
=∂
∂y( 12 x^2 y^2 + 14 y)= 24 x^2 y+ 14