I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 9.2 : Tangents with Parametric Equations
For problems 1 – 3 compute \(\displaystyle \frac{{dy}}{{dx}}\) and \(\displaystyle \frac{{{d^2}y}}{{d{x^2}}}\) for the given set of parametric equations.
- \(x = 7{t^2} - 9t\hspace{0.5in}y = {t^6} + 2{t^2}\)
- \(x = \tan \left( {2t} \right) - 12\hspace{0.5in}y = 3\sin \left( {2t} \right) + \sec \left( {2t} \right) + 4t\)
- \(x = \ln \left( {3{t^2}} \right) + 8t\hspace{0.5in}y = \ln \left( {{t^4}} \right) - 6\ln \left( {{t^2}} \right)\)
For problems 4 – 7 find the equation of the tangent line(s) to the given set of parametric equations at the given point.
- \(x = {t^3} + \cos \left( {\pi \,t} \right)\hspace{0.25in}y = 4t + \sin \left( {2t + 6} \right)\) at \(t = - 3\)
- \(x = {t^2} + 2t - 1\hspace{0.25in}y = {t^3} + 7{t^2} + 8t\) at \(t = 1\)
- \(x = 6 - {{\bf{e}}^{{t^{\,3}} - 9t}}\hspace{0.25in}y = {t^3} + 3{t^2} - 18t + 2\) at \(\left( {5,2} \right)\)
- \(x = {t^2} + 5t - 6\hspace{0.25in}y = {t^2} + 2t - 8\) at \(\left( { - 6,7} \right)\)
For problems 8 and 9 find the values of t that will have horizontal or vertical tangent lines for the given set of parametric equations.
- \(x = {t^3} - 5{t^2} + t + 1\hspace{0.25in}y = {t^4} + 8{t^3} + 3{t^2}\)
- \(x = 7{t^2} + {{\bf{e}}^{2 - {t^{\,2}}}}\hspace{0.25in}y = 10\sin \left( {\displaystyle \frac{1}{2}t} \right) - 1\)