Tangent line to the curve

Question

Write the equations off the tangent line and normal line to the curve: $$x = e^t\cos(t)$$ $$y = e^t\sin(t)$$ at the point A(1,0). Do I take $x(t_0) = 0$ and $y(t_0) = 1$? I solved it this way and I got the tangent line: $$(x-e\cos(1))/(e(\sin(1)-\cos(1))) = (y-0)/1$$ Is this correct or do I have to find $t_0$ in another way?

Answer 1

First you have to find the parameter $t$ for which $$c(t)=e^t\cdot\begin{pmatrix}\cos(t)\\\sin(t)\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}.$$ An easy calculation shows that $t=0$. As $$\dot c(t)=e^t\cdot\begin{pmatrix}\cos(t)-\sin(t)\\\sin(t)+\cos(t)\end{pmatrix}$$ the tangent vector in $t=0$ is $$\dot c(0)=e^0\cdot\begin{pmatrix}1+0\\0+1\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix}$$ the tangent's equation is $$\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\0\end{pmatrix}+s\begin{pmatrix}1\\1\end{pmatrix} $$ or $y=x-1$.

Answer 2

For your second question, define $$c(t)=\begin{pmatrix} \frac{t+2}{3t+4}\\\frac{4t+3}{2t+1} \end{pmatrix}.$$ Compute $$\dot c(t)=-2\begin{pmatrix} \frac{1}{(3t+4)^2}\\\frac{1}{(2t+1)^2} \end{pmatrix}.$$ The tangent line passes through the origin iff $\dot c(t)$ is parallel to $c(t)$. From here you'll arrive in solving $$(t+2)(3t+4)=(4t+3)(2t+1),$$ which has the solutions $t=\pm1$.