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Question

In the earlier example, we calculated the thickness of a glacier assuming the friction at the base of the glacier is constant. In this exercise, we consider cases where the friction varies along the length of the glacier.

(a) Solve the glacier thickness differential equation, and use the equation

TdTdx=τρgT \frac{d T}{d x}=\frac{\tau}{\rho g}

for T(x) with τ(x)=0.3x(1000x)\tau(x)=0.3 x(1000-x) N/m2^2 and initial condition T(0)=0.

(b) Sketch the graph of T for 0x10000 \leq x \leq 1000 m.

Solution

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We know, the model for a glacier thickness T(x)T(x), xx is the distance in meters from the front of the glacier, is given by the differential equation

TdTdx=τρgT\, \frac {dT } {dx }=\frac { \tau} { \rho g}

where τ\tau is the friction at the base of the glacier, ρ\rho is the ice density and gg is acceleration due to gravity.

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