Problem 8 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(7-16\) . If you have a grapher, you may want to graph these curves to see what they look like. \(y=x^{3 / 2}\) from \(x=0\) to \(x=4\)

Step-by-Step Solution

Answer

The curve's length is approximately 9.1133 units.

Step 1: Understanding the Problem

We need to find the length of the curve described by the function \(y = x^{3/2}\) between \(x = 0\) and \(x = 4\). The curve's length can be found using the formula for the arc length of a curve.

Step 2: Recall the Arc Length Formula

The arc length \(L\) of a function \(y = f(x)\) over the interval \([a, b]\) is given by:\[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]Here, \(a = 0\) and \(b = 4\).

Step 3: Differentiate the Function

Differentiate \(y = x^{3/2}\) with respect to \(x\):\[\frac{dy}{dx} = \frac{3}{2}x^{1/2}\]This derivative represents the slope of the tangent to the curve at any point \(x\).

Step 4: Substitute the Derivative into the Arc Length Formula

Substitute \(\frac{dy}{dx} = \frac{3}{2}x^{1/2}\) into the arc length formula:\[L = \int_{0}^{4} \sqrt{1 + \left(\frac{3}{2}x^{1/2}\right)^2} \, dx = \int_{0}^{4} \sqrt{1 + \frac{9}{4}x} \, dx\]

Step 5: Simplify the Integral

Simplify the expression inside the square root:\[\sqrt{1 + \frac{9}{4}x} = \sqrt{\frac{4 + 9x}{4}} = \frac{\sqrt{4 + 9x}}{2}\]Thus, the arc length integral becomes:\[L = \frac{1}{2} \int_{0}^{4} \sqrt{4 + 9x} \, dx\]

Step 6: Evaluate the Integral

To solve \(\int \sqrt{4 + 9x} \, dx\), let \(u = 4 + 9x\), which implies \(du = 9\, dx\) or \(dx = \frac{du}{9}\). Change the limits accordingly. At \(x = 0\), \(u = 4\); at \(x = 4\), \(u = 40\).Now, integrate:\[L = \frac{1}{18} \int_{4}^{40} u^{1/2} \, du = \frac{1}{18} \left(\frac{2}{3} u^{3/2} \right) \bigg|_{4}^{40}\]

Step 7: Compute the Definite Integral

Compute the values of the definite integral:\[L = \frac{1}{27} \left((40)^{3/2} - (4)^{3/2}\right) = \frac{1}{27} (254.04 - 8)\]Calculating gives:\[L = \frac{1}{27} \times 246.04 \]\[L \approx 9.1133\]

Step 8: Final Step: Conclusion

The length of the curve \(y = x^{3/2}\) from \(x = 0\) to \(x = 4\) is approximately \(9.1133\).

Key Concepts

Differentiation

Differentiation is a key concept in calculus that helps us understand how functions change. It involves finding the derivative of a function, which represents the rate at which the function's value changes with respect to change in its input. In our problem, given the function \(y = x^{3/2}\), we need to find its derivative. This involves applying the power rule of differentiation, which says that if you have a power function of the form \(x^n\), its derivative is \(nx^{n-1}\).

This means for \(y = x^{3/2}\), the derivative \(\frac{dy}{dx} = \frac{3}{2}x^{1/2}\). This tells us how steep the curve is at any point \(x\). The derivative is crucial, as it is used in the arc length formula to determine precisely how the curve's path extends over an interval.

Definite Integral

The definite integral is a powerful tool in calculus used to calculate things like area, volume, and in this case, the length of a curve. In our problem, the definite integral is used to sum up small distances along the curve \(y = x^{3/2}\) from \(x = 0\) to \(x = 4\).

The integral captures the accumulated value over this interval, represented by \(L = \int_{0}^{4} \sqrt{1 + \left(\frac{3}{2}x^{1/2}\right)^2} \, dx\).

Here is what happens step-by-step:

• First, we substitute our previously found derivative \(\frac{dy}{dx}\) into the arc length formula.
• Next, we simplify the expression under the square root: \(\sqrt{1 + \frac{9}{4}x}\).
• Then we evaluate the definite integral, which gives us the total length of the curve.

Understanding the definite integral allows us to solve for real-world geometrical measures involving continuous curves.

Function of a Curve

A function of a curve describes the path formed by plotted points where the output value changes as the input value changes along an x-y plane. The function \(y = x^{3/2}\) creates a curve with a distinct shape.

When we plot this function, each x-value corresponds to a point on the curve, shaping the path from \(x = 0\) to \(x = 4\). The curve can be visually complex because it relies on the square root and the power of three multiplied by half. As the x-values increase, so does the value of \(y\), which means the curve begins at the origin when \(x = 0\) and rises more steeply with higher values of \(x\).

The function defines exactly what the path looks like, and by using calculus, specifically through differentiation and integration, we derive not just how the function behaves, but also measurable properties such as length.