Why a Parametric Curve Has a Slope Formula
A plane curve can be described by giving its horizontal and vertical coordinates as functions of a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). The parameter determines a point \((x(t),y(t))\) on the curve. It need not be the horizontal coordinate itself.
When \(x\) changes as \(t\) changes, the curve’s slope measures how quickly \(y\) changes relative to \(x\), rather than relative to \(t\). The chain rule makes that comparison precise. It leads to the parametric slope formula, provided the horizontal coordinate is changing at a nonzero rate.
The denominator condition matters: division by \(dx/dt\) is valid only when \(dx/dt\ne 0\). The formula describes a slope with respect to \(x\), so it requires \(x\) to change locally as the parameter changes.
Deriving the Formula from the Chain Rule
Suppose that near \(t_0\), the curve can be written as a graph \(y=Y(x)\), while \(x=x(t)\). Then the parametrized vertical coordinate is the composite function \(y(t)=Y(x(t))\). By the chain rule,
If \(dx/dt\ne 0\), divide both sides by \(dx/dt\). The result is \(dY/dx=(dy/dt)/(dx/dt)\), which is the parametric slope formula.
Why may the curve be treated locally as a graph when \(x'(t_0)\ne 0\)? Because the derivative \(x'\) is continuous near \(t_0\) for the usual smooth parametrizations considered in calculus. If \(x'(t_0)\) is nonzero, it keeps the same sign in a sufficiently small interval around \(t_0\). The mean value theorem then implies that \(x(t)\) is strictly increasing or strictly decreasing on that interval. Thus nearby parameter values give distinct horizontal coordinates, and \(y\) can be regarded locally as a function of \(x\).
The quotient also follows directly from secant slopes. For a small nonzero change \(h\) in the parameter, the secant slope between the curve points at \(t_0\) and \(t_0+h\) is \(\Delta y/\Delta x\), as long as \(\Delta x\ne 0\). Since \(x'(t_0)\ne 0\), for all sufficiently small nonzero \(h\), \(\Delta x\ne 0\). Dividing the increments by \(h\) gives
As \(h\) tends to zero, the numerator tends to \(y'(t_0)\) and the denominator tends to \(x'(t_0)\). Because the latter limit is nonzero, the quotient tends to \(y'(t_0)/x'(t_0)\). The tangent slope is the limit of these secant slopes, so it is exactly the value given by the formula.
Worked Examples
Worked Example: A Polynomial Parametrization
Let \(x(t)=t^2+1\) and \(y(t)=2t-1\). Find the curve’s slope at \(t=2\), then check the result using a local Cartesian description.
First calculate the parameter derivatives:
At \(t=2\), the point is \((x(2),y(2))=(5,3)\), and
Since \(dx/dt=4\ne 0\), the formula applies:
To check, near \(t=2\) we have \(t>0\), so \(x=t^2+1\) gives \(t=\sqrt{x-1}\). Therefore the local graph is \(y=2\sqrt{x-1}-1\), whose derivative is \(1/\sqrt{x-1}\). At \(x=5\), this derivative is \(1/\sqrt{4}=1/2\), agreeing with the parametric calculation.
Worked Example: A Circular Parametrization
Let \(x(t)=\cos t\) and \(y(t)=\sin t\). Find the slope at \(t=\pi/4\), then check it from the local equation of the circle.
Differentiate each coordinate with respect to \(t\):
At \(t=\pi/4\), the point is \((\sqrt{2}/2,\sqrt{2}/2)\), and
The horizontal-coordinate derivative is nonzero. Hence
For the check, the parametrization satisfies \(x^2+y^2=1\). Near this point \(y>0\), so the local graph is \(y=\sqrt{1-x^2}\). Differentiating gives \(dy/dx=-x/\sqrt{1-x^2}\). At \(x=\sqrt{2}/2\), this is \(-(\sqrt{2}/2)/(\sqrt{2}/2)=-1\), as required.
Worked Example: Checking with Secant Slopes
Let \(x(t)=t^3-3t\) and \(y(t)=t^2+1\). Find the slope at \(t=2\), and verify it by taking a limit of secant slopes.
Differentiate and evaluate:
Since \(9\ne0\), the slope is
For an independent check, replace \(t\) with \(2+h\). Direct expansion gives
Thus, for nonzero \(h\) close to zero, the secant slope from the point at \(t=2\) is
Taking \(h\) to zero gives \(4/9\), which verifies the calculated tangent slope.
When the Formula Does Not Apply
The parametric slope formula is a quotient, not a rule that can be used without checking its denominator. If \(dx/dt=0\) at the parameter value, the expression \((dy/dt)/(dx/dt)\) is undefined there. This does not by itself determine the curve’s tangent behavior; it means only that this particular formula cannot be used at that value. A zero denominator must not be replaced by an arbitrary number or treated as an ordinary finite slope.
Also keep the order of the quotient straight. The numerator is the rate of change of the vertical coordinate, and the denominator is the rate of change of the horizontal coordinate. Reversing them gives a different quantity, not the slope \(dy/dx\).
Worked Example: Avoiding a Reversed Quotient
Let \(x(t)=2t+1\) and \(y(t)=3t^2\). Find the slope at \(t=1\), and identify a common incorrect calculation.
Differentiate the coordinates:
At \(t=1\), \(dx/dt=2\) and \(dy/dt=6\). The correct quotient is therefore
A common wrong calculation reverses the quotient and gives \(2/6=1/3\). To check the correct value, solve \(x=2t+1\) for \(t=(x-1)/2\). Then \(y=3(x-1)^2/4\), so \(dy/dx=3(x-1)/2\). At \(t=1\), \(x=3\), and this derivative is \(3(3-1)/2=3\). The reversed quotient does not agree.
Using the Formula Reliably
For a parametrization \(x=x(t)\), \(y=y(t)\), first find both derivatives with respect to the same parameter. Evaluate them at the parameter value of interest, check that \(dx/dt\ne0\), and only then form the quotient. The point itself is found by evaluating \(x(t)\) and \(y(t)\) at that parameter value.
Find \(dx/dt\) and \(dy/dt\) as functions of \(t\).
Compute the point and both coordinate rates at that same value.
Confirm that \(dx/dt\ne0\) before dividing.
The value \((dy/dt)/(dx/dt)\) is the tangent slope with respect to \(x\).
The essential idea is that a parametrization gives two rates measured against \(t\). Their ratio removes the parameter’s rate of progress and compares the vertical change directly with the horizontal change. The chain rule establishes the formula, while the limiting secant-slope argument explains why that quotient is the tangent slope.
Check Your Understanding
Use the derivative quotient carefully, and state the condition that allows division in each calculation.
- For \(x(t)=t+2\) and \(y(t)=t^2\), find the slope at \(t=1\).
- In one sentence, explain how the chain rule gives the parametric slope formula.
- For \(x(t)=\sin t\) and \(y(t)=\cos t\), write the formula for the slope at a parameter value \(t\), where \(\cos t\ne0\).
- What does \(dx/dt=0\) tell you about whether the quotient formula can be used at that parameter value?
- Why does the limit of \(\Delta y/\Delta x\) give \(y'(t_0)/x'(t_0)\) when \(x'(t_0)\ne0\)?
