Question

For what values of t does the curve given by the parametric equations x=t^3-t^2-1 and y=t^4+2t^2-8t have a vertical tangent?

Answer 1

@SithsAndGiggles

Answer 2

A vertical tangent is achieved when the derivative is infinite and the function is continuous at that point.
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2-2t}{4t^3+4t-8}\]
This will attain an infinite value when the denominator is zero.

Answer 3

So we set 4t^3+4t-8=0 and find the values of t, right?

Answer 4

I think you got opposite because 4t^3+4t-8 should be the numerator instead of the denominator.

Answer 5

Oh, sorry, you're right, they're reversed. Solve for \(3t^2-2t=0\), which is much easier by hand.

Answer 6

Got it, thanks a lot.

Answer 7

yw