Question

If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer 1

Taking
`y=y(x)` and differentiating both sides with respect to
`x` yields


`(\mathrm d)/(\mathrm dx)\bigg[3x^2+y^2\bigg]=(\mathrm d)/(\mathrm dx)\bigg[7\bigg]\implies 6x+2y(\mathrm dy)/(\mathrm dx)=0`

Solving for the first derivative, we have


`(\mathrm dy)/(\mathrm dx)=-\frac{3x}y`

Differentiating again gives


`(\mathrm d)/(\mathrm dx)\bigg[6x+2y(\mathrm dy)/(\mathrm dx)\bigg]=(\mathrm d)/(\mathrm dx)\bigg[0\bigg]\implies 6+2\left((\mathrm dy)/(\mathrm dx)\right)^2+2y(\mathrm d^2y)/(\mathrm dx^2)=0`

Solving for the second derivative, we have


`(\mathrm d^2y)/(\mathrm dx^2)=-\frac{3+\left((\mathrm dy)/(\mathrm dx)\right)^2}y=-\frac{3+(9x^2)/(y^2)}y=-(3y^2+9x^2)/(y^3)`

Now, when
`x=1` and
`y=2`, we have


`(\mathrm d^2y)/(\mathrm dx^2)\bigg|_(x=1,y=2)=-(3\cdot2^2+9\cdot1^2)/(2^3)=\frac{21}8\approx2.63`