Find the exact length of the curve defined by: $36y^2 = (x^2 - 4)^3, \quad 5 \leq x \leq 9, \quad y \geq 0$

Neetesh Kumar | November 08, 2024

Calculus Homework Help

Arc Length Question
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Step-by-step solution:

We will use the arc length formula to find the exact length of the curve.
The arc length $L$ of a curve defined by a function $y = f(x)$ from $x = a$ to $x = b$ is given by:

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step 1: Express $y$ in Terms of $x$

First, we need to isolate $y$ from the equation given.

Starting with the equation: $36y^2 = (x^2 - 4)^3$

Divide both sides by 36: $y^2 = \frac{(x^2 - 4)^3}{36}$

Taking the square root, we get: $y = \frac{1}{6} (x^2 - 4)^{3/2}$

Step 2: Compute the Derivative $\frac{dy}{dx}$

Next, we need to compute the derivative of $y$ with respect to $x$:

Using the chain rule: $\frac{dy}{dx} = \frac{1}{6} \cdot \frac{3}{2} (x^2 - 4)^{1/2} \cdot \frac{d}{dx}(x^2 - 4)$ $= \frac{1}{6} \cdot \frac{3}{2} (x^2 - 4)^{1/2} \cdot 2x$ $= \frac{x}{2} (x^2 - 4)^{1/2}$

Step 3: Compute $\left(\frac{dy}{dx}\right)^2$

Now we compute $\left(\frac{dy}{dx}\right)^2$: $\left(\frac{dy}{dx}\right)^2 = \left(\frac{x}{2} (x^2 - 4)^{1/2}\right)^2$ $= \frac{x^2}{4} (x^2 - 4)$

Step 4: Set Up the Arc Length Integral

We can now substitute $\left(\frac{dy}{dx}\right)^2$ into the arc length formula:

$L = \int_5^9 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$ $= \int_5^9 \sqrt{1 + \frac{x^2}{4} (x^2 - 4)} \, dx$ $= \int_5^9 \sqrt{1 + \frac{x^4 - 4x^2}{4}} \, dx$ $= \int_5^9 \sqrt{\frac{4 + x^4 - 4x^2}{4}} \, dx$

$L = \int_5^9 \frac{\sqrt{(x^2 - 2)^2}}{2} \, dx = \int_5^9 \frac{(x^2 - 2)}{2} \, dx = \int_5^9 (\frac{x^2}{2} - 1) \, dx$

Step 5: Simplify the Integral

$\bigg[ \frac{x^3}{6} - x \bigg]_5^9 = (\frac{9^3}{6} - 9) - (\frac{5^3}{6} - 5) = \frac{290}{3}$

Final Answer:

The exact length of the curve is = $\boxed{\frac{290}{3}}$ or $\boxed{96.6667}$

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