What a Parameter Changes
A parameter selects a member of a family of curves. In this tutorial, \(a\) is a real number that is held fixed while \(x\) and \(y\) vary along a chosen curve. Differentiating with respect to \(x\) therefore does not differentiate \(a\).
This distinction matters: the parameter changes the shape of the curve, but it is not another variable tracing a direction along that curve. We will use the family
to see how the parameter affects slopes and curvature. The resulting formulas have conditions: a formula for \(dy/dx\) requires a local graph \(y=y(x)\), and divisions by \(a\) require \(a\ne 0\). Geometric curvature, in contrast, can still be defined at points where the tangent is vertical.
Local Graphs and Curvature
Write a curve as \(F(x,y)=0\), where \(F\) has continuous second partial derivatives in a neighborhood of the point under consideration. If \(F_y\ne 0\) at that point, the implicit function theorem gives a local \(C^2\) graph \(y=y(x)\). The \(C^2\) condition is important: it ensures that the second derivative used to compute curvature exists and is continuous locally.
For a \(C^2\) graph \(y=y(x)\), its signed curvature with the usual graph orientation is \(y''/(1+(y')^2)^{3/2}\). Its geometric, unsigned curvature is therefore
The absolute value makes geometric curvature nonnegative. This graph formula is not directly usable when the curve has a vertical tangent and \(y'\) is not finite; an implicit formula will handle that case as well.
Differentiate the Family
For \(F(x,y)=x^2+ay^2-1\), with \(a\) fixed, the partial derivatives are \(F_x=2x\) and \(F_y=2ay\). At a point where \(ay\ne0\), we have \(F_y\ne0\), so the curve is locally a \(C^2\) graph \(y=y(x)\). Implicit differentiation gives
The slope formula requires \(a\ne0\) and \(y\ne0\). Differentiate the first-derivative equation once more, remembering that \(a\) remains constant:
Here and below, \(y'\) denotes \(dy/dx\). Substituting \(y'=-x/(ay)\), then using \(x^2+ay^2=1\), simplifies the second derivative:
These formulas describe a graph only where \(ay\ne0\). At points with \(y=0\) and \(a\ne0\), the curve may still be regular, but it cannot be represented there as a differentiable graph \(y(x)\). We will use curvature in a form that does not require that graph orientation.
Why the Implicit Curvature Formula Works
For a \(C^2\) curve \(F(x,y)=0\) at a regular point, its geometric curvature can be written in terms of the partial derivatives:
To derive this when \(F_y\ne0\), use \(y=y(x)\). Differentiating \(F(x,y(x))=0\) once gives \(y'=-F_x/F_y\). Differentiating again gives
where the second partial derivatives are evaluated along the graph. Substituting \(y'=-F_x/F_y\) into this equation yields
Also, \(1+(y')^2=(F_x^2+F_y^2)/F_y^2\). Substitution into \(\kappa=|y''|/(1+(y')^2)^{3/2}\) cancels the factors of \(|F_y|^3\) and proves the stated formula. If \(F_y=0\) at a regular point, then \(F_x\ne0\); using a local graph \(x=x(y)\) gives the same formula. Thus it also applies at vertical tangents, provided the stated \(C^2\) and regularity conditions hold.
For the parameter family, \(F_{xx}=2\), \(F_{xy}=0\), and \(F_{yy}=2a\). Therefore
The final equality uses \(x^2+ay^2=1\). This curvature formula requires a regular point; for this family, the gradient is \((2x,2ay)\). The formula is useful even when the slope formula is not, as long as the point is regular.
Worked Examples
Worked Example: A Slope and Curvature on an Ellipse
Let \(a=4\), and consider the point \((\sqrt{3}/2,1/4)\). First check that it is on the curve:
Because \(ay=4(1/4)=1\ne0\), the point lies on a local graph \(y(x)\). Its slope is
The second derivative is
For the curvature, \(x^2+a^2y^2=3/4+16(1/16)=7/4\), so
Check using the graph formula: \(1+(y')^2=1+3/4=7/4\), hence \(|y''|/(1+(y')^2)^{3/2}=4/(7/4)^{3/2}=32/(7\sqrt7)\), in agreement.
Worked Example: Curvature at a Vertical Tangent
Set \(a=4\) and take the point \((1,0)\). It lies on the curve because \(1^2+4(0)^2=1\). The gradient there is \((2,0)\ne(0,0)\), so the point is regular. Since \(F_y=2ay=0\), the slope formula \(-x/(ay)\) does not apply. The implicit curvature formula does apply:
To check the vertical-tangent interpretation, solve locally for \(x\) as a function of \(y\): \(x=\sqrt{1-4y^2}\) near \(y=0\). Then \(dx/dy=-4y/\sqrt{1-4y^2}\), which is \(0\) at \(y=0\). The tangent in the \((x,y)\)-plane is therefore vertical. Using the curvature formula for this graph orientation gives curvature \(4\) at \(y=0\), matching the implicit calculation.
Worked Example: A Negative Parameter
Let \(a=-2\), and choose \((x,y)=(\sqrt3,1)\). The curve condition holds because \(3-2=1\). Since \(ay=-2\ne0\), this point has a local graph \(y(x)\). Its slope is
Its second derivative is
For curvature, \(x^2+a^2y^2=3+4=7\), so
Check from the graph values: \(1+(y')^2=1+3/4=7/4\), and \(|y''|/(1+(y')^2)^{3/2}=(1/4)/(7/4)^{3/2}=2/(7\sqrt7)\). Both the curve equation and curvature calculation are consistent.
Special Parameter Values and Curve Types
The parameter controls what real curve the equation describes. Its cases should be considered separately because the slope formula divides by \(a\).
| Parameter | Curve | Geometric consequence |
|---|---|---|
| \(a>0\) | An ellipse, with semiaxes \(1\) and \(1/\sqrt a\). | When \(a=1\), it is the unit circle and its curvature is constant, equal to \(1\). |
| \(a=0\) | The two vertical lines \(x=1\) and \(x=-1\). | The slope formula divided by \(a\), so it does not apply. Each line has curvature \(0\). |
| \(a<0\) | A hyperbola. | The slope and curvature formulas apply wherever their stated conditions hold; the curvature varies with the point. |
For \(a=1\), the curve equation gives \(x^2+y^2=1\). The curvature formula becomes \(\kappa=1/(x^2+y^2)^{3/2}=1\) at every point. Thus the unit circle is the special member of this family with constant curvature. For \(a=0\), the original equation is \(x^2=1\), not a formula for \(y\) as a function of \(x\). The two components are straight lines, so their curvature is zero; this conclusion does not come from substituting \(a=0\) into the slope formula.
For \(a\ne0\), the family is regular at every point on the curve. Indeed, a singular point would require \(F_x=2x=0\) and \(F_y=2ay=0\), so \(x=y=0\), which cannot satisfy \(x^2+ay^2=1\). When \(a=0\), points on the two lines are also regular, since \(F_x=2x=\pm2\). Regularity alone is not enough for the curvature theorem: the \(C^2\) condition on \(F\) must also hold. Here \(F=x^2+ay^2-1\) is a polynomial, so it is \(C^2\) for every fixed real \(a\).
A Common Parameter Mistake
A frequent error is to omit the parameter when differentiating the term \(ay^2\). For example, at \(a=4\) one might incorrectly write
This treats \(ay^2\) as though its derivative were \(2yy'\), dropping the factor \(a\). At the point \((\sqrt3/2,1/4)\), that incorrect formula gives \(y'=-2\sqrt3\). The correct derivative, with \(a=4\) fixed, is
Substitution checks the corrected result directly: \(2x+2ayy'=\sqrt3+2(4)(1/4)(-\sqrt3/2)=\sqrt3-\sqrt3=0\). The erroneous value does not make this differentiated equation equal zero.
Why These Formulas Matter
Keeping parameter-dependent formulas general makes it possible to compare an entire family of curves rather than solve each case from scratch. The slope formula identifies the direction of a local graph, the second derivative records how that graph bends, and the implicit curvature formula measures bending without requiring the curve to be a graph \(y(x)\). The parameter cases also show why checking conditions is part of the calculation: a formula valid for \(a\ne0\) cannot simply be used at \(a=0\), and a vertical tangent requires a different graph orientation or an implicit curvature formula.
Check Your Understanding
Use the hypotheses and formulas in this tutorial to answer the following questions.
- For fixed \(a\), differentiate \(x^2+ay^2=1\) and state the conditions under which the resulting formula for \(y'\) applies.
- At \(a=2\) and \((x,y)=(0,1/\sqrt2)\), find \(y'\), \(y''\), and the geometric curvature. Check that the point satisfies the curve equation.
- Why can the slope formula fail at \((1,0)\) when \(a>0\), while the implicit curvature formula remains usable?
- What curve does the equation describe when \(a=0\), and why is it invalid to use the formula \(-x/(ay)\) in that case?
- For which parameter value is the curve a circle, and what is its curvature at every point?
