# Tangent and concavity of parametric equations

#### Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

#### Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

#### Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

##### Intros
###### Lessons
1. Tangent and Concavity of Parametric Equations Overview
##### Examples
###### Lessons
1. Find $\;\frac{dy}{dx}\;$ and $\;\frac{d^2y}{dx^2}$
1. $x=t-t^2$, $y=3+t$
2. $x=e^t$, $y=e^{-t}$
2. Questions Regarding to Tangents and Concavity
Find the tangent to the cycloid $x=r(\theta - \sin \theta)$, $y=r(1-\cos \theta)$ when $\theta = \frac{\pi}{4}$ and $r$ > $0$. Determine the concavity for all values of $\theta$. (Do not eliminate the parameter)
1. Find the point of the parametric curve $x=t^2+1$ and $y=t^3+t^2$, in which the tangent is horizontal.
1. Find the tangent to the curve $x=3 \cos t$, $y=4 \cos t$ by:
1. Eliminating the parameter.
2. Without eliminating the parameter.
###### Topic Notes
In this lesson, we will focus on finding the tangent and concavity of parametric equations. Just like how we can take derivatives of Cartesian equations, we can also do it for parametric equations. First, we will learn to take the derivatives of parametric equations. Then we will look at an application which involves finding the tangents and concavity of a cycloid. After, we will look at special cases of finding a point with a horizontal tangent. Lastly, we will compare the difference of finding tangents by eliminating and without eliminating the parameter.
We can find the tangent (or derivative) without having to eliminate the parameter $t$ by using the equation:
$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \;$ where $\;\frac{dx}{dt} \neq0$
The horizontal tangent occurs when $\;\frac{dy}{dt} =0\;$ given that $\;\frac{dx}{dt} \neq0$.
The vertical tangent occurs when $\;\frac{dx}{dt} =0\;$ given that $\;\frac{dy}{dt} \neq0$.

To find the concavity (or second derivative), we use the following equation:
$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$