Question

# Find the second derivative of y with respect to x from the parametric equations given. x=t^{2}-1, y = t^{2}+t

Equations
Find the second derivative of y with respect to x from the parametric equations given.
$$\displaystyle{x}={t}^{{{2}}}-{1},{y}={t}^{{{2}}}+{t}$$

2021-04-28
Step 1:Given
$$\displaystyle{x}={t}^{{{2}}}-{1},{y}={t}^{{{2}}}+{t}$$
Step 2:Solution
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={2}{t},{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={2}{t}+{1}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}}}{{{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}}}}$$
$$\displaystyle={\frac{{{2}{t}+{1}}}{{{2}{t}}}}$$
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}={\frac{{{2}{t}{\left({2}\right)}-{\left({2}{t}+{1}\right)}{\left({2}\right)}}}{{{4}{t}^{{{2}}}}}}$$
$$\displaystyle={\frac{{-{2}}}{{{4}{t}^{{{2}}}}}}=-{\frac{{{1}}}{{{2}{t}^{{{2}}}}}}$$
$$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={\frac{{{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}}}{{{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}}}}$$
$$\displaystyle={\frac{{{\frac{{-{1}}}{{{2}{t}^{{{2}}}}}}}}{{{2}{t}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}{t}^{{{3}}}}}}$$