Question

The partial differentiation of excess hydrostatic pressure \overline{u} as a function of r and θ with respect to x is given by _______

(a) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ\)

(b) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ\)

(c) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} sinθ\)

(d) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ\)

I got this question in examination.

The doubt is from Consolidation Equation in Polar Coordinates topic in division One Dimensional Consolidation of Geotechnical Engineering I

Answer 1

The correct choice is (a) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ\)

Explanation: Partially differentiating the excess hydrostatic pressure \overline{u} with respect to x,

\(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} \frac{∂r}{∂x}+\frac{∂\overline{u}}{∂θ}\frac{∂θ}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.\)

∴ \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.\)