Expert answers to "At what value of t does the curve..."

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

Bevan Mcdonald

Answered question

2020-10-19

At what value of t does the curve

x = 2 t - 3 t^{2}, y = t^{2} - 3 t

have a vertical tangent?

Answer & Explanation

lamusesamuset

Skilled2020-10-20Added 93 answers

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

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