Find the slope of the tangent line to the curve given parametrically

Question

The parametric curves I'm working with are:

$ x=t^2-t$ and $y=\sqrt{t+2}$

I need to find the slope of the tangent at t = 2

I found the derivative wrt t for both:

$ dx/dt = 2t-1$, $dy/dt={1\over 2\sqrt{t+2}}$

The answer I got is:

$y={\sqrt{3}\over 2}x +2-\sqrt{3}$

Can someone please verify if my answer is right?

Answer 1

Your answer is not correct. Your derivatives wrt $t$ are correct, so $\frac{dy}{dx} |_{t=2} = \frac{1/(2 \sqrt{2+2})}{2(2) - 1} = \frac{1}{12}$. Can you continue?

Answer 2

Let $\gamma(t) = (x(t),y(t)) = (t^{2} - t,\sqrt{t+2})$. Then the tangent vector at $t = 2$ is given by: \begin{align*} \gamma'(t) = \left(2t - 1,\frac{1}{2\sqrt{t+2}}\right) \Rightarrow \gamma'(2) = \left(3,\frac{1}{4}\right) \end{align*}

Consequently, the tangent line at $(2,2)$ can be described as: \begin{align*} s(t) & = (2,2) + t\gamma'(2)\\\\ & = (2,2) + t\left(3,\frac{1}{4}\right)\\\\ & = \left(2 + 3t, 2 + \frac{t}{4}\right) \end{align*}

Eliminating the parameter $t$, we get \begin{align*} \frac{x-2}{3} = 4(y-2) & \Longleftrightarrow 12y - x - 4 = 0\\\\ & \Longleftrightarrow y = \frac{x}{12} - \frac{1}{3} \end{align*}

Therefore the slope of the tangent line is given by $1/12$.

Hopefully this helps!