For a fixed positive constant

MATHEMATICS · Multi-part · AP · Calculus BC · Applications of Integration · Applications of Integration

Problem

For a fixed positive constant kk, consider the curve y=1kcosh⁡(kx)y=\dfrac{1}{k}\cosh(kx) on the interval 0≤x≤b0\le x\le b, where b>0b>0. (a) Show that 1+(dydx)2=cosh⁡(kx)\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\cosh(kx). Present a clear chain of equalities that leads to your conclusion. (b) Derive a closed‐form expression for the arc length L(k,b)L(k,b) of the curve on [0,b][0,b] in terms of kk and bb. Express your final answer without an integral symbol. (c) Evaluate L(2,ln⁡3)L(2,\ln 3) exactly, where k=2k=2 and b=ln⁡3b=\ln 3. Show all intermediate steps. (d) Determine lim⁡k→0+L(k,b)\displaystyle \lim_{k\to 0^{+}} L(k,b) for a fixed b>0b>0 and interpret the limiting value in the context of the family of curves y=1kcosh⁡(kx)y=\dfrac{1}{k}\cosh(kx).

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