PARTIAL DIFFERENTIATION
5.19 The cost always includes 2αh, which can therefore be ignored in the optimisation.
With Lagrange multiplierλ,sinθ=λw/(4β)andβsecθ−^12 λwtanθ=λh, leading
to the stated results.
5.21 The envelope of the linesx/a+y/(c−a)−1 = 0, asais varied, is
√
x+√y=√
c.
Area =c^2 /6.
5.23 (a) Usingα=cotθ,whereθis the initial angle a jet makes with the vertical, the
equation isf(z, ρ, α)=z−ρα+[gρ^2 (1+α^2 )/(2v^20 )], and setting∂f/∂α= 0 gives
α=v^20 /(gρ). The water bell has a parabolic profilez=v^20 /(2g)−gρ^2 /(2v 02 ).
(b) Settingz= 0 gives the minimum diameter as 2v^20 /g.
5.25 Show that (∂G/∂P)T=Vand (∂G/∂T)P=−S. From each result, obtain an
expression for∂^2 G/∂T ∂Pand equate these, giving (∂V /∂T)P=−(∂S/∂P)T.
5.27 Find expressions for (∂S/∂V)Tand (∂S/∂T)V, and equate∂^2 S/∂V ∂Twith
∂^2 S/∂T ∂V.U(V,T)=cT−aV−^1.
5.29 dI/dy=−Im[
∫∞
0 exp(−xy+ix)dx]=−^1 /(1 +y(^2) ). IntegratedI/dyfrom 0 to∞.
I(∞)=0andI(0) =J.
5.31 Integrate the RHS of the equation by parts, before differentiating with respect
toy. Repeated application of the method establishes the result for all orders of
derivative.
5.33 I(0) = 0; use Leibnitz’ rule.
5.35 Writex(t)=−cost
∫t
0 sinξf(ξ)dξ−sint
∫π
tcosξf(ξ)dξand differentiate each
term as a product to obtaindx/dt.Obtaind^2 x/dt^2 in a similar way. Note that
integrals that have equal lower and upper limits have value zero. The value of
x(π)is
∫π
0 sinξf(ξ)dξ.