Question

Are length Find the length of the curve \(y=x^{5 / 4}\) on the interval \([0,1] .\) (Hint: Write the are length integral and let \(u^{2}=1+\left(\frac{5}{4}\right)^{2} \sqrt{x}\)

Short answer

The length of the curve \(y = x^{5/4}\) on the interval \([0, 1]\) is \(L = \frac{25}{20}\) or \(1.25\).

Step-by-step solution

Find the derivative of the function

First, we need to find the derivative, \(f'(x)\), of the given function, \(f(x) = x^{5/4}\). Using the power rule, we can find the derivative: \(f'(x) = \frac{d}{dx}(x^{5/4}) = \frac{5}{4} x^{\frac{5}{4} - 1} = \frac{5}{4} x^{\frac{1}{4}}\) Now that we have the derivative, we can move on to the next step.

Write the arc length integral

Using the arc length formula, we can set up the integral to find the length of the curve on the interval \([0, 1]\). The integral looks like this: \(L = \int_0^1 \sqrt{1 + (\frac{5}{4} x^{\frac{1}{4}})^2} dx\)

Use the given hint to make a substitution

From the hint, we have that \(u^2 = 1 + (\frac{5}{4} x^{\frac{1}{4}})^2\). Let us rewrite this as \(u = \sqrt{1 + (\frac{5}{4} x^{\frac{1}{4}})^2}\). Now, differentiate both sides with respect to \(x\) to find the differential \(du\) in terms of \(dx\). We get: \(\frac{du}{dx} = \frac{5}{8x^\frac{3}{4}} \Rightarrow du = \frac{5}{8x^\frac{3}{4}} dx\) Then, to find the corresponding limits of integration, substitute the original limits, \(x = 0\) and \(x = 1\), into the expression for \(u\): - \(u(0) = \sqrt{1 + (\frac{5}{4} (0)^{\frac{1}{4}})^2} = 1\) - \(u(1) = \sqrt{1 + (\frac{5}{4} (1)^{\frac{1}{4}})^2} = \sqrt{1 + (\frac{5}{4})^2} = \sqrt{\frac{41}{16}}\) Now, substitute \(u\) and \(du\) into our integral and change the limits of integration: \(L = \int_1^{\sqrt{\frac{41}{16}}} \frac{8}{5} u du\)

Evaluate the integral

Now we need to evaluate the integral which is straightforward: \(L = \frac{8}{5}\left[ \frac{1}{2}u^2\right]_1^{\sqrt{\frac{41}{16}}} = \frac{8}{5} \cdot \frac{1}{2} (\frac{41}{16} - 1)\) \( = \frac{4}{5} (\frac{25}{16}) = \frac{25}{20}\) So, the length of the curve \(y = x^{5/4}\) on the interval \([0, 1]\) is \(L = \frac{25}{20}\) or \(1.25\).

Key concepts

Integral Calculus and Arc Length

Integral calculus is a part of mathematics that concerns itself with finding quantities like area, volume, and, as in our example, the length of a curve. To find the arc length of a curve represented by a function, we use a specific integral formula:
\[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \]
Here, \( L \) is the arc length, \( a \) and \( b \) are the bounds of the interval, and \( f'(x) \) is the derivative of the function whose arc length we want to measure. The arc length formula derives from the Pythagorean theorem, accounting for the infinitesimally small segments of the curve and summing them up using integration. For our specific function \( y = x^{5/4} \), the arc length integral is written for the interval \([0,1]\) and involves a square root that contains the square of the function's derivative.

The Power Rule for Derivatives

The power rule is a fundamental rule in differential calculus used to find the derivative of a function in the form of \( x^n \), where \( n \) is any real number. The rule states that:
\[ \frac{d}{dx}[x^n] = nx^{n-1} \]
In the context of calculating arc length, the derivative of the function helps us measure the rate at which \( y \) changes with respect to \( x \), providing us with the slope at every point along the curve. This slope is squared and added to 1 under the square root in the arc length formula. For the function given in our exercise, \( y = x^{5/4} \), applying the power rule gives us its derivative, \( f'(x) = \frac{5}{4} x^{1/4} \), which is then inserted into the arc length formula to continue the problem-solving process.

Substitution Method in Integration

When an integral has a complicated function within a root or another complex expression, the substitution method, sometimes called 'u-substitution', is often employed to simplify the integration process. This technique involves substituting a part of the integral with a new variable, \( u \), effectively transforming the integral into a simpler form that is easier to evaluate.
In our exercise, we simplify the arc length integral by letting \( u \) represent the entire square root, so \( u = \sqrt{1 + (\frac{5}{4} x^{1/4})^2} \). Differentiating both sides of this equation with respect to \( x \) allows us to find a relationship between \( dx \) and \( du \), which we can then use to change the integral's variable of differentiation from \( x \) to \( u \). After substitution, we also change the limits of integration, using the relationship to express the original \( x \) limits in terms of \( u \). The integral becomes easier to manage once we apply this method, as shown in the steps leading up to the evaluation of the integral and finding the arc length.