Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

Consider the curve x(t) = t^3 − 4t and y(t) = t^2. Determine where the curve has a horizontal tangent.

OpenStudy (anonymous):

I got (0,0)

OpenStudy (zehanz):

You get a horizontal tangent when dy/dt=0 AND dx/dt <> 0 (not zero). dy/dt=2t=0 if t=0. dx/dt=3t²-4. If t=0, then dx/dt=-4, so that is ok. If t=0 there is a horizontal tangent. x(0)=0, y(0)=0, so it is (0, 0). Well done!

OpenStudy (anonymous):

Awesome! Thank you so much!

OpenStudy (zehanz):

YW! You said you got (0, 0). Did you find that in the same way I did?

OpenStudy (anonymous):

I found dy/dx. I did this by getting dy/dt=2t and dx/dt=3t^2-4

OpenStudy (anonymous):

Then I did (dy/dt)/(dx/dt)=dy/dx=2t/(3t^2-4)

OpenStudy (anonymous):

then I solved 2t=0

OpenStudy (anonymous):

made sure t=0 was in my domain

OpenStudy (anonymous):

then plugged it into x(t) and y(t) to get my point

OpenStudy (anonymous):

So, yes. I think it's the same.

OpenStudy (zehanz):

:D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!