At what value of t does the curve x = 2t - 3t^2, y = t^2 - 3t have a vertical tangent?

Bevan Mcdonald 2020-10-19 Answered
At what value of t does the curve \(x = 2t - 3t^2, y = t^2 - 3t\) have a vertical tangent?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

lamusesamuset
Answered 2020-10-20 Author has 27040 answers

Step 1 The given parametric equations are \(x = 2t − 3t2\)
\(y = t2 − 3t\) Vertical tangent: It occurs when derivative is undefined for given parametric equations. Step 2 Solve the equations to find derivative with respect to t \(\frac{dx}{dt} = 2 - 6t\)
\({dy}{dt} = 2t - 3\)
\(\frac{dy}{dx} = \frac{dy}{dt} = (\frac{dt}{dx})\)
\(= \frac{2t-3}{2-6t}\) Now, the derivative is undefined when denominator of the above equation becomes zero i.e. \(2 − 6t = 0\)
\(2 = 6t\)
\(t = \frac{2}{6} = \frac{1}{3}\) Answer: Hence, for the given parametric equations , the vertical tangent is at \(t = \frac{1}{3}\) and option (a) is correct answer.

Not exactly what you’re looking for?
Ask My Question
17
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

...