Calculus Notes on Implicit Differentiation of Circle Equations

Mathematical Definition of the Implicit Function

  • The study of the equation a2+y2=30a^2 + y^2 = 30 focuses on the relationship between two variables, where one is typically treated as the independent variable (in this case, aa) and the other as the dependent variable (represented by yy).
  • This equation defines a circle in the cartesian plane when plotted. The standard form of a circle centered at the origin is x2+y2=r2x^2 + y^2 = r^2.
  • Comparing the given equation to the standard form reveals that the radius squared is r2=30r^2 = 30, meaning the radius of this circle is approximately 30≈5.477\sqrt{30} \approx 5.477.
  • The objective is to determine the second derivative of yy with respect to aa, denoted as d2yda2\frac{d^2y}{da^2}, through the method of implicit differentiation.

Implicit Differentiation: First Derivative Calculation

  • To find the first derivative, differentiate both sides of the equation with respect to the independent variable aa.
  • Apply the power rule to the term a2a^2:   dda(a2)=2a\frac{d}{da}(a^2) = 2a
  • Apply the chain rule to the term y2y^2, treating yy as a function of aa:   dda(y2)=2y⋅dyda\frac{d}{da}(y^2) = 2y \cdot \frac{dy}{da}
  • Differentiate the constant on the right side of the equation:   dda(30)=0\frac{d}{da}(30) = 0
  • Combine these results into a single differential equation:   2a+2y⋅dyda=02a + 2y \cdot \frac{dy}{da} = 0
  • Isolate the first derivative term by subtracting 2a2a from both sides:   2y⋅dyda=−2a2y \cdot \frac{dy}{da} = -2a
  • Divide both sides by 2y2y to solve for the first derivative:   dyda=−2a2y\frac{dy}{da} = -\frac{2a}{2y}
  • Simplify the fraction by canceling common terms:   dyda=−ay\frac{dy}{da} = -\frac{a}{y}

Higher-Order Differentiation: Second Derivative Calculation

  • To determine the second derivative, differentiate the first derivative expression dyda=−ay\frac{dy}{da} = -\frac{a}{y} with respect to aa.
  • Utilize the quotient rule for differentiation, defined as ddx(uv)=u′v−uv′v2\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}.
  • Define the components for the quotient rule:   u=−au = -av=yv = y
  • Calculate the derivatives of the components:   u′=dda(−a)=−1u' = \frac{d}{da}(-a) = -1v′=dydav' = \frac{dy}{da}
  • Substitute these components into the quotient rule formula:   d2yda2=(−1)(y)−(−a)(dyda)y2\frac{d^2y}{da^2} = \frac{(-1)(y) - (-a)(\frac{dy}{da})}{y^2}
  • Simplify the expression in the numerator:   d2yda2=−y+a⋅dyday2\frac{d^2y}{da^2} = \frac{-y + a \cdot \frac{dy}{da}}{y^2}

Final Simplification and Substitution

  • To obtain the final form of the second derivative, substitute the value discovered for the first derivative, dyda=−ay\frac{dy}{da} = -\frac{a}{y}, back into the equation:   d2yda2=−y+a(−ay)y2\frac{d^2y}{da^2} = \frac{-y + a(-\frac{a}{y})}{y^2}
  • Simplify the term in the numerator by multiplying:   d2yda2=−y−a2yy2\frac{d^2y}{da^2} = \frac{-y - \frac{a^2}{y}}{y^2}
  • Eliminate the complex fraction by finding a common denominator in the numerator:   d2yda2=−y2−a2yy2\frac{d^2y}{da^2} = \frac{\frac{-y^2 - a^2}{y}}{y^2}
  • Further simplify the division by multiplying the denominator:   d2yda2=−y2−a2y3\frac{d^2y}{da^2} = \frac{-y^2 - a^2}{y^3}
  • Factor out the negative sign:   d2yda2=−a2+y2y3\frac{d^2y}{da^2} = -\frac{a^2 + y^2}{y^3}
  • Refer back to the original equation, where a2+y2=30a^2 + y^2 = 30, to substitute for the expression in the numerator:   d2yda2=−30y3\frac{d^2y}{da^2} = -\frac{30}{y^3}
  • The final result establishes that the second derivative of the given function is the constant value −30-30 divided by the cube of the dependent variable.