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Magazine Chapter 6: Problem 2
Find the lengths of the curves. $$y=x^{3 / 2} \quad \text { from } \quad x=0 \text { to } x=4$$
Short Answer
Expert verified
The length of the curve is approximately 4.531 units.
Step by step solution
01
Understand the Formula for Arc Length
To find the length of a curve given by the function \( y = f(x) \) from \( x = a \) to \( x = b \), use the formula for arc length: \( L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).
02
Differentiate the Function
Given \( y = x^{3/2} \), we need to find \( \frac{dy}{dx} \). Using the power rule for differentiation, \( \frac{dy}{dx} = \frac{3}{2}x^{1/2} \).
03
Square the Derivative
Next, square the derivative to substitute into the arc length formula: \( \left( \frac{dy}{dx} \right)^2 = \left( \frac{3}{2}x^{1/2} \right)^2 = \frac{9}{4}x \).
04
Set Up the Integral
Substitute the squared derivative into the arc length formula to get the integral: \( L = \int_{0}^{4} \sqrt{1 + \frac{9}{4}x} \, dx \).
05
Simplify the Integrand
The integrand is \( \sqrt{1 + \frac{9}{4}x} \). Simplifying this for integration, we integrate \( \int \sqrt{1 + \frac{9}{4}x} \, dx \) from 0 to 4.
06
Solve the Integral
This can be solved using a substitution method. Let \( u = 1 + \frac{9}{4}x \) which implies \( du = \frac{9}{4} \, dx \). Essentially, dx becomes \( \frac{4}{9} du \), and the limits of integration change from \( x = 0 \), \( u = 1 \) to \( x = 4 \), \( u = 10 \). With these transformations, the integral becomes \( \int \sqrt{u} \, \frac{4}{9} \, du \).
07
Evaluate the Integral
The integral \( \int \sqrt{u} \, du \) is \( \frac{2}{3}u^{3/2} \). Substituting back, \( L = \frac{4}{9} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{1}^{10} \).
08
Calculate the Result
Substitute and evaluate: \( L = \frac{4}{27} \left[ 10^{3/2} - 1^{3/2} \right] \). Compute numerically: \( 10^{3/2} = \sqrt{1000} = 31.622 \), so \( L \approx \frac{4}{27} \times 30.622 \).
09
Final Calculation
Calculate the final length: \( L \approx \frac{4}{27} \times 30.622 \approx 4.531 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function is changing at any point. When you're given a function, differentiation allows you to find the derivative. The derivative is essentially the slope of the tangent line to the function at any given point. - For a function like \( y = x^{3/2} \), using differentiation will help us understand how steep the curve is at different points. - In the exercise, we differentiate the function to help set up the calculation of the curve's arc length.To differentiate \( y = x^{3/2} \), we apply the power rule, which is one of the simplest and most widely used rules in differentiation.
Integration
Integration is the process of finding the accumulated value of a function over an interval. It can be seen as the reverse process of differentiation. With integration, we can find areas under curves and solve problems involving accumulation. - In this exercise, the goal is to find the arc length of a curve described by the function \( y = x^{3/2} \) between specific points.- We integrate the function to add up infinitely small lengths along the curve, from one point to another.The integral we set up, \( L = \int_{0}^{4} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \), uses the arc length formula which involves integration over the range \( 0 \) to \( 4 \).
Substitution method
The substitution method in integration is a technique used to simplify the integral by changing variables. This makes it easier to evaluate. It's like changing the view to see the problem more clearly. - In arc length calculation, we substitute for a more straightforward integration. - In this example, we substitute \( u = 1 + \frac{9}{4}x \) to simplify the integral \( \int \sqrt{1 + \frac{9}{4}x} \, dx \).By making this substitution, the problem becomes evaluating \( \int \sqrt{u} \, \frac{4}{9} \, du \). This process involves changing both the function and the limits of integration to reflect the new variable.
Power rule in calculus
The power rule is a simple differentiation rule used to calculate the derivative of a function in the form of \( x^n \). It's one of the basic building blocks of calculus, making it faster to differentiate certain kinds of functions than doing them from scratch.- For a given function \( y = x^{3/2} \), the power rule allows finding its derivative with ease. - According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \).Applying this rule to \( y = x^{3/2} \), we find \( \frac{dy}{dx} = \frac{3}{2}x^{1/2} \). This simplicity in finding derivatives helps in setting up the necessary calculations to find the arc length.
