We are tasked with finding the value of $r$ for the tangent line to the equation

$x^2 + y^2 = (3x^2 + 8y^2 - 5x)^2$

at the point $\left(0, \frac{1}{8}\right)$, where the tangent line is given by:

$y = \frac{1}{8} + r(x - 0).$

Steps:

1. Differentiate both sides of the given equation implicitly with respect to $x$. This will help us find the slope $r$ of the tangent line.

   Given equation: $x^2 + y^2 = \left(3x^2 + 8y^2 - 5x\right)^2$

2. Use implicit differentiation on both sides.

   • The derivative of the left-hand side $x^2 + y^2$ is: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 2x + 2y \frac{dy}{dx}$

   • The derivative of the right-hand side $\left(3x^2 + 8y^2 - 5x\right)^2$ will require the chain rule: $\frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right)^2 = 2 \cdot \left(3x^2 + 8y^2 - 5x\right) \cdot \frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right)$

     Now, differentiate $3x^2 + 8y^2 - 5x$: $\frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right) = 6x + 16y \frac{dy}{dx} - 5$

3. After differentiation, substitute the point $\left(0, \frac{1}{8}\right)$ into the resulting equation to solve for $\frac{dy}{dx}$, which will give the value of $r$.

I'll work out the details for this differentiation step-by-step.

Differentiating and Substituting

Left-hand side:

$2x + 2y \frac{dy}{dx}$

At $x = 0$ and $y = \frac{1}{8}$:

$2(0) + 2 \left(\frac{1}{8}\right) \frac{dy}{dx} = \frac{1}{4} \frac{dy}{dx}$

Right-hand side:

$2 \cdot (3x^2 + 8y^2 - 5x) \cdot (6x + 16y \frac{dy}{dx} - 5)$

At $x = 0$ and $y = \frac{1}{8}$, we substitute into $(3x^2 + 8y^2 - 5x)$ and find:

$3(0)^2 + 8\left(\frac{1}{8}\right)^2 - 5(0) = \frac{8}{64} = \frac{1}{8}.$

Now, substitute into the derivative term $6x + 16y \frac{dy}{dx} - 5$:

$6(0) + 16\left(\frac{1}{8}\right) \frac{dy}{dx} - 5 = 2 \frac{dy}{dx} - 5.$

Thus, the right-hand side becomes:

$2 \cdot \frac{1}{8} \cdot \left(2 \frac{dy}{dx} - 5\right) = \frac{1}{4} \left(2 \frac{dy}{dx} - 5\right).$

Equating Both Sides:

We now equate the two sides:

$\frac{1}{4} \frac{dy}{dx} = \frac{1}{4} \left(2 \frac{dy}{dx} - 5\right).$

Simplifying:

$\frac{dy}{dx} = 2 \frac{dy}{dx} - 5$

$0 = \frac{dy}{dx} - 5$

$\frac{dy}{dx} = 5.$

Thus, the slope of the tangent line is $r = 5$.

Final Answer:

The value of $r$ is $\boxed{5}$.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the geometric interpretation of the slope of a tangent line?
2. How does implicit differentiation differ from explicit differentiation?
3. Can the chain rule be applied to functions other than polynomials? How?
4. What are some real-life applications of finding tangent lines to curves?
5. How would the process change if the equation had been parametric?

Tip: Always double-check substitution points in differentiation to ensure accuracy in complex expressions.