Find parametric equations for the tangent line to the curve with parametric equations at the point with

Question

Find parametric equations for the tangent line to the curve with parametric equations    at the point with

Quizplus · Accepted Answer

To find the tangent line to a curve with parametric equations $x=f(t)$ and $y=g(t)$ at a point $(x_0,y_0)$ on the curve, we compute the derivative of $y$ with respect to $x$ and evaluate it at $t_0$ to get the slope of the tangent line: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g'(t)}{f'(t)}.$$ Then we use the point-slope form of the equation of a line to get the equation of the tangent line: $$y-y_0=m(x-x_0),$$ where $m=\frac{g'(t_0)}{f'(t_0)}.$

Using these formulas for the given parametric equations, we have $$x=2t-1,\quad y=t^2+1,$$ so $$\frac{dx}{dt}=2,\quad \frac{dy}{dt}=2t.$$ At $t=-1$, we have $x=-3$ and $y=2$, so $(x_0,y_0)=(-3,2)$. Evaluating the slope at $t=-1$, we get $$m=\frac{g'(-1)}{f'(-1)}=\frac{2(-1)}{2}=-1.$$ Therefore, the equation of the tangent line is $$y-2=-1(x+3),$$ or $$y=-x-1.$$ Converting this to parametric form, we get $$x=t-3,\quad y=-t-1.$$ Comparing these equations to the answer choices, we see that only choice (B) matches.

Find Parametric Equations for the Tangent Line to the Curve