Calculating tangent line at $t=2$

Question

Calculate the tangent line at $t=2$ of the curve $$x(t)=t^2+2t+4,\ y(t)=1+te^t$$

How would you go about determining this? So far I have

$$x'(t)=2t + 2\\ y'(t) = te^t+e^t$$

Where do I go from here to get the tangent line?

Answer 1

The following expression is how you can find the slope of a parametric curve:

$$ \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)},\ \frac{dx}{dt}\ne0. $$

When you have found the slope of the curve at $t=2$, use the equation of a straight line to form the actual line:

$$ y-y_0=m(x-x_0). $$

Use your original parametric equation to find the point $(x_0,y_0)$ at $t=2$.

Answer 2

Hint:

A directing vector of the tangent to the curve at the point $M_2$ with coordinates$(x(2),y(2))$ is the vector $$\vec v=(x'(2),y'(2)),$$ hence a point $M(x,y)$ is on this tangent line if and only if the vectors $\overrightarrow{M_2M}$ and $\vec v$ are collinear, i.e. the determinant of their coordinates is $0$.