Problem 10 - 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. \(x=\left(y^{3 / 2} / 3\right)-y^{1 / 2}\) from \(y=1\) to \(y=9\) (Hint: \(1+(d x / d y)^{2}\) is a perfect square.)

Step-by-Step Solution

Answer

The length of the curve is \( \frac{32}{3} \).

Step 1: Differentiate the function

First, we need to find the derivative of the function with respect to y. The given function is \( x = \frac{y^{3/2}}{3} - y^{1/2} \). Differentiate this function with respect to y:\[\frac{d x}{d y} = \frac{d}{d y} \left( \frac{y^{3/2}}{3} \right) - \frac{d}{d y} \left( y^{1/2} \right)\]Using the power rule, the derivatives are:\[\frac{d x}{d y} = \frac{1}{3} \times \frac{3}{2} y^{1/2} - \frac{1}{2} y^{-1/2} \]This simplifies to:\[\frac{d x}{d y} = \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}\]

Step 2: Set up the arc length integral

Next, use the formula for arc length in terms of y:\[L = \int_{y_0}^{y_1} \sqrt{1 + \left( \frac{d x}{d y} \right)^2} \, dy\]Substitute \( \frac{d x}{d y} \) into the equation:\[\frac{d x}{d y} = \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}\]\[L = \int_{1}^{9} \sqrt{1 + \left( \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2} \right)^2} \, dy\]

Step 3: Simplify the expression under the square root

Let's simplify the expression inside the square root:\[\left( \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2} \right)^2 = \frac{1}{4}(y - 2 + y^{-1})\]Adding 1 to this expression:\[1 + \frac{1}{4}(y - 2 + y^{-1}) = \frac{1}{4}(4 + y - 2 + y^{-1}) = \frac{1}{4}(y + 2 + y^{-1})\]Since it's given that this is a perfect square, we suspect:\[\left( \frac{1}{2}(y^{1/2} + y^{-1/2}) \right)^2 = \frac{1}{4}(y + 2 + y^{-1})\]

Step 4: Evaluate the integral

Substitute the simplified perfect square back into the integral:\[L = \int_{1}^{9} \left( \frac{1}{2}y^{1/2} + \frac{1}{2}y^{-1/2} \right) \, dy\]Simplify and evaluate the integral:\[L = \frac{1}{2} \int_{1}^{9} y^{1/2} \, dy + \frac{1}{2} \int_{1}^{9} y^{-1/2} \, dy\]Calculate both integrals:\[= \frac{1}{2} \left[ \frac{2}{3}y^{3/2} \right]_{1}^{9} + \frac{1}{2} \left[ 2y^{1/2} \right]_{1}^{9}\]\[= \frac{1}{2} \left( \frac{2}{3}(27 - 1) + 2(3 - 1) \right)\]Evaluate the expressions inside:\[L = \frac{1}{2} \left( \frac{52}{3} + 4 \right)\] \[L = \frac{1}{2} \left( \frac{64}{3} \right) = \frac{32}{3}\]

Step 5: Conclude the arc length

Therefore, the length of the curve from \( y = 1 \) to \( y = 9 \) is \( \frac{32}{3} \).

Key Concepts

Differentiation

Differentiation is a crucial concept in calculus, allowing us to find the rate at which a function changes. In this exercise, the goal is to differentiate the function \( x = \frac{y^{3/2}}{3} - y^{1/2} \) in terms of \( y \). By differentiating, we obtain \( \frac{dx}{dy} \), which represents the slope of the tangent to the curve at any point on it.

The differentiation process involves applying the power rule, which is a fundamental technique in calculus. The power rule states that for any function \( y^n \), its derivative is \( ny^{n-1} \).

For our function, differentiate each term separately:

• The derivative of \( \frac{y^{3/2}}{3} \) is \( \frac{1}{3} \times \frac{3}{2}y^{1/2} = \frac{1}{2}y^{1/2} \).
• The derivative of \( y^{1/2} \) is \( \frac{1}{2}y^{-1/2} \).

By combining these, the differentiated equation results in \( \frac{dx}{dy} = \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2} \). Understanding this helps in setting up the arc length integral for the curve.

Integral Calculus

Integral calculus is another fundamental branch of calculus. It's used to determine the integral or the accumulation of quantities, such as areas under a curve or, in this case, the length of a curve. After differentiation, we used integral calculus to find the arc length of the curve specified by the given function.

To calculate the arc length of a curve described by \( x = f(y) \) from \( y = a \) to \( y = b \), the formula used is:

• \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{d x}{d y} \right)^2} \, dy \]

In our exercise, that integral became \( L = \int_{1}^{9} \sqrt{1 + \left( \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2} \right)^2} \, dy \). Integrals like this one help us not only find the arc length but also provide a deeper understanding by analyzing how the curve behaves between two points.

Power Rule

The power rule is a widely used formula in calculus to find the derivative of powers of variables. It simplifies the process of differentiation when dealing with algebraic expressions.

According to the power rule, if \( y^n \) is a term in your function, its derivative is \( ny^{n-1} \). Understanding and applying this rule is key to correctly differentiating functions and subsequently solving calculus problems, as seen in this exercise.

Here is how we applied the power rule:

• For \( y^{3/2} \), you derive \( \frac{3}{2}y^{1/2} \).
• For \( y^{1/2} \), you derive \( \frac{1}{2}y^{-1/2} \).

The power rule assists in managing complex expressions and ensures the differentiation of polynomial terms is handled correctly, making it easier to set up and solve integrals that depend on these derivatives.

Perfect Square

A perfect square is a number or expression that is the square of another number or expression. Recognizing perfect squares can greatly simplify mathematical calculations, particularly when dealing with integrals or equations.

In the context of this exercise, the expression \( 1 + \left( \frac{dx}{dy} \right)^2 \) is a perfect square. This simplifies the integrand significantly, allowing us to write it as \( \left( \frac{1}{2}(y^{1/2} + y^{-1/2}) \right)^2 \).

Identifying the perfect square helps in reducing the complexity of the square root in the integral, leading to an easier calculation. It avoids unnecessary complications and transforms the integral into a simpler, more manageable form:

• \( L = \int_{1}^{9} \left( \frac{1}{2}y^{1/2} + \frac{1}{2}y^{-1/2} \right) \, dy \)

This technique not only enhances computational efficiency but also aids in better understanding and solving integral-based problems.