matthewphs857
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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.

Respuesta :

LammettHash
Taking [tex]y=y(x)[/tex] and differentiating both sides with respect to [tex]x[/tex] yields

[tex]\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]

Solving for the first derivative, we have

[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y[/tex]

Differentiating again gives

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

Solving for the second derivative, we have

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

Now, when [tex]x=1[/tex] and [tex]y=2[/tex], we have

[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63[/tex]

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