Problem Overview
We calculate the first derivative d y d x \frac{dy}{dx} d x d y d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y
x 3 − y 2 = 4 x^3 - y^2 = 4 x 3 − y 2 = 4
evaluated at the baseline configuration
( x , y ) = ( 2 , 2 ) (x, y) = (2, 2) ( x , y ) = ( 2 , 2 ) .
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 3 ] − d d x [ y 2 ] = d d x [ 4 ] \frac{d}{dx}[x^3] - \frac{d}{dx}[y^2] = \frac{d}{dx}[4] d x d [ x 3 ] − d x d [ y 2 ] = d x d [ 4 ] 3 x 2 − 2 y d y d x = 0 ⟹ d y d x = 3 x 2 2 y 3x^2 - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x^2}{2y} 3 x 2 − 2 y d x d y = 0 ⟹ d x d y = 2 y 3 x 2 (This rule is defined in Implicit Differentiation .)
Evaluate at ( 2 , 2 ) (2, 2) ( 2 , 2 )
d y d x ∣ ( 2 , 2 ) = 3 ( 2 ) 2 2 ( 2 ) = 12 4 = 3 \left.\frac{dy}{dx}\right|_{(2,2)} = \frac{3(2)^2}{2(2)} = \frac{12}{4} = 3 d x d y ( 2 , 2 ) = 2 ( 2 ) 3 ( 2 ) 2 = 4 12 = 3 Step 2: Differentiate a Second Time Using the Quotient Rule
Apply the Quotient Rule to d y d x = 3 x 2 2 y \frac{dy}{dx} = \frac{3x^2}{2y} d x d y = 2 y 3 x 2
d 2 y d x 2 = d d x [ 3 x 2 ] ⋅ ( 2 y ) − ( 3 x 2 ) ⋅ d d x [ 2 y ] ( 2 y ) 2 = ( 6 x ) ( 2 y ) − ( 3 x 2 ) ( 2 d y d x ) 4 y 2 = 12 x y − 6 x 2 d y d x 4 y 2 \frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[3x^2] \cdot (2y) - (3x^2) \cdot \frac{d}{dx}[2y]}{(2y)^2} = \frac{(6x)(2y) - (3x^2)\left(2\frac{dy}{dx}\right)}{4y^2} = \frac{12xy - 6x^2\frac{dy}{dx}}{4y^2} d x 2 d 2 y = ( 2 y ) 2 d x d [ 3 x 2 ] ⋅ ( 2 y ) − ( 3 x 2 ) ⋅ d x d [ 2 y ] = 4 y 2 ( 6 x ) ( 2 y ) − ( 3 x 2 ) ( 2 d x d y ) = 4 y 2 12 x y − 6 x 2 d x d y Step 3: Back-Substitute d y d x = 3 x 2 2 y \frac{dy}{dx} = \frac{3x^2}{2y} d x d y = 2 y 3 x 2
d 2 y d x 2 = 12 x y − 6 x 2 ( 3 x 2 2 y ) 4 y 2 = 12 x y − 18 x 4 2 y 4 y 2 = 12 x y − 9 x 4 y 4 y 2 \frac{d^2y}{dx^2} = \frac{12xy - 6x^2\left(\frac{3x^2}{2y}\right)}{4y^2} = \frac{12xy - \frac{18x^4}{2y}}{4y^2} = \frac{12xy - \frac{9x^4}{y}}{4y^2} d x 2 d 2 y = 4 y 2 12 x y − 6 x 2 ( 2 y 3 x 2 ) = 4 y 2 12 x y − 2 y 18 x 4 = 4 y 2 12 x y − y 9 x 4 Step 4: Clear Fractions
Multiply numerator and denominator by y y y
d 2 y d x 2 = 12 x y 2 − 9 x 4 4 y 3 \frac{d^2y}{dx^2} = \frac{12xy^2 - 9x^4}{4y^3} d x 2 d 2 y = 4 y 3 12 x y 2 − 9 x 4 (This technique is detailed in Higher-Order Implicit Differentiation .)
Step 5: Evaluate at ( 2 , 2 ) (2, 2) ( 2 , 2 )
Substitute x = 2 x = 2 x = 2 y = 2 y = 2 y = 2
d 2 y d x 2 ∣ ( 2 , 2 ) = 12 ( 2 ) ( 2 2 ) − 9 ( 2 4 ) 4 ( 2 3 ) = 12 ( 2 ) ( 4 ) − 9 ( 16 ) 4 ( 8 ) = 96 − 144 32 = − 48 32 = − 1.5 dm − 1 \left.\frac{d^2y}{dx^2}\right|_{(2,2)} = \frac{12(2)(2^2) - 9(2^4)}{4(2^3)} = \frac{12(2)(4) - 9(16)}{4(8)} = \frac{96 - 144}{32} = \frac{-48}{32} = -1.5\text{ dm}^{-1} d x 2 d 2 y ( 2 , 2 ) = 4 ( 2 3 ) 12 ( 2 ) ( 2 2 ) − 9 ( 2 4 ) = 4 ( 8 ) 12 ( 2 ) ( 4 ) − 9 ( 16 ) = 32 96 − 144 = 32 − 48 = − 1.5 dm − 1 Engineering Conclusion
At ( 2 , 2 ) (2, 2) ( 2 , 2 ) 3 3 3 − 1.5 dm − 1 -1.5\text{ dm}^{-1} − 1.5 dm − 1 concave downward , causing the tire to gain negative camber as the chassis rolls into a turn. In high-performance automotive engineering, negative camber counters lateral tire rollover, maintaining maximum tire contact patch area against the road and preventing breakaway skids during high-speed maneuvering.