Finding whether a parametric curve has a well defined tangent at the origin

Question

Does the curve $$ x(t) = 5t^3, y(t) = 4t - sin(4t), t \in \Bbb{R}$$ Have a well defined tangent line at the origin?

Since $\frac{dy}{dx} = \frac{0}{0} $ When t = 0 (i.e at the origin) my guess is that the curve does not have a well defined tangent at the origin, however this is not correct apparently.

Answer 1

use that $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Answer 2

$\frac{dy}{dx}=\frac{y'}{x'}=\frac{0}{0}$, You can actually compute the limit of the tangent lines: $\begin{align} \lim_{t \to 0}\frac{y'(t)}{x'(t)}=\\ =\lim_{t \to 0} \frac{4-4\cos(4t)}{30t^2}=\\ =\lim_{t \to 0} \frac{16}{15}\frac{4\left(1-\cos(4t)\right)}{\left(4t\right)^2}=\\ =\lim_{t \to 0}\frac{\sin(4t)}{30t}=\\ =\frac{64}{30}=\frac{32}{15} \end{align}$

If you are not comfortable with The L'Hospital's theorem, consider directly

$\begin{align} \lim_{x \to 0}\frac{1-\cos(x)}{x^2}=\\ =\lim_{x \to 0}\frac{2\sin^2\left(\frac{x}{2}\right)}{x^2}= \\=\lim_{\frac{x}{2} \to 0}\frac{2\sin\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}=\\ =\frac{1}{2} \end{align}$

And apply it to the third passage of the computation above with the substitution $4t=z$

Actually, an equivalent (even if a bit lengthy) computation might be carried out with $\begin{align}\lim_{x \to 0}y'(x)=\\ =\lim_{x \to 0}\frac{d}{dx}\left(4\left(\frac{x}{5}\right)^{\frac{1}{3}}-4\sin\left(\left(\frac{x}{5}\right)^{\frac{1}{3}}\right)\right)=\\ =\lim_{h \to 0}\frac{\left(4\left(\frac{h}{5}\right)^{\frac{1}{3}}-4\sin\left(\left(\frac{h}{5}\right)^{\frac{1}{3}}\right)\right)}{h}=\\ =\frac{32}{15} \end{align}$,

(the term $y(0)$ vanishes since it equals 0, and the result is obtained applying The L'Hospital theorem or, equivalently, substituting back $5t^3=x$)

Answer 3

The curve has a very well defined tangent.

EDIT1

(After hasty post!)

By L'Hospital's Rule applied twice due to $\frac{0}{0}$ persisting after first application the Rule

$$ \frac{dy}{dx}=\frac{\dot y}{\dot x}= \frac{4-4 \cos \,t}{15\, t^2} =\frac{4}{15} \frac{1-\cos 4t}{t^2}=\frac{4}{15} \frac{4 \sin 4t}{2t} \ = \frac{4}{15} \cdot 2\cdot 4 =\frac{32}{15} $$

is the slope of tangent at origin with a second order contact.

Answer 4

Using the equivalence $1-\cos{t} \sim \frac{t^2}{2}, \ t\to{0},$ we have $$y'(0) = \lim\limits_{t\to{0}}\dfrac{y'(t)}{x'(t)} = \lim\limits_{t\to{0}}\dfrac{4-4 \cos{4t}}{15t^2} = \lim\limits_{t\to{0}}\dfrac{32t^2}{15t^2} = \dfrac{32}{15}.$$

Answer 5

$\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} = \lim_\limits{t\to 0}\frac {4(1-\cos 4t)}{15t^2} = \frac {64}{30}$

$t = \left(\frac {x}{5}\right)^{\frac 13}\\ y = 4\left(\frac {x}{5}\right)^{\frac 13} - \sin (4\left(\frac {x}{5}\right)^{\frac 13})\\ \frac {dy}{dx} = \lim_\limits{h\to 0} \frac{y(h)-y(0)}{h}$

Which is defined and indeed equals $\frac {32}{15}$