Explanation Given: T d T d x = τ ρ g Here τ is the friction at the base of the glacier, ρ is the ice density and g is the acceleration due to gravity τ ( x ) = 0.3 x ( 1000 − x ) N/m 2 T ( 0 ) = 0 Calculation: Let us separate the differential with respect to their variable. T d T d x = 0.3 x ( 1000 − x ) N/m 2 ρ g ( ρ g ) T d T = 0.3 x ( 1000 − x ) d x ( ρ g ) T d T = ( 300 x − 0.3 x 2 ) d x Here ρ and g are constant To determine (b) To graph: T for 0 ≤ x ≤ 1000 m

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

21st Edition

ISBN: 9781319438333

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.1, Problem 44E

To determine

(a)

To solve:

The glacier thickness differential equation for T(x) with τ(x)=0.3x(1000−x)N/m2 and initial condition T(0)=0.

To determine

(b)

To graph:

T for 0≤x≤1000m

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Chapter 9.1, Problem 43E

Chapter 9.1, Problem 45E

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(a) To solve: The glacier thickness differential equation for T ( x ) with τ ( x ) = 0.3 x ( 1000 − x ) N/m 2 and initial condition T ( 0 ) = 0 .