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### Section 9.2 : Tangents with Parametric Equations

1. Compute $$\frac{{dy}}{{dx}}$$ and $$\frac{{{d^2}y}}{{d{x^2}}}$$ for the following set of parametric equations.

$x = 4{t^3} - {t^2} + 7t\hspace{0.5in}\,\,y = {t^4} - 6$

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Start Solution

The first thing we’ll need here are the following two derivatives.

$\frac{{dx}}{{dt}} = 12{t^2} - 2t + 7\hspace{0.5in}\frac{{dy}}{{dt}} = 4{t^3}$ Show Step 2

The 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 3

For 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 4

The second derivative is then,

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}} = \frac{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^2}}}}}{{12{t^2} - 2t + 7}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{48{t^4} - 16{t^3} + 84{t^2}}}{{{{\left( {12{t^2} - 2t + 7} \right)}^3}}}}}$