Question

find an equation for the line tangent to the parametric curve at the given value ot t.

$x=t+2,y=e^t,t=0$

Short answer

The tangent line at $t=0$ for the parametric equations is $y=x-1$.

Step-by-step solution

Given information

The curve is$x=t+2,y=e^t,t=0$

Calculation

Consider the parametric curves $x=t+2,y=e^t$at $t=0$.

The objective is to find the equation of a tangent line for the given parametric equations

The formula to find the tangent line equation is $y-y_1=m\left(x-x_1\right)$.

First find the slope of the parametric curves by finding the derivative of the parametric curves.

For that we use the formula $\frac{dy}{dx}=\frac{\frac{dy}{dx}}{\frac{dx}{dt}}$.

Now take the parametric equation $x=t+2$.

Differentiate the curve with respect to t.

Now substitute the values of $\frac{dx}{dt},\frac{dy}{dt}$in the slope formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

By substituting the values we get,

localid="1652450504962" $\frac{dy}{dx}=\frac{1}{e^t}\left[Since\frac{dx}{dt}=1,\frac{dy}{dt}=e^t\right]$

The slope when $t=0$is as follows,

$$\begin{gathered} \frac{dy}{d{x}_1=0}=\frac{1}{e^0} \\ \frac{dy}{d{x}_t=0}=1 \end{gathered}$$

Thus, the slope of the parametric equation is localid="1652450741668" $\frac{dy}{dx}=m=1$.

Further simplification

The point (x, y) When t=0 is,

$$\begin{gathered} (x,y)=\left(t+2,e^t\right) \\ (x,y)=\left(0+2,e^0\right)[\text{since by substituting}t=0] \\ (x,y)=(2,1)\left[\text{since}{e}^0=1\right] \end{gathered}$$

Now the slope point formula is $y-y_1=m\left(x-x_1\right)$

The point is $(2,1)$and the slope $m=1$.

$$\begin{gathered} y-1=1(x-2)\left[sincey-y_1=m\left(x-x_1\right)\right] \\ y-1=x-2 \end{gathered}$$

Add I on both sides of the equation.

$$\begin{gathered} y-1+1=x-2+1 \\ y=x-1 \end{gathered}$$

Therefore, the tangent line at $t=0$ for the parametric equations is $y=x-1$.