The parametric equations of a curve are given by:
$$p(t) = (x(t),y(t)) = (5+\cos(3t), 4e^{3t})$$
Find the tangent vector when $t = 4$ .
My question is to solve this do i just take the derivative of $5+\cos(3t)$ and $4e^{3t}$ and then just plug in $4$ ?
$\endgroup$ 21 Answer
$\begingroup$The tangent line at $t = 0$ (convenience) has parametric equation: $T(t) = p(0)+ tp'(0)$. You have $p(0) = (6,4), p'(0) = (0,12)\implies T(t) = (6,4)+t(0,12) = (6,4+12t)$, with $t \in \mathbb{R}$ . I hope this helps. I mistakenly did the case $t = 0$ since it is popular, but you can see how I did it and adjust it to your case $t = 4$. For the tangent vector only at say $t = 4$, it is just $v(4) = p'(4) = (x'(4),y'(4))= (-3\sin(12), 12e^{12})$ .
$\endgroup$
