Problem I'm trying to solve:
$x=e^{2t}-t$, $y=2e^{3t}-2t$ are parametric equations of a curve. Find the value(s) of $t$ at the point(s) of the curve where the normal has a gradient of $-\frac{1}{2}$.
My Attempt:
My thoughts were just to apply the definition of derivative so
$$\begin{align} \lim_{h\to0}\frac{\bigg(2e^{3(t+h)}-2(t+h)\bigg)-\bigg(2e^{3t}-2t\bigg)}{\bigg(e^{2(t+h)}-(t+h) \bigg)-\bigg(e^{2t}-t \bigg)} &=\lim_{h\to0}\frac{\bigg(2e^{3t+3h}-2t-2h\bigg)-2e^{3t}+2t}{\bigg(e^{2t+2h}-t-h \bigg)-e^{2t}+t} \\ &=\lim_{h\to0}\frac{2e^{3t+3h}-2h-2e^{3t}}{e^{2t+2h}-h -e^{2t}} \\ &=2 \cdot\lim_{h\to0}\frac{e^{3t+3h}-h-e^{3t}}{e^{2t+2h}-h -e^{2t}} \end{align}$$
Then applying l'Hôpital's rule we have
$$\begin{align}2 \cdot\lim_{h\to0}\frac{e^{3t+3h}-h-e^{3t}}{e^{2t+2h}-h -e^{2t}} \end{align}=2\cdot \lim_{h\to 0}\frac{3e^{3t+3h}-1}{2e^{2t+2h}-1}=2\cdot \frac{3e^{3t}-1}{2e^{2t}-1}$$
And this derivative is equal to the negative reciprocal of the normal so we have:
$$\begin{align} \Leftrightarrow \;& 2=2\cdot \frac{3e^{3t}-1}{2e^{2t}-1} \\ \Leftrightarrow \;& 1=\frac{3e^{3t}-1}{2e^{2t}-1} \\ \Leftrightarrow \; &2e^{2t}-1= 3e^{3t}-1 \\ \Leftrightarrow \; &0=3e^{3t}-2e^{2t}\end{align}$$
By letting $a=e^t$ we end up with a cubic equation;
$$\begin{align} & 0=3e^{3t}-2e^{2t} \\ \Leftrightarrow \; &0=3a^3-2a^2 \\ \Leftrightarrow \; &0=(a^2)(3a-2) \end{align}$$
So $a=0,\frac{2}{3}$ and we're left with $t=\ln(\frac{2}{3})$.
Is this correct? I'm mainly just making sure since I've never done 'tangent/normal' questions with parametric equations.
$$\dfrac{dy}{dt}=6e^{3t}-2$$ $$\dfrac{dx}{dt}=2e^{2t}-1$$ So, the slope of the curve is $$m=\dfrac{dy}{dx}=\dfrac{6e^{3t}-2}{2e^{2t}-1}$$ Hence, the slope of the normal is $$-\dfrac{1}{m}=\dfrac{1-2e^{2t}}{6e^{3t}-2}$$ Given, $$\dfrac{1-2e^{2t}}{6e^{3t}-2}=\dfrac{-1}{2}$$ $$2-4e^{2t}=2-6e^{3t}$$ $$4e^{2t}=6e^{3t}$$ $$e^{t}=\dfrac{2}{3}$$ $$t=\ln{\dfrac{2}{3}}$$ Yes, your answer is correct.