Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 16A / b

Table of contents
- Solution b

Solution

Using

sinh (a + ib) = sinh a cos b + i cosh a sin b

in the quotient from part (a), multiplying numerator and denominator by the complex conjugate of the denominator, and taking the real part after multiplication by

e^{iω t}

gives

\frac{u ( y , t )}{U _{0}} = \frac{cos ω t [ cosh Δ _{+} cos Δ _{-} - cosh Δ _{-} cos Δ _{+} ]}{cosh ( 2Δ ) - cos ( 2Δ )} + \frac{sin ω t [ sinh Δ _{+} sin Δ _{-} - sinh Δ _{-} sin Δ _{+} ]}{cosh ( 2Δ ) - cos ( 2Δ )},

where

Δ_{\pm} = Δ (1 \pm y) .

The denominator follows from

2∣ sinh ((1 + i) Δ) ∣^{2} = cosh (2Δ) - cos (2Δ) .

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