# Let a curve be given by x = 2 t^3 - 3 t^2 + 1, y = 2 t^3 - 9 t^2 + 12 t - 2. a) Find the slope...

## Question:

Let a curve be given by {eq}x = 2 t^3 - 3 t^2 + 1, \\ y = 2 t^3 - 9 t^2 + 12 t - 2. {/eq}

a) Find the slope of the tangent line to this curve at the point {eq}(0, 3). {/eq}

b) Find the values of the t where the tangent line is vertical, and then horizontal. Make use of L'Hopital's Rule.

## Tangents and Asymptotes of a Parametrically Defined Function

First we are given a curve in parametric form. Using the formula for the derivative of a parametrically defined function we find the slope of the tangent line to the curve at a given point in part (a). Then in part (b), using the derivative function again, we find the value of the parameter t for which we have a vertical and a horizontal tangent line or asymptote. The concepts used are first derivatives and point slope form of the equation of a line from Calculus and Algebra.

## Answer and Explanation:

(a) First we note that the value of {eq}t=1 {/eq} gives us the point of tangent {eq}(x(1),y(1))=(2-3+1,2-9+12-2)=(0,3). {/eq}

The slope of the tangent at the above point can be found by finding the value of {eq}\displaystyle \frac {dy}{dx} \; when \; t = 1. {/eq}

To that end, we calculate

{eq}\displaystyle \frac {dy}{dx} = \frac {dy/dt}{dx/dt} = \frac {6t^2-18t+12}{6t^2-6t} = \frac {6(t-1)(t-2)}{6t(t-1)} \qquad (1) {/eq}

From (1) the slope of the tangent line at (0, 3) is undefined when t = 1.

(b) The tangent line is horizontal when {eq}\displaystyle \frac {dy}{dx}=0 \implies \frac {dy}{dt}=0. {/eq}

From part (a) this happens when t = 1 and t = 2.

Similarly the tangent line is vertical when {eq}\displaystyle \frac {dy}{dx}=undefined \implies \frac {dx}{dt}=0. {/eq}

From part (a) this happens when t = 0 and t = 1.

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