x=t^2+t+1 (d x)/(d t)=2 t+1 y=t^2-t+1 (d y)/(d t)=2 t-1 (d y)/(d x)=(2 t-1)/(2 t+1) ⇒y-(t^2-t+1)

step 1 Hello everyone in this question we will be learning about some basic concept of the tangent and

X=t^2+t+1 (d x)/(d t)=2 t+1 y=t^2-t+1 (d y)/(d t)=2 t-1 (d...

$x=t^{2}+t+1$ $\frac{d x}{d t}=2 t+1$ $y=t^{2}-t+1$ $\frac{d y}{d t}=2 t-1$ $\frac{d y}{d x}=\frac{2 t-1}{2 t+1}$ $\Rightarrow y-\left(t^{2}-t+1\right)=\frac{2 t-1}{2 t+1}\left(x-\left(t^{2}+t+1\right)\right)$ $\Rightarrow\left(t-t^{2}\right)=\left(\frac{2 t-1}{2 t+1}\right)\left(-\left(t^{2}+t\right)\right)$ $\Rightarrow 2 t^{2}+t-2 t^{3}-t^{2}=t^{2}+t-2 t^{3}-2 t^{2}$ $\Rightarrow 2 t^{2}=0$ $\Rightarrow t=0$

$x=t^{2}+t+1$
$\frac{d x}{d t}=2 t+1$
$y=t^{2}-t+1$
$\frac{d y}{d t}=2 t-1$
$\frac{d y}{d x}=\frac{2 t-1}{2 t+1}$
$\Rightarrow y-\left(t^{2}-t+1\right)=\frac{2 t-1}{2 t+1}\left(x-\left(t^{2}+t+1\right)\right)$
$\Rightarrow\left(t-t^{2}\right)=\left(\frac{2 t-1}{2 t+1}\right)\left(-\left(t^{2}+t\right)\right)$
$\Rightarrow 2 t^{2}+t-2 t^{3}-t^{2}=t^{2}+t-2 t^{3}-2 t^{2}$
$\Rightarrow 2 t^{2}=0$
$\Rightarrow t=0$

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of one or more independent parameters instead of a direct relation between the coordinates. This representation can simplify the description and analysis of curves, especially when the curve does not pass the vertical line test or has complex geometry that is difficult to express in the form y = f(x).

Differentiation of Parametric Functions

When curves are given parametrically, the derivative dy/dx is obtained by differentiating both y and x with respect to the parameter and then dividing the derivatives. This process allows the calculation of the slope of the tangent line to the curve without having to explicitly solve for y in terms of x.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. In the context of parametric equations, it is applied to relate the derivatives of x and y with respect to the parameter, thereby determining the rate of change of y with respect to x.

Tangent Line to a Parametric Curve

The tangent line at a given point on a parametric curve can be obtained by using the derivative dy/dx, which represents the slope at that point. The equation of the tangent line is then determined by applying the point-slope form, utilizing the coordinates from the parametric expressions and the evaluated derivative.

Elimination of the Parameter

Eliminating the parameter involves removing the independent parameter from a set of parametric equations to establish a direct relationship between x and y. This technique is useful for solving equations and analyzing the curve’s properties without the extra complexity introduced by the parameter.

Transcript

Please give Ace some feedback