Problem Overview
We calculate the second derivative d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y
x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25
evaluated at the coordinate point
( x , y ) = ( 3 , 4 ) (x, y) = (3, 4) ( x , y ) = ( 3 , 4 ) to determine localized curvature and concavity under thermal expansion stress.
Step-by-Step Mathematical Solution
Step 1: Find the First Derivative d y d x \frac{dy}{dx} d x d y
Differentiate both sides with respect to x x x
d d x [ x 2 ] + d d x [ y 2 ] = d d x [ 25 ] \frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25] d x d [ x 2 ] + d x d [ y 2 ] = d x d [ 25 ] 2 x + 2 y d y d x = 0 ⟹ d y d x = − x y 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} 2 x + 2 y d x d y = 0 ⟹ d x d y = − y x (This rule is defined in Implicit Differentiation .)
Step 2: Differentiate a Second Time Using the Quotient Rule
Apply the Quotient Rule to d y d x = − x y \frac{dy}{dx} = -\frac{x}{y} d x d y = − y x
d 2 y d x 2 = d d x [ − x y ] = − [ d d x [ x ] ⋅ y − x ⋅ d d x [ y ] y 2 ] = − y − x d y d x y 2 = − y + x d y d x y 2 \frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\left[ \frac{\frac{d}{dx}[x] \cdot y - x \cdot \frac{d}{dx}[y]}{y^2} \right] = -\frac{y - x\frac{dy}{dx}}{y^2} = \frac{-y + x\frac{dy}{dx}}{y^2} d x 2 d 2 y = d x d [ − y x ] = − [ y 2 d x d [ x ] ⋅ y − x ⋅ d x d [ y ] ] = − y 2 y − x d x d y = y 2 − y + x d x d y Step 3: Back-Substitute d y d x = − x y \frac{dy}{dx} = -\frac{x}{y} d x d y = − y x
d 2 y d x 2 = − y + x ( − x y ) y 2 = − y − x 2 y y 2 \frac{d^2y}{dx^2} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^2} = \frac{-y - \frac{x^2}{y}}{y^2} d x 2 d 2 y = y 2 − y + x ( − y x ) = y 2 − y − y x 2 Step 4: Clear Fractions and Substitute Original Equation Constraint
Multiply numerator and denominator by y y y
d 2 y d x 2 = − y 2 − x 2 y 3 = − ( x 2 + y 2 ) y 3 \frac{d^2y}{dx^2} = \frac{-y^2 - x^2}{y^3} = \frac{-(x^2 + y^2)}{y^3} d x 2 d 2 y = y 3 − y 2 − x 2 = y 3 − ( x 2 + y 2 )
Because the curve is constrained by
x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 :
d 2 y d x 2 = − 25 y 3 \frac{d^2y}{dx^2} = -\frac{25}{y^3} d x 2 d 2 y = − y 3 25 (This technique is detailed in Higher-Order Implicit Differentiation .)
Step 5: Evaluate at ( 3 , 4 ) (3, 4) ( 3 , 4 )
Substitute y = 4 y = 4 y = 4
d 2 y d x 2 ∣ ( 3 , 4 ) = − 25 4 3 = − 25 64 ≈ − 0.3906 cm − 1 \left.\frac{d^2y}{dx^2}\right|_{(3,4)} = -\frac{25}{4^3} = -\frac{25}{64} \approx -0.3906\text{ cm}^{-1} d x 2 d 2 y ( 3 , 4 ) = − 4 3 25 = − 64 25 ≈ − 0.3906 cm − 1 Engineering Conclusion
Because d 2 y d x 2 = − 25 64 < 0 \frac{d^2y}{dx^2} = -\frac{25}{64} < 0 d x 2 d 2 y = − 64 25 < 0 concave downward at ( 3 , 4 ) (3, 4) ( 3 , 4 ) κ = ∣ y ′ ′ ∣ ( 1 + ( y ′ ) 2 ) 3 / 2 \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} κ = ( 1 + ( y ′ ) 2 ) 3/2 ∣ y ′′ ∣ d 2 y d x 2 = − 25 64 \frac{d^2y}{dx^2} = -\frac{25}{64} d x 2 d 2 y = − 64 25