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Let a curve be given in the parametrized form by:

$r(t) = (2\cos t, 2\sin t), 0 \leq t \leq 2\pi$

Find the equations of the tangents to the curve at each of its points $(X_0, Y_0)$.

Having gone through some text, it never really directly approaches a problem such as this. I have read through some articles on this website, and it provides varying solutions to this type of question and as such I can't tell what method is correct.

From the methods I have used for tangent vectors at a certain point:

$x = X_0 - 2s\sin t$ is the tangent equation for $X_0$ where $X_0 = 2\cos t$ for some $t$

$y = Y_0 + 2s\cos t$, for $Y_0$ where $Y_0= 2\sin t$ for the same $t$

If anyone could clarify if I am approaching this the right way or provide the correct method but not the answer then that would be appreciated.

mwtmurphy
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    You appear to be using the letter $t$for two different purposes, i.e. For parametrizing the curve and parametrizing the line. You need two different letters, so that for example $x=X_0-2\lambda\sin t$ etc... – David Quinn Jul 28 '15 at 22:24
  • Ah yes I notice that now, thanks for correcting me. – mwtmurphy Jul 28 '15 at 23:38

1 Answers1

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The vector that locates the loci of points to the circle is given parametrically by

$$\vec r(t)=\hat x 2\cos t+\hat y2\sin t$$

At point $t_0$, we denote the vector $\vec r_0=\vec r(t_0)$. The unit tangent $\hat u(t_0)$ to the curve at $t_0$ is

$$\begin{align} \hat u(t_0)&=\left.\frac{\frac{d\vec r(t)}{dt}}{\left|\frac{d\vec r(t)}{dt}\right|}\right|_{t=t_0}\\\\ &=-\hat x\sin t_0+\hat y\cos t_0 \end{align}$$

The parametric equation of a line tangent to the circle at $\vec r_0$ is

$$\begin{align} \vec R(s)&=\vec r_0+s\vec u_0\\\\ &=\hat x (2\cos t_0-s\sin t_0)+\hat y(2\sin t_0+s\cos t_0) \end{align}$$

$$\bbox[5px,border:2px solid #C0A000]{X(s)=2\cos (t_0)-\sin (t_0)\,s}$$

$$\bbox[5px,border:2px solid #C0A000]{Y(s)=2\sin (t_0)+\cos (t_0)\,s}$$

Eliminating $s$ gives the equation of the line as

$$\bbox[5px,border:2px solid #C0A000]{y_0Y=4-x_0X}$$

Mark Viola
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