Q 13. - Calculus

Question

Explain how we can find the locations at which a parametric curve determined by $x = x\left(t\right)$ and $y = y\left(t\right)$ has horizontal or vertical tangent lines.

Step-by-Step Solution

Answer

$\underset{t\to 4_0}{\lim }\frac{\sin \theta }{1-\cos \theta }=\infty$

Step 1: Given information

$x = x\left(t\right)$ and $y = y\left(t\right)$

Step 2: Concept

A collection of quantities is defined as a function of one or more independent variables called parameters in a parametric equation.

Step 3: Calculation

Consider the parametric curves $x=x\left(t\right), y=y\left(t\right)$

The goal is to demonstrate how to locate horizontal and vertical tangent lines.

The parametric equations are functions of $t$ then the derivative is given by,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

When the numerator is zero and the denominator is non-zero, the horizontal tangent line is possible.

That means when $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\ne 0$

At a point $t=t_0$, $\underset{t\to c_0}{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=0$

The numerator is non zero and the denominator is zero to determine the vertical tangent line.

That means $\frac{dx}{dt}=0$ and $\frac{dy}{dt}\ne 0$

At a point $t=t_0$

$\underset{t\to m_0}{\lim }\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\infty$Take the example of $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\sin \theta }{1-\cos \theta }$

Horizontal tangent: $\frac{dy}{dt}=\sin \theta =0$

$\underset{t\to 4_e}{\lim }\frac{\sin \theta }{1-\cos \theta }=0$

Step 4: Calculation

Vertical tangent: $\frac{dx}{dt}=1-\cos \theta =0$

$\underset{t\to 4_0}{\lim }\frac{\sin \theta }{1-\cos \theta }=\infty$

Hence the explanation.