Q. 41 - Calculus

Question

Find an equation for the line tangent to the parametric curve at the given value of t 

$x=2 t-1, y=3 t+5, t=-1$

Step-by-Step Solution

Answer

The tangent line at $t=-1$ for the parametric equations is $y-2=\frac{3}{2}(x+3)$

Step 1: Given information

The parametric curve is $x=2 t-1, y=3 t+5, t=-1$

Step 2: Calculation

Consider the parametric curves $x=2 t-1, y=3 t+5$ at $t=-1$.

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

Now take the parametric equation $x=2 t-1$.

Differentiate the curve with respect to t.

Then,

$$\begin{gathered} \frac{dx}{dt}=\frac{d(2t-1)}{dt}\text{ since }x=2t-1 \\ \frac{dx}{dt}=2·\frac{dt}{dt}-\frac{d1}{dt} \\ \frac{dx}{dt}=2·1-0\text{ since }\frac{d1}{dt}=0 \\ \frac{dx}{dt}=2 \end{gathered}$$

Now take the parametric equation $y=3 i+5$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(3t+5)\text{ since }y=3t+5 \\ \frac{dy}{dt}=3\frac{d}{dt}t+\frac{d}{dt}5 \end{gathered}$$

The slope of the parametric equation is$\frac{dy}{dx}=m=\frac{3}{2}$.

Step 3: Further Calculation

The point $(x, y)$When $t=-1$ is,

$$\begin{gathered} (x,y)=(2t-1,3t+5) \\ (x,y)=\left(2\right(-1)-1,3(-1)+5) [\text{ since by substituting }t=-1] \\ (x,y)=(-3,2)\text{ simplify } \end{gathered}$$

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

The point is $(-3,2)$ and the slope is $m=\frac{3}{2}$.

$$\begin{gathered} y-2=\frac{3}{2}(x-(-3\left)\right)\left[\text{ since }y-y_1=m\left(x-x_1\right)\right] \\ y-2=\frac{3}{2}(x+3) \end{gathered}$$

Therefore, the tangent line at $t=-1$ for the parametric equations is $y-2=\frac{3}{2}(x+3)$