# Finding the equation for a line tangent to a parametric curve

I have the parametric equation

$x = 2t - 1$

$y = 3t + 5$

$t = -1$ (defined as $t_0$)

I am trying to find the line tangent to it.

My book says if $x'(t_0) \not = 0$ then you can use the equation m = $\frac{y'(t_0)}{x'(t_0)}$ to find the tangent line.

I am having confusion, however, because I was under the impression that x'(-1) is equal to the derivative of x evaluated at -1, (2(-1) - 1)' , which would be 0.

The answer to $x'(t_0)$ is supposed to be 2, which I guess is (2t - 1)'

Is this a typo, or am I having a serious brain fog on how to solve derivatives?

• It's the derivative (evaluate first) of $x(t)$ evaluated (evaluate second) at $t_0$. Commented Apr 22, 2016 at 1:48
• Your parametric "curve" is already a line...(Eliminate $t$ and see what you get) Commented Apr 22, 2016 at 1:53
• Since $x'(t) = 2$ is constant, $x'(-1) = 2$. Similarly, $y'(t) = 3$ is a constant, so $y'(-1) = 3$. Commented Apr 22, 2016 at 10:38

Given: $$x(t)=2t-1,y(t)=3t+5, t=-1$$ Find Derivatives of $x(t)$ and $y(t)$: $$\frac{\mathrm{d} x}{\mathrm{d} t}=2$$ $$\frac{\mathrm{d} y}{\mathrm{d} t}=3$$ Use Derivatives of the Parametric Equations to find $\frac{\mathrm{d} y}{\mathrm{d} x}$: $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}=\frac{3}{2}$$ Evaluate Parametric Equations at t=-1: $$x(-1)=-3$$ $$y(-1)=2$$ Write Tangent Line: $$y-2=\frac{3}{2}(x+3)$$ As desired.