Parametric Eqns, Tangent Line Eqn: x = t^3+6t+1, y = 2t-t^2

[Magugag]

Hello all! Joined specifically to get help with the problem, hope I phrase things correctly.

I was given the parametric equation:

x = t³+6t+1 and y = 2t-t²

I was then asked to eliminate the parameter, and to find where the tangent line is vertical or horizontal.

After some work, I figured that I had to use the quadratic equation on y = 2t-t² in order to isolate t, and in doing so I got t = 1 + or - (1-y)^1/2

This is where things get sticky for me. It's unlike any parametric equation I've worked with in my course so far, as those are usually fairly straightforward to isolate for t and then to take the derivative of. I'm not certain how to proceed, or how to take the derivative of an equation that includes a + or - sign. Is there some trick I'm missing, or is my math way off?

Any help is deeply appreciated, and thanks for your time and attention.

Also, sorry if there's a better way to do some of these symbols, I'm still learning the forum tools.

 

[tkhunny]

That is a good dilemma, seeing that you now have a piecewise definition. Very good work. Now what? Does the problem statement DEMAND that you use this new definition to compute derivatives?

Why not just ignore the piecewise definition and use y'(t) and x'(t) to answer the other parts?

 

[Magugag]

That is a good dilemma, seeing that you now have a piecewise definition. Very good work. Now what? Does the problem statement DEMAND that you use this new definition to compute derivatives?

Why not just ignore the piecewise definition and use y'(t) and x'(t) to answer the other parts?

...Oh. After rereading the question, no, I do not need to use the isolated function to answer the other parts of the problem. That's so obvious that I can't believe I missed it. You're absolutely right and I should be able to use the derivative of y(t) and x(t) to get the answers I need.

Thanks a lot, that was what I needed!