# For the given parametric curve, compute the following questions: x(t) = 5\cos t - 4 y(t) =...

## Question:

For the given parametric curve, compute the following questions:

{eq}x(t) = 5\cos t - 4\\ y(t) = -2\sin t + 3\\ 0 \leq t \leq 2\pi {/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 a Parametric Equation:

We're given a set of parametric equations. We can eliminate the parameter *t* from this equation by applying the following trigonometric identity.

{eq}\begin{align} \sin^2\theta+\cos^2\theta&=1\\ \end{align} {/eq}

We can apply the following formula to find the equation of the slope of the tangent line.

{eq}\begin{align} \frac{dy}{dx} &= \left[\displaystyle \frac{ \frac{dy}{dt} } { \frac{dx}{dt} } \right] \\ \end{align} {/eq}

The solution to the equation of a tangent for horizontal tangent can be used to find corresponding the set of coordinates.

## Answer and Explanation:

a.

{eq}\begin{align} x(t) &= 5\cos t - 4\\ \Rightarrow \cos t &= \frac{ x+4 }{ 5}\\ y(t) &= -2\sin t + 3\\ \Rightarrow \sin t &= \frac{ 3-y }{ 2}\\ \sin^2(t)+\cos^2(t) &= \left[ \frac{ 3-y }{ 2} \right]^2+ \left[ \frac{ x+4 }{ 5}\right]^2\\ \Rightarrow 1 &= \left[ \frac{ 3-y }{ 2} \right]^2+ \left[ \frac{ x+4 }{ 5}\right]^2\\ \Rightarrow \left[ \frac{ 3-y }{ 2} \right]^2+ \left[ \frac{ x+4 }{ 5}\right]^2 &=1 & \left[\text{ This equation relates x and y. } \right]\\ \end{align} {/eq}

Therefore, the curve with an equation which relates x and y is:

{eq}\displaystyle \boxed{\color{blue} { \left[ \frac{ 3-y }{ 2} \right]^2+ \left[ \frac{ x+4 }{ 5}\right]^2 =1 }} {/eq}

b.

{eq}\begin{align} x(t) = 5\cos t - 4\\ \frac{dx}{dt} &= \frac{d}{dt} \left[ 5\cos t - 4 \right]\\ &= - 5\sin t - 0 \\ &= - 5\sin t \\ y(t) = -2\sin t + 3\\ \frac{dy}{dt} &= \frac{d}{dt} \left[ -2\sin t + 3 \right]\\ &= -2\cos t + 0\\ &= -2\cos t\\ \frac{dy}{dx} &= \left[\displaystyle \frac{ \frac{dy}{dt} } { \frac{dx}{dt} } \right] \\ &= \frac{ -2\cos t } { - 5\sin t } \\ &= \frac{ 2 } { 5 } \cot t \\ t = \frac \pi 6\\ \frac{dy}{dx}|_{ t = \frac \pi 6 } &= \frac{ 2 } { 5 } \cot \frac \pi 6\\ &= \frac{ 2 } { 5 } \sqrt 3 \\ \end{align} {/eq}

For horizontal tangents, the slope of a curve is zero.

{eq}\begin{align} \frac{dy}{dx} &= \frac{ 2 } { 5 } \cot t & \left[\text{ This is the equation of slope of the curve.} \right]\\ \frac{dy}{dx} &=0 & \left[\text{ Slope is to be zero for horozontal tangents } \right]\\ \\ \frac{ 2 } { 5 } \cot t &=0\\ \Rightarrow \cot t &= 0\\ \Rightarrow \cot t &= \cot \frac{\pi}{2} \; \; , \; \cot \frac{3\pi}{2} & \left[\text{ These are the solutoin in the given domain}\; 0 \leq t \leq 2\pi \right]\\ \Rightarrow t &= \frac{\pi}{2} \; \; , \; \frac{3\pi}{2} \\ x(t) = 5\cos t - 4\\ y(t) = -2\sin t + 3\\ \end{align} {/eq}

Now, we will find the x and y coordinates at the above-calculated values of *t*.

Therefore, the curve with an equation which relates x and y is:

{eq}\begin{align} x(t) &= 5\cos t - 4\\ y(t) &= -2\sin t + 3\\ t &= \frac{\pi}{2} \; \; , \; \frac{3\pi}{2} \\ : t &= \frac{\pi}{2} \\ x(t) &= 5\cos t - 4\\ \Rightarrow x( \frac{\pi}{2} ) &= 5\cos \frac{\pi}{2} - 4\\ &= 0 - 4\\ &= -4\\ y(t) &= -2\sin t + 3\\ \Rightarrow y( \frac{\pi}{2} ) &= -2\sin \frac{\pi}{2} + 3\\ &= -2 + 3\\ &= 1\\ \Rightarrow (x, y)&= (-4, 1)\\ : t &= \frac{3\pi}{2} \\ x(t) &= 5\cos t - 4\\ \Rightarrow x( \frac{3\pi}{2} ) &= 5\cos \frac{3\pi}{2} - 4\\ &= 0 - 4\\ &= -4\\ y(t) &= -2\sin t + 3\\ \Rightarrow y( \frac{3\pi}{2} ) &= -2\sin \frac{3\pi}{2} + 3\\ &= -2(-1) + 3\\ &= 2+3\\ &=5\\ \Rightarrow (x, y)&= (-4, 5)\\ \end{align} {/eq}

Therefore, the pair (x,y) where the curve has a horizontal tangent line are:

{eq}\displaystyle \boxed{\color{blue} { \matrix{ \begin{align} (x, y)&= (-4, 1)\\ (x, y)&= (-4, 5)\\ \end{align}} }} {/eq}

#### Learn more about this topic:

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11