Problem 56 - Calculus for Biology and Medicine

Question

Find the length of the curve $$ 2 y^{2}=3 x^{3} $$ from \(x=0\) to \(x=1\).

Step-by-Step Solution

Answer

The length of the curve from \(x=0\) to \(x=1\) is approximately 1.656.

Step 1: Understand the Curve Length Formula

The length of a curve defined as a function, say \( f(x) \), from \( x = a \) to \( x = b \) is calculated using the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx. \]For the given curve \( 2y^2 = 3x^3 \), we need to first express \( y \) in terms of \( x \).

Step 2: Express y in terms of x

Given \( 2y^2 = 3x^3 \), solve for \( y \): \[ y^2 = \frac{3}{2}x^3 \]\[ y = \pm \sqrt{\frac{3}{2}} x^{3/2}. \] Choose \( y = \sqrt{\frac{3}{2}} x^{3/2} \) for deriving \( \frac{dy}{dx} \).

Step 3: Differentiate y with respect to x

Differentiate \( y = \sqrt{\frac{3}{2}} x^{3/2} \) with respect to \( x \):\[ \frac{dy}{dx} = \frac{3}{2} \sqrt{\frac{3}{2}} x^{1/2}. \]

Step 4: Substitute into the Curve Length Formula

Substitute \( \frac{dy}{dx} = \frac{3}{2} \sqrt{\frac{3}{2}} x^{1/2} \) into the curve length formula:\[ L = \int_{0}^{1} \sqrt{1 + \left( \frac{3}{2} \sqrt{\frac{3}{2}} x^{1/2} \right)^2} \, dx. \]Simplify inside the square root:\[ L = \int_{0}^{1} \sqrt{1 + \frac{9}{4} \cdot \frac{3}{2} x} \, dx. \]\[ L = \int_{0}^{1} \sqrt{1 + \frac{27}{8} x} \, dx. \]

Step 5: Evaluate the Integral

Use substitution to evaluate the integral. Let \( u = 1 + \frac{27}{8}x \), which gives \( du = \frac{27}{8} dx \) and thus \( dx = \frac{8}{27} du \).Change the limits of integration: when \( x = 0, u = 1 \); when \( x = 1, u = \frac{35}{8} \).Then the integral transforms to:\[ L = \int_{1}^{\frac{35}{8}} \sqrt{u} \cdot \frac{8}{27} \, du. \]Evaluate the integral:\[ L = \frac{8}{27} \int_{1}^{\frac{35}{8}} u^{1/2} \, du. \]\[ L = \frac{8}{27} \left[ \frac{u^{3/2}}{3/2} \right]_{1}^{\frac{35}{8}}. \]\[ L = \frac{8}{27} \cdot \frac{2}{3} \left[ \left( \frac{35}{8} \right)^{3/2} - 1 \right]. \]

Step 6: Final Evaluation

Calculate \( \left( \frac{35}{8} \right)^{3/2} \), simplify, and find the final result:\[ L = \frac{16}{81} \left[(\frac{35}{8})^{3/2} - 1\right]. \]This evaluates to approximately \( 1.656 \) (rounded to three decimal places).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. In this exercise, differentiation plays a crucial role in computing the length of a curve. When given a curve defined by an equation like \(2y^2 = 3x^3\), the goal is to express one variable in terms of the other, allowing us to find the derivative.We started by expressing \(y\) in terms of \(x\): \(y = \sqrt{\frac{3}{2}} x^{3/2}\). With this expression, you can differentiate \(y\) with respect to \(x\), resulting in \(\frac{dy}{dx} = \frac{3}{2} \sqrt{\frac{3}{2}} x^{1/2}\). This derivative represents the slope or the rate of change of \(y\) with respect to \(x\) at any point on the curve.

• **Why do we differentiate?** Differentiation is used here to find the slope of the tangent to the curve, a crucial component in the curve length formula.
• **Application:** Differentiation helps us understand how steep or flat the curve is at different points, which is essential for accurate length calculation.

Integration

Integration is a core tool in calculus for finding areas, volumes, and in this case, the length of a curve. To compute the length of the given curve from \(x = 0\) to \(x = 1\), we use a specific integration formula:\[L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]This formula calculates the length by integrating the square root of one plus the square of the derivative, \(\frac{dy}{dx}\), over the interval from \(a\) to \(b\).Integration is performed step by step:

• Substitute the expression for \(\frac{dy}{dx}\) into the formula.
• Simplify the expression inside the square root.
• Evaluate the resulting integral using suitable techniques, like substitution, to find the actual length.

The integral \[L = \int_{0}^{1} \sqrt{1 + \frac{27}{8}x} \, dx\] simplifies using substitution, allowing us to evaluate it over the adjusted limits. This calculated integral gives the length of the curve, which is approximately 1.656 units.

Change of Variables

The technique of changing variables, also known as substitution, simplifies complicated integrals and makes them possible to solve. In this exercise, substitution is used to transform a complex integral into a simpler, more manageable form.During the curve length calculation, after setting up the integral \(L = \int_{0}^{1} \sqrt{1 + \frac{27}{8} x} \, dx\), we used substitution:

• **Substitute:** Let \(u = 1 + \frac{27}{8}x\). This substitution alters the variable of integration from \(x\) to \(u\), making the integral easier to solve.
• **Differentiate:** Compute \(du = \frac{27}{8} dx\), rearranging gives \(dx = \frac{8}{27} du\).
• **Adjust Limits:** Change the limits of integration according to \(u\). When \(x = 0\), \(u = 1\); when \(x = 1\), \(u = \frac{35}{8}\).

Substituting variables simplifies the integration process, changing the expression into \(\int_{1}^{\frac{35}{8}} \sqrt{u} \cdot \frac{8}{27} \, du\). This new integral is straightforward to solve, providing the necessary expression to determine the curve's length. Changing variables provides clarity and ease in handling otherwise complex integral expressions.