Say that we have the parametrized curve $x=e^{3t}, y=te^{-t}$. What would be the slope of this at the point $(1,0)$ and also on which points on the curve would the curve be horizontal?
What I have done:
For the first question, I used $\frac{dy/dt}{dx/dt}$ and found the slope to be 1/3. Is that right? I don’t know how to do the second part though. Can someone help me?
You are correct. The slope $\frac{dy}{dx}$ of a parametric function $\left(x(t), y(t)\right)$ is $\frac{dy/dt}{dx/dt}$.
$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{d}{dt}te^{-t}}{\frac{d}{dt}e^{3t}} = \frac{e^{-t}(1-t)}{3e^{3t}} = \frac{1-t}{3e^{4t}} $$
To find the slope at $(1,0)$ you would need to find the value of $t$ that satisfies $\left(e^{3t}, te^{-t}\right) = (1,0)$, and then evaluate $\frac{dy/dt}{dx/dt}$ at that $t$.
$$e^{3t} = 1 \text{ and } te^{-t} = 0$$
Clearly there is only one solution, $t=0$.
So then $\frac{dy/dt}{dx/dt}|_{(0)}$ becomes
$$\frac{1-t}{3e^{4t}}|_{(0)} = \frac{1-0}{3e^{0}} = \frac{1}{3}$$
For the second part, the curve is horizontal when $\frac{dy}{dx} = \frac{1-t}{3e^{4t}} = 0$, or when $1-t = 0$.
The curve is horizontal when $t=1$, which is at the point $\left(e^3,e^{-1}\right)$.