y=-t+e^a t=0 x=t+e^a (d y)/(d t)=-1+a e^u r (d x)/(d t)=1+a e^u (d y)/(d x)=(a e^a t-1)/(a e^a t

step 1 So these two equations are given. So first we will find D -Y by D -T. This is equal to minus 1 p

Y=-t+e^a t=0 x=t+e^a (d y)/(d t)=-1+a e^u r (d x)/(d t)=1+a...

$y=-t+e^{a t}=0$ $x=t+e^{a}$ $\frac{d y}{d t}=-1+a e^{u r}$ $\frac{d x}{d t}=1+a e^{u}$ $\frac{d y}{d x}=\frac{a e^{a t}-1}{a e^{a t}+1}=0$ $\Rightarrow \mathrm{e}^{a t}=\frac{1}{a}$ $\Rightarrow a t=-\ln a$ $\mathrm{t}=-\frac{\ln \mathrm{a}}{\mathrm{a}}$ $\mathrm{x}=-\frac{\ln \mathrm{a}}{\mathrm{a}}+\frac{1}{\mathrm{a}}=\frac{\ln \mathrm{e} / \mathrm{a}}{\mathrm{a}}=\frac{2}{\mathrm{a}}$

$y=-t+e^{a t}=0$
$x=t+e^{a}$
$\frac{d y}{d t}=-1+a e^{u r}$
$\frac{d x}{d t}=1+a e^{u}$
$\frac{d y}{d x}=\frac{a e^{a t}-1}{a e^{a t}+1}=0$
$\Rightarrow \mathrm{e}^{a t}=\frac{1}{a}$
$\Rightarrow a t=-\ln a$
$\mathrm{t}=-\frac{\ln \mathrm{a}}{\mathrm{a}}$
$\mathrm{x}=-\frac{\ln \mathrm{a}}{\mathrm{a}}+\frac{1}{\mathrm{a}}=\frac{\ln \mathrm{e} / \mathrm{a}}{\mathrm{a}}=\frac{2}{\mathrm{a}}$

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of an independent parameter. This framework allows one to describe curves that may not be functions in the traditional y=f(x) sense, and enables the analysis of curves with respect to their parameter, leading to insights about the geometric and dynamic nature of the curve.

Differentiation of Parametric Functions

When a curve is defined parametrically, the derivative of y with respect to x is obtained by dividing the derivative of y with respect to the parameter by the derivative of x with respect to the same parameter. This technique is fundamental in determining the slope of the tangent to the curve at any point and is instrumental in identifying critical points where the behavior of the curve changes.

Critical Points in Curves

Critical points on a curve occur where the derivative (slope) is zero or undefined, indicating possible local extrema, points of inflection, or singularities. Analyzing these points helps in understanding the overall shape and key features of the curve, such as peaks, valleys, or turning points.

Exponential Functions

Exponential functions, characterized by a constant base raised to a variable exponent, exhibit unique properties such as continuous growth or decay. Their differentiable nature makes them significant in calculus, especially when combined with other functions, where they require careful consideration during differentiation and solving equations.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are used to solve equations involving exponentials. They play a crucial role in calculus when dealing with problems that require the manipulation and inversion of exponential expressions in order to isolate variables or solve for parameters.

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