Question

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

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asked 2021-04-26
Find the second derivative of y with respect to x from the parametric equations given.
\(\displaystyle{x}={t}^{{{2}}}-{1},{y}={t}^{{{2}}}+{t}\)

Answers (1)

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