# For the given parametric curve, compute the following questions: x(t) = \tan t + 1 y(t) = \sec^2t...

## Question:

For the given parametric curve, compute the following questions:

{eq}x(t) = \tan t + 1\\ y(t) = \sec^2t - 3 {/eq}

a. Express the curve with an equation which relates x and y.

b. Find the slope of the tangent line to the curve at the point {eq}t = \frac \pi 6 {/eq} .

c. State the pair (x,y) where the curve has a horizontal tangent line.

## Tangent to Parametric Curve

The question presents us with a parametric curve. Using Trigonometric Identities we convert the parametric curve to Cartesian form in x and y. Then using the first derivative from Calculus, we find the slope of the tangent line to the curve at a given point x. Finally we find the point (x, y) on the curve where the graph of the function has a horizontal tangent with slope zero.

## Answer and Explanation:

a. First we have the graph in parametric form as follows:

{eq}x(t)=\tan t+1 \implies \tan t = x-1 \qquad (1) {/eq}

and

{eq}y(t)=\sec^2 t-3 \implies \sec^2 t = y+3 \qquad (2) {/eq}

From (1)

{eq}1+\tan^2 t = 1+(x-1)^2=1+x^2-2x+1=x^2-2x+2 \qquad (3) {/eq}

From (2) and (3), since {eq}1+\tan^2 t = \sec^2 t {/eq} from Trigonometry, we have

{eq}y+3=x^2-2x+2 \implies y(x)=x^2-2x-1 \qquad (4) {/eq}

The curve that relates x to y is {eq}y=y(x)=x^2-2x-1. {/eq}

b. The x-coordinate of the point when {eq}x=\pi/6 {/eq} can be obtained from the parametric form as {eq}x=\tan(\pi/6)+1=1.58. {/eq}

Then the slope of the tangent line at this x-point can be obtained from (4) above as follows:

{eq}slope=m=y'(1.58)= \left. (2x-2) \right\vert_{x=1.58} = 2(1.58)-2=1.16. {/eq}

The slope of the tangent line at this t-value is 1.16.

c. To find the point where the curve has a horizontal tangent line with slope zero we use equation (4) again to solve y'(x) = 0 which gives us the equation {eq}2x-2=0 \implies 2x=2 \implies x=1. {/eq}

The corresponding y-value can be obtained from (4) as {eq}y(1)=1^2-2(1)-1=1-2-1=-2. {/eq}

The (x, y) point where the tangent line is horizontal is (1, -2).