Q.12 - Calculus

Question

Suppose a parametric curve is given by parametric equations $x=x\left(t\right), y=y\left(t\right)$for $t$in some interval L. How can we find the slope of the parametric curve at some point $\left(x\left(t_0\right),y\left(t_0\right)\right)$? What is the equation of the tangent line to the parametric curve at the point $t_0$?

Step-by-Step Solution

Answer

Thus, the slope and the tangent line equations are 

$m=\frac{y^{\text{'}}\left(t_e\right)}{x^{\text{'}}\left(t_e\right)},y-y\left(t_e\right)=m\left(x-x\left(t_e\right)\right)\text{. }$

Step: 1 Given information

Consider the parametric equations$x=x\left(t\right), y=y\left(t\right)$, for t  in some interval $I$

Step: 2 Calculation

The objective is to find the slope and the tangent line for the parametric curve at $t_{0}x=x\left(t\right),y=y\left(t\right)$are the parametric equations for some $t$

Let's$t_0$ take any point in the interval $I$

Then the points at $t_0$ are $x=x\left(t_0\right),y=y\left(t_e\right)$.

At $t_0,x^1\left(t_0\right)·y^1\left(t_0\right)$ exists.

Step 3: Further Calculation

If $x^{\text{'}}\left(t_0\right)\ne 0,$

We define the slope of the parametric curve at the point $\left(x\left(t_0\right),y\left(t_e\right)\right)$ is $m=\frac{y^1\left(t_e\right)}{x^1\left(t_e\right)}$

If $x^l\left(t_0\right)=0$.

the slope to be, $m=\underset{t\to 6}{\lim }\frac{y^{\text{'}}\left(t_0\right)}{x^1\left(t_0\right)}$

If the slope $m$is defined, the tangent line to the parametric curve at the point $t_0$is given by the equation.

$y-y\left(t_6\right)=m\left(x-x\left(t_6\right)\right)$