Chapter 10: Problem 28
Find the lengths of the curves $$x=(2 t+3)^{3 / 2} / 3, \quad y=t+t^{2} / 2, \quad 0 \leq t \leq 3$$
Short Answer
Expert verified
The length of the curve is 10.5 units.
Step by step solution
01
Parameterize the curve
The curve is given by the parametric equations \( x(t) = \frac{(2t+3)^{3/2}}{3} \) and \( y(t) = t + \frac{t^2}{2} \) for the parameter \( t \) ranging from 0 to 3.
02
Derivatives with respect to t
Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Use the power and chain rules for differentiation:\[ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{(2t+3)^{3/2}}{3} \right) = \frac{1}{3} \cdot \frac{3}{2}(2t+3)^{1/2} \cdot 2 = (2t+3)^{1/2} \]\[ \frac{dy}{dt} = \frac{d}{dt} \left( t + \frac{t^2}{2} \right) = 1 + t \]
03
Integrate for curve length
The length of the curve is given by the integral: \[ L = \int_{0}^{3} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]Substitute the derivatives:\[ L = \int_{0}^{3} \sqrt{(2t+3) + (1+t)^2} \, dt \]This simplifies to:\[ L = \int_{0}^{3} \sqrt{2t+3 + 1 + 2t + t^2} \, dt = \int_{0}^{3} \sqrt{t^2 + 4t + 4} \, dt \]Realize \( t^2 + 4t + 4 = (t+2)^2 \), so it simplifies further to:\[ L = \int_{0}^{3} |t+2| \, dt \]
04
Evaluate the integral
Since \( 0 \leq t \leq 3 \), \( t+2 \geq 0 \) throughout, so \( |t+2| = t+2 \). Integrate:\[ L = \int_{0}^{3} (t+2) \, dt \]\[ L = \left[ \frac{t^2}{2} + 2t \right]_{0}^{3} \]\[ L = \left( \frac{3^2}{2} + 2 \cdot 3 \right) - \left( \frac{0^2}{2} + 2 \cdot 0 \right) \]\[ L = \frac{9}{2} + 6 = \frac{21}{2} = 10.5 \]
05
Conclusion: Length of the curve
The entire process shows that the length of the curve from the parameter for \( t = 0 \) to \( t = 3 \) is \( 10.5 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Curve Length
The concept of curve length is essential when working with parametric equations. A curve on a plane is often represented by a set of parametric equations that define the path of the curve through the coordinate plane. The length of such a curve can be determined using calculus methods. To find the length of a curve given by parametric functions, we utilize the curve length formula:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\] This formula integrates the square root of the sum of the squares of the derivatives of the parametric functions with respect to the parameter, usually denoted by \( t \).
- \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the derivatives of \( x \) and \( y \) with respect to \( t \).
- \( a \) and \( b \) are the bounds for the parameter.
Parametric Equations
Parametric equations are a powerful way to represent curves. Unlike Cartesian equations which link \( x \) and \( y \) directly, parametric equations express both \( x \) and \( y \) in terms of a separate parameter, often \( t \).For the curve in the given exercise, the parametric equations are:
- \( x(t) = \frac{(2t+3)^{3/2}}{3} \)
- \( y(t) = t + \frac{t^2}{2} \)
- The parameter \( t \) typically ranges over an interval, dictating the start and end of the path we're interested in — in this exercise, from 0 to 3.
- One major advantage of parametric equations is their ability to represent curves that cannot be expressed easily as functions \( y = f(x) \).
Integration
Integration is a fundamental calculus technique used to calculate areas, volumes, and curve lengths among other things. In the context of finding the length of a parametric curve, integration helps to accumulate small segments along the curve to find the total distance.When applying the curve length formula, the integration process involves:
- Substituting the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the length formula.
- Simplifying the integrand, which in this exercise becomes \( \sqrt{(t+2)^2} \).
- Understanding that \( |t+2| \) simplifies to \( t+2 \) because \( t \) is always non-negative in the given range.
Differentiation
Differentiation is a key calculus operation used to find the rate of change of a function. In the context of parametric equations, differentiation allows us to determine how each coordinate, \( x \) and \( y \), varies with respect to the parameter \( t \).To differentiate parametric equations, as in this exercise, one generally applies the rules for derivatives including:
- The power rule, which helps when differentiating terms like \( (2t+3)^{3/2} \).
- The chain rule, which is used in cases where a function is composed of another function, like \( (2t+3)^{3/2} \).
- \( \frac{dx}{dt} = (2t+3)^{1/2} \)
- \( \frac{dy}{dt} = 1 + t \)