Find y'_x for the parametric curve x = cos t, y = sin t
Find $y'_{x}$ for the parametric curve $x = \cos t$, $y = \sin t$.
Explanation of Method
The unit circle, traced clockwise from the top.
Step 1: Differentiate with respect to the parameter
\[ x'_{t} = -\sin t, \qquad y'_{t} = \cos t. \]
Step 2: Apply the parametric derivative formula
The chain rule gives $y'_{t} = y'_{x}\,x'_{t}$, so dividing (where $x'_{t} \neq 0$): \[y'_{x} = \frac{y'_{t}}{x'_{t}} = \frac{\cos t}{-\sin t} = -\cot t.\]
Step 3: Express in terms of $x$ and $y$
Since $x = \cos t$ and $y = \sin t$: $y'_{x} = -\frac{\cos t}{\sin t} = -\frac{x}{y}$.
Step 4: Verification
The curve is the unit circle $x^{2}+y^{2} = 1$. Differentiating that relation implicitly: $2x + 2y\,y' = 0 \;\Longrightarrow\; y' = -\frac{x}{y} \;\;\checkmark$.
\[t = \frac{\pi}{4}:\quad \bigl(x,y\bigr) = \left(\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2}\right),\quad y'_{x} = -\cot\tfrac{\pi}{4} = -1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'_{x} = -\cot t = -\dfrac{x}{y}, \qquad \sin t \neq 0\,}\]
Remark
Remark. The tangent at $(x,y)$ has slope $-x/y$, which is the negative reciprocal of the radius slope $y/x$: the tangent to a circle is perpendicular to the radius, recovered here by pure computation. The formula breaks down at $t = 0,\pi$, where the tangent is vertical — these are exactly the points where $x'_{t} = 0$. Compare Problem 154, which is the same circle with the roles of the coordinates swapped.