So I have a question that goes like this:

The parametric equations of a curve are $$x = \frac{1}{\cos^3t}, y = \tan^3t$$ where $0 \le t \lt \frac{1}{2}\pi$.

(i) Show that $\frac{dy}{dx} = \sin t $

(ii) Hence show that the equation of the tangent to the curve at the point with parameter t is $y = x\sin t - \tan t$.

(i) is easy enough, but I have no idea where to even start with (ii), considering I don't know what the value of $t$ at this point is.


2 Answers 2


Since you think (i) is easy enough, you should know what does the result in (i) means. It actually tells you the slope in (ii), that is to say, the slope of the tangent line to the curve is actually $\dfrac{dy}{dx}$, which equals to $\sin t$. Then for a line going through the point $(x(t),y(t))$ with slope $\sin t$, we can write the line equation as $$ \frac{y-y(t)}{x-x(t)}=\sin t $$ Thus $$ y-\tan ^3t=\sin t\left( x-\frac{1}{\cos ^3t} \right) \\ y=x\sin t+\tan ^3t-\frac{\sin t}{\cos ^3t} = x \sin t-\tan t $$

As desired.

Notice that we are calculating for a general $t$, so we don't need to know what the value of $t$. We calculate for general points $(x(t),y(t))$.

  • $\begingroup$ Ah ok, I just never had to solve for a general $t$ before. Thanks a lot for the explanation! $\endgroup$ Dec 19, 2020 at 3:34
  • $\begingroup$ My pleasure. :D $\endgroup$
    – FFjet
    Dec 19, 2020 at 3:34

Use $\frac{y-y(t)}{x-x(t)}=\frac{dy}{dx}\Big|_t$. You don't need to know the value of $t$, we can calculate the tangent for a general $t$.


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