Problem 43 - Calculus Early Transcendentals

Question

Find the exact length of the curve. $$x=\frac{t}{1+t^{\prime}}, \quad y=\ln (1+t), \quad 0 \leqslant t \leqslant 2$$

Step-by-Step Solution

Answer

The length of the curve is \( \ln(3) \).

Step 1: Parametric Equations of the Curve

We are given the parametric equations for the curve: \( x = \frac{t}{1+t} \) and \( y = \ln(1+t) \). We need to find the length of this curve for values of \( t \) ranging from 0 to 2.

Step 2: Find Derivatives with Respect to \( t \)

To use the formula for arc length, we first find the derivatives of \( x \) and \( y \) with respect to \( t \):For \( x \):\[ \frac{dx}{dt} = \frac{(1+t)(1) - t(1)}{(1+t)^2} = \frac{1}{(1+t)^2} \]For \( y \):\[ \frac{dy}{dt} = \frac{d}{dt}[\ln(1+t)] = \frac{1}{1+t} \]

Step 3: Set Up the Arc Length Integral

The arc length \( L \) of a curve given by parametric equations is calculated using the formula:\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \]Substitute the derivatives we found:\[ L = \int_{0}^{2} \sqrt{\left( \frac{1}{(1+t)^2} \right)^2 + \left( \frac{1}{1+t} \right)^2 } \, dt \]

Step 4: Simplify the Expression Inside the Integral

Inside the square root, simplify:\[ \left( \frac{1}{(1+t)^2} \right)^2 = \frac{1}{(1+t)^4} \]\[ \left( \frac{1}{1+t} \right)^2 = \frac{1}{(1+t)^2} \]Thus:\[ \sqrt{\frac{1}{(1+t)^4} + \frac{1}{(1+t)^2}} = \sqrt{\frac{1 + (1+t)^2}{(1+t)^4}} = \sqrt{\frac{1 + 1 + 2t + t^2}{(1+t)^4}} \]

Step 5: Further Simplification

Upon further simplification, we see:\[ 1 + (1+t)^2 = 2 + 2t + t^2 = (1+t)^2 \]Therefore the square root expression becomes:\[ \frac{1}{(1+t)^2} \]Now, our integral is:\[ L = \int_{0}^{2} \frac{1}{1+t} \, dt \]

Step 6: Integrate and Evaluate

The integral \( \int \frac{1}{1+t} \, dt \) is straightforward, giving us \( \ln|1+t| \):\[ L = \left[ \ln(1+t) \right]_{0}^{2} \]Evaluate this from 0 to 2:\[ L = \ln(1+2) - \ln(1+0) = \ln(3) - \ln(1) = \ln(3) \]

Step 7: Conclusion

Thus, the exact length of the curve is \( \ln(3) \).

Key Concepts

Parametric Equations

In calculus, curves are often represented using parametric equations, which define coordinates via a third variable known as the parameter. This is useful for curves where a single function cannot describe both x and y coordinates. Parametric equations involve:

• A parameter, typically 't', which differs from x or y.
• Two separate equations: one for x in terms of t (x = f(t)) and one for y (y = g(t)).

For instance, in the given exercise, the parametric equations are:

• \( x = \frac{t}{1+t} \)
• \( y = \ln(1+t) \)

Parametric equations allow the representation of more complex curves, adding flexibility over standard functions. This versatility comes in handy for calculating properties such as the arc length of curves which might otherwise be difficult to integrate or describe through simple y=f(x) forms.

Calculus

Calculus is a critical branch of mathematics that deals with continuous change. It is particularly adept at solving problems that involve rates of change and the accumulation of quantities. The two main areas of calculus are:

• Differential Calculus: Focuses on finding the rate of change (derivative) of a function with respect to one of its variables.
• Integral Calculus: Concerned with the accumulation of quantities, such as areas under a curve, using integrals.

To solve the given problem, we utilize both derivatives and integrals. First, we find the derivatives of the parametric equations with respect to t. These derivatives help us develop an expression for the arc length integral. Derivatives tell us about the instantaneous rate of change, while the overall curvature represented by the length is found using integration over a defined interval. This integration helps us capture the essence of the curve’s arc length along the parameterized path.

Integral Evaluation

Evaluating an integral often involves finding the exact area under a curve over a specified interval. Integrals can be indefinite (without specified limits) or definite (with limits). In our case, we deal with a definite integral to find the arc length of the parametric curve.
The formula for computing the arc length of a curve given by parametric equations is:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt\]This formula essentially accumulates infinitesimally small distances along the curve.
Once the integral is set up, simplifying it can involve laborious algebra. In the original solution, we discovered the simplification of the expression inside the integral to reduce complexity.
Ultimately, evaluating the integral involves basic integration techniques. For this problem, integrating \(\int \frac{1}{1+t} \, dt\) results in a natural logarithm—specifically, \(\ln(1+t)\), which is then evaluated over the given interval: \[L = \left[ \ln(1+t) \right]_{0}^{2}\]Upon evaluation, the final result, \( \ln(3) \), gives the exact arc length of the curve, demonstrating the analytic power of integral calculus.