Problem 69 - Calculus of a Single Variable

Question

Arc Length and Area Let \(C\) be the curve given by \(f(x)=\cosh x\) for \(0 \leq x \leq t,\) where \(t>0 .\) Show that the arc length of \(C\) is equal to the area bounded by \(C\) and the \(x\) -axis. Identify another curve on the interval \(0 \leq x \leq t\) with this property.

Step-by-Step Solution

Answer

The arc length \(s\) of the curve \(C\) defined by \(f(x) = \cosh x\) is equal to the area \(A\) bounded by this curve and the \(x\)-axis and both are equal to \(\sinh t\). Another curve with this property is \(g(x) = e^x\) over the interval \(0 \leq x \leq t\). All three on the interval result as \(e^t - 1\).

Step 1: Compute the Derivative of \(f(x)\)

We need to find the derivative \(f'(x)\) of the function, which in this case is \(f'(x) = \sinh x\). This will be used in the next step to compute the arc length.

Step 2: Calculate the Arc Length

The formula for the arc length of a curve defined by \(y=f(x)\) on the interval \([a,b]\) is given by: \[ s = \int_{a}^{b} \sqrt{1+(f'(x))^2} dx \] where \(f'(x)\) is the derivative of \(f(x)\). In this case \(f'(x) = \sinh x\), so we can substitute it to get the arc length formula: \[ s = \int_{0}^{t} \sqrt{1+(\sinh x)^2} dx \] which simplifies to: \[ s = \int_{0}^{t} \cosh x dx \] and integrating this expression from 0 to t, we get \(s = \sinh t - \sinh 0 = \sinh t\).

Step 3: Compute Area Under the Curve

The area \(A\) under the curve \(y = f(x)\) between \(0 \leq x \leq t\) is given by the integral: \[ A = \int_{0}^{t} f(x) dx \] Substituting \(f(x) = \cosh x\) we get: \[ A = \int_{0}^{t} \cosh x dx \] which upon integrating gives \(A = \sinh t - \sinh 0 = \sinh t\). You can see that \(A = s\), thus proving that the arc length of \(C\) equals the area bounded by \(C\) and the \(x\)-axis.

Step 4: Find Another Curve that Satisfies the Given Property

Another curve that will have the same property might be \(g(x) = e^x\) on the interval \(0 \leq x \leq t\). Here the derivative \(g'(x) = e^x = g(x)\), which makes the calculation of the arc length and the area under the curve very similar, and the results come out to be identical.