Find slope of parametric equation
I am looking for the slope of the parametric equation:
$$x=e^{3t}-2e^{-3t}$$ $$y=e^{3t}+5e^{-3t}$$
Find slope when $t=2$.
As of now I have figured the slope of $\frac{dy}{dx}$ to be:
$$\frac{e^{12}-5}{e^{12}-2}$$
at $t=2$ however I think I went about this problem the wrong way. To explain how I came to that slope is I calculated the derivative of y over the derivative of x and simplified.
Could someone tell me if I went about this the wrong way?
EDIT: Also I don't know if this is relevant to the problem, but I found a point at:
($e^6-2e^{-6}$ , $e^6-5e^{-6}$) at t=2.
$\endgroup$ 21 Answer
$\begingroup$Use $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Now, $\displaystyle y=e^{3t}-2e^{-3t}\implies \frac{dy}{dt}=3e^{3t}-2\cdot(-3e^{-3t})=3(e^{3t}+2e^{-3t})$
Similarly, find $\displaystyle \frac{dx}{dt}$
Finally divide and put $t=2$
$\endgroup$ 4
