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Chapter 11: Problem 27

Find the lengths of the curves. $$ x=t^{2} / 2, \quad y=(2 t+1)^{3 / 2} / 3, \quad 0 \leq t \leq 4 $$

Short Answer

Expert verified
The length of the curve is 12 units.

Step by step solution

01

Understand the Problem

We need to find the length of the curve defined by the parametric equations \(x = \frac{t^2}{2}\) and \(y = \frac{(2t+1)^{3/2}}{3}\) for the parameter \(t\) ranging from 0 to 4.
02

Formula for Arc Length of Parametric Curve

The formula for the length of a parametric curve \(x = f(t), y = g(t)\) from \(t = a\) to \(t = b\) is: \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\] We will use this formula to compute the curve's length from \(t = 0\) to \(t = 4\).
03

Compute Derivatives

First, compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):- \(\frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^2}{2}\right) = t\)- \(\frac{dy}{dt} = \frac{d}{dt}\left(\frac{(2t+1)^{3/2}}{3}\right) = (2t+1)^{1/2}\) after applying the chain rule.
04

Set Up the Integral

Substitute \(\frac{dx}{dt} = t\) and \(\frac{dy}{dt} = (2t+1)^{1/2}\) into the formula:\[L = \int_{0}^{4} \sqrt{t^2 + (2t+1)} \, dt\]Simplify under the square root: \(t^2 + (2t+1) = t^2 + 2t + 1 = (t+1)^2\).
05

Evaluate the Integral

Since \(\sqrt{(t+1)^2} = |t+1|\) and \(t+1\) is positive on \([0, 4]\), we have:\[L = \int_{0}^{4} (t+1) \, dt\]Compute the integral:\[\int_{0}^{4} (t+1) \, dt = \left[ \frac{t^2}{2} + t \right]_{0}^{4} = \left( \frac{4^2}{2} + 4 \right) - \left( \frac{0^2}{2} + 0 \right) = 8 + 4 = 12\]
06

Conclusion

With all calculations complete, the length of the curve from \(t = 0\) to \(t = 4\) is 12.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations provide a powerful way to describe curves in a plane. Instead of expressing each coordinate separately as a function of another variable, parametric equations represent both coordinates鈥攈ere, x and y鈥攁s functions of a third variable, often denoted as t (time).
  • For the given problem, the curve is described by the equations:
  • \( x = \frac{t^2}{2} \)
  • \( y = \frac{(2t+1)^{3/2}}{3} \)
Through these equations, as the parameter t ranges over an interval (e.g., from 0 to 4), we trace out a path in the xy-plane, effectively describing the entire curve geometrically.
This approach is incredibly useful in calculus, as it allows for more complex curves to be studied and analyzed using calculus techniques.
Integral Calculus
Integral calculus is primarily concerned with the accumulation of quantities and the areas under curves. One of its main applications is calculating arc lengths, which is notably useful when dealing with parametric equations.
The formula for finding the arc length of a parametric curve between two points is given by:
  • \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \)
  • This involves the integration of the square root of the sum of the squares of the derivatives of x and y with respect to t, over the interval \([a, b]\).
The arc length integral effectively sums up tiny segments of the curve (which are nearly straight) to find the total length. By working through integration, notably over the interval from t=0 to t=4 in the given exercise, we accumulate the complete length of the curve.
Derivative Computation
Derivative computation is essential in understanding how a function behaves as its input changes. For parametric equations, differentiating each component with respect to the parameter t is crucial.
  • The derivative \( \frac{dx}{dt} \) for \( x = \frac{t^2}{2} \) computes to t.
  • The derivative \( \frac{dy}{dt} \), with careful application of the chain rule, becomes \((2t+1)^{1/2}\).
These derivatives tell us how fast and in what direction each coordinate is changing with respect to t.
In the arc length formula, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are essential as they form the basic building blocks of the length computation.
They are squared, indicative of Pythagoras' theorem's fundamental role in calculating distances in two dimensions.

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Most popular questions from this chapter

Volume The region bounded on the left by the \(y\) -axis, on the right by the hyperbola \(x^{2}-y^{2}=1,\) and above and below by the lines \(y=\pm 3\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid.

Use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(11.15 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad 0 \leq t \leq \pi $$

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ y^{2}=4 x, \quad \text { left } 2, \text { down } 3 $$

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1, \quad y=2$$

Volume Find the volume of the solid generated by revolving the region enclosed by the ellipse \(9 x^{2}+4 y^{2}=36\) about the (a) \(x\) -axis, (b) \(y\) -axis.
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