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Consider the curve x(t) = t^3 − 4t and y(t) = t^2. Determine where the curve has a horizontal tangent.
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I got (0,0)
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!
Awesome! Thank you so much!
YW! You said you got (0, 0). Did you find that in the same way I did?
I found dy/dx. I did this by getting dy/dt=2t and dx/dt=3t^2-4
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Then I did (dy/dt)/(dx/dt)=dy/dx=2t/(3t^2-4)
then I solved 2t=0
made sure t=0 was in my domain
then plugged it into x(t) and y(t) to get my point
So, yes. I think it's the same.
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