To find dy dx in terms of t, one must first appreciate that many curves are most naturally described by a pair of parametric equations where both x and y are expressed as functions of a third variable, usually denoted t. In such a representation, the independent variable t parameterizes the curve, and the slope of the tangent line at any point can be obtained by differentiating y with respect to x. This process involves computing the derivatives of x and y with respect to t and then combining them using the chain rule, yielding the expression dy/dx as a function of t.
Understanding Parametric Equations
A parametric equation system consists of two (or more) equations that define the coordinates of a point on a curve in terms of a parameter. For a plane curve, the typical form is
[ x = f(t), \qquad y = g(t), ]
where t belongs to some interval. The parameter t can represent time, an angle, or any other quantity that varies along the curve. Because both x and y depend on t, the usual derivative dy/dx is not directly available
Computing (\displaystyle \frac{dy}{dx}) from Parametric Forms
Once the parametric description (x = f(t)) and (y = g(t)) is established, the slope of the tangent line at a given parameter value (t) follows directly from the chain rule. Because (x) and (y) are both functions of (t),
[ \frac{dy}{dx}= \frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}, \qquad\text{provided } \frac{dx}{dt}\neq 0 . ]
If (\displaystyle \frac{dx}{dt}=0) while (\displaystyle \frac{dy}{dt}\neq 0), the tangent is vertical and the derivative is undefined (or “infinite”). Conversely, when both numerator and denominator vanish, further analysis (often using limits or higher‑order derivatives) is required to determine the curve’s behavior at that point That alone is useful..
Example 1 – Simple polynomial curve
Consider the parametric curve
[ x = t^{2}, \qquad y = t^{3}, \qquad t\in\mathbb{R}. ]
Differentiating with respect to (t),
[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^{2}. ]
Hence
[ \frac{dy}{dx}= \frac{3t^{2}}{2t}= \frac{3}{2}t \quad (t\neq 0). ]
At (t=0) we have (dx/dt=0) and (dy/dt=0); the curve passes through the origin with a horizontal tangent after a closer inspection of higher‑order terms.
Example 2 – Trigonometric curve
Let
[ x = \cos t, \qquad y = \sin t, \qquad 0\le t<2\pi . ]
Then
[ \frac{dx}{dt}= -\sin t, \qquad \frac{dy}{dt}= \cos t, ]
so
[ \frac{dy}{dx}= \frac{\cos t}{-\sin t}= -\cot t . ]
This reproduces the familiar slope of the unit circle at each point, confirming that the parametric method works for curves that are not functions (y=f(x)) in the usual sense.
Example 3 – A curve with a vertical tangent
Take
[ x = t^{3}, \qquad y = t^{2}, \qquad t\in\mathbb{R}. ]
Here
[ \frac{dx}{dt}=3t^{2}, \qquad \frac{dy}{dt}=2t. ]
Thus
[ \frac{dy}{dx}= \frac{2t}{3t^{2}} = \frac{2}{3t}, \qquad t\neq 0. ]
At (t=0) we have (dx/dt=0) while (dy/dt=0) as well, so the standard formula fails. A limit as (t\to0) shows that the curve approaches the origin with a vertical tangent, a fact that can be verified by eliminating the parameter ((y^{2}=x^{2/3})).
Practical Tips
- Check the denominator. Before applying the formula, verify that (\displaystyle \frac{dx}{dt}\neq0). If it is zero, examine the numerator to determine whether the tangent is vertical or the point is singular.
- Simplify the expression. Often the ratio (\frac{dy/dt}{dx/dt}) can be simplified using algebraic or trigonometric identities, making further analysis easier.
- Use limits for indeterminate forms. When both derivatives vanish, compute (\displaystyle \lim_{t\to t_{0}}\frac{dy/dt}{dx/dt}) or apply L’Hôpital’s rule to the original parametric functions.
- Interpret geometrically. The sign of (\displaystyle \frac{dy}{dx}) tells you whether the curve is rising ((>0)) or falling (\
