Answer
(a)
(i) $f'(L) = -\frac{1}{2L^2}~\sqrt{\frac{T}{\rho}}$
(ii) $f'(T) = \frac{1}{4L}\sqrt{\frac{1}{T\rho}}$
(iii) $f'(\rho) = -\frac{1}{4L}\cdot~\frac{\sqrt{T}}{\rho^{3/2}}$
(b)
(i) When the length is decreased, the pitch increases.
(ii) When the tension is increased, the pitch increases.
(iii) When the linear density is increased, the pitch decreases.
Work Step by Step
(a) $f = \frac{1}{2L}~\sqrt{\frac{T}{\rho}}$
(i) $f'(L) = -\frac{1}{2L^2}~\sqrt{\frac{T}{\rho}}$
(ii) $f'(T) = (\frac{1}{2L})~(\frac{1}{2}\sqrt{\frac{1}{T\rho}})$
$f'(T) = \frac{1}{4L}\sqrt{\frac{1}{T\rho}}$
(iii) $f'(\rho) = (\frac{1}{2L})~(-\frac{1}{2}\frac{\sqrt{T}}{\rho^{3/2}})$
$f'(\rho) = -\frac{1}{4L}\cdot~\frac{\sqrt{T}}{\rho^{3/2}}$
(b)
(i) $f'(L) = -\frac{1}{2L^2}~\sqrt{\frac{T}{\rho}} \lt 0$
When the length is decreased, the pitch increases.
(ii) $f'(T) = \frac{1}{4L}\sqrt{\frac{1}{T\rho}} \gt 0$
When the tension is increased, the pitch increases.
(iii) $f'(\rho) = -\frac{1}{4L}\cdot~\frac{\sqrt{T}}{\rho^{3/2}} \lt 0$
When the linear density is increased, the pitch decreases.
