Consider the equation $$1.51x^2 +y^2 = 1 + 0.71x^2y^2.$$ In this question you will calculate the curvature, $\rho$. Evaluate the derivative at the point described — you should get decimal numbers $$x= -1.92, y= -1.68.$$
I'm having trouble finding the second derivative of the above equation provided. When finding the first derivative i got up to: $$y' = \frac{0.71xy^2 - 3.02x}{2y-0.71x^2y}.$$ I then substituted the first $x$- and $y$-value and got $y'=1.88105$.
However I'm struggling to find the second derivative to then be able to calculate the curvature.
I have had two other previous questions with different values ($1.51$ and $0.71$), and I used the quotient rule to find the second derivative however both attempts, both derivative values were wrong ($y'$ and $y''$), as well as the rho value.
Any help would be appreciated, thank you :)
Your original equation is
$$\alpha x^2 + y^2 = 1 + \beta x^2 y^2$$
where I have substituted variables $\alpha = 1.51$ and $\beta = 0.71$ for the parameters in place of given values for ease of solving. Differentiating with respect to $x$ gives
$$2 \alpha x + 2 y y' = 2 \beta x y^2 + 2 \beta x^2 y y'$$
Solving for $y'$ gives
$$y' = \frac{(\beta y^2 - \alpha) x}{(1 - \beta x^2) y}$$
Differentiating the equation obtained before solving for $y'$ gives
$$2 \alpha + 2 (y')^2 + 2 y y'' = 2 \beta y^2 + 4 \beta x y y' + 4 \beta x y y' + 2 \beta x^2 (y')^2 + 2 \beta x^2 y y''$$
Solving for $y''$ gives
$$y'' = \frac{\beta y^2 + 4 \beta x y y' + (\beta x^2 - 1) (y')^2 - \alpha}{(1 - \beta x^2) y}$$
Substituting values, for $x = -1.92$, we obtain $y \approx -1.680$ (taking the negative root as given), $y' \approx -0.3495$ and $y'' \approx -0.9235$. I will leave it to you to calculate the curvature.