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Section 9.2 : Tangents with Parametric Equations
1. Compute dydx and d2ydx2 for the following set of parametric equations.
x=4t3−t2+7ty=t4−6Show All Steps Hide All Steps
Start SolutionThe first thing we’ll need here are the following two derivatives.
dxdt=12t2−2t+7dydt=4t3 Show Step 2The first derivative is then,
\frac{{dy}}{{dx}} = \frac{{\displaystyle \,\,\frac{{dy}}{{dt}}\,\,}}{{\displaystyle \,\,\frac{{dx}}{{dt}}\,\,}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{t^3}}}{{12{t^2} - 2t + 7}}}} Show Step 3For the second derivative we’ll now need,
\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\left( {\frac{{4{t^3}}}{{12{t^2} - 2t + 7}}} \right) = \frac{{\left( {12{t^2}} \right)\left( {12{t^2} - 2t + 7} \right) - 4{t^3}\left( {24t - 2} \right)}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}}}} Show Step 4The second derivative is then,
\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}} = \frac{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}}}}{{12{t^2} - 2t + 7}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^3}}}}}