Why Concavity on an Implicit Curve Needs Care
An equation involving \(x\) and \(y\) may describe a curve without giving \(y\) explicitly as a formula in \(x\). Even so, near many points on the curve, it is possible to view \(y\) as a function of \(x\). At such a point, implicit differentiation can give both the slope \(y'\) and the second derivative \(y''\).
The second derivative measures how the slope changes as \(x\) changes. Its sign tells us whether the graph bends upward or downward locally, provided that the curve forms a sufficiently smooth function of \(x\) near the point. The essential steps are to find the first derivative, differentiate again, and only then substitute the coordinates and slope of the specified point.
Implicit Differentiation Along a Curve
Write the curve as \(F(x,y)=0\), where \(F\) is differentiable. If, near a point on the curve, \(y\) depends on \(x\), differentiating the equation with respect to \(x\) gives
Here \(F_x\) and \(F_y\) denote partial derivatives, and the factor \(y'\) appears because \(y\) changes with \(x\). If \(F_y\ne 0\), this equation can be solved for the slope:
The condition \(F_y\ne 0\) matters: it makes this particular equation solvable for \(y'\). It also ensures, under the usual smoothness assumptions, that the curve can locally be represented as a smooth function \(y(x)\). If \(F_y=0\), division by \(F_y\) is not valid, and this formula cannot be used at that point.
Why the Second-Derivative Formula Works
To find \(y''\), differentiate the first-derivative equation along the curve. For a twice continuously differentiable \(F\), differentiating \(F_x+F_y y'=0\) gives
The first two terms come from differentiating \(F_x\). For \(F_y y'\), the product rule gives \((F_y)'y'+F_y y''\), and the chain rule gives \((F_y)'=F_{yx}+F_{yy}y'\). Since the mixed partial derivatives agree when they are continuous, \(F_{xy}=F_{yx}\), so the equation becomes
Solving for \(y''\) is valid when \(F_y\ne 0\). This formula is not a new rule separate from implicit differentiation: it follows by applying the chain rule and product rule to the first-derivative equation.
A Reliable Calculation Sequence
In practical problems, it is often easiest to differentiate the original equation twice and keep the resulting equation in terms of \(x\), \(y\), \(y'\), and \(y''\). Then find \(y'\) at the point and substitute all known values. This avoids substituting the point too early, before the derivative equations have been obtained.
Substitute its coordinates into the original equation to confirm it lies on the curve.
Differentiate implicitly once, then use the point to calculate \(y'\).
Use the product and chain rules, keeping track of every factor of \(y'\).
Substitute the point and slope to solve for \(y''\), then state what its sign says locally.
Worked Examples
Worked Example: A Circle at a Point in the First Quadrant
Find \(y''\) and describe the local concavity of \(x^2+y^2=25\) at \((3,4)\).
Step 1: Check the point. Substitution gives \(3^2+4^2=9+16=25\), so \((3,4)\) lies on the circle.
Step 2: Find \(y'\). Differentiate both sides with respect to \(x\):
Divide by \(2\), then solve for \(y'\): \(x+y y'=0\), so \(y'=-x/y\). At \((3,4)\), \(y'=-3/4\).
Step 3: Find \(y''\). Differentiate \(x+y y'=0\). The product rule gives
Substitute \(y=4\) and \(y'=-3/4\):
Thus \(4y''=-25/16\), so \(y''=-25/64\). Check by substitution: \(1+9/16+4(-25/64)=25/16-25/16=0\). The second derivative is negative, so this local branch of the circle bends downward at \((3,4)\).
Worked Example: A Curve with Positive Second Derivative
Find the second derivative and local bending of \(xy+y^2=6\) at \((1,2)\).
Step 1: Check the point. \(1\cdot2+2^2=2+4=6\), so the point satisfies the equation.
Step 2: Find \(y'\). Differentiate \(xy+y^2=6\), using the product rule on \(xy\):
At \((1,2)\), this becomes \(2+y'+4y'=0\), or \(5y'=-2\). Therefore \(y'=-2/5\).
Step 3: Differentiate again. Differentiating each term of the first-derivative equation gives
Collect like terms to obtain \(2y'+2(y')^2+(x+2y)y''=0\). Substitute \(x=1\), \(y=2\), and \(y'=-2/5\):
Since \(-4/5+8/25=-20/25+8/25=-12/25\), we get \(5y''=12/25\), and hence \(y''=12/125\). Check: \(2(-2/5)+2(4/25)+5(12/125)=-20/25+8/25+12/25=0\). The second derivative is positive, so the curve bends upward locally at the point.
Worked Example: A Curve with Negative Second Derivative
Find \(y''\) and describe the local bending of \(x^2+xy+y^2=7\) at \((1,2)\).
Step 1: Check the point. \(1^2+(1)(2)+2^2=1+2+4=7\), as required.
Step 2: Find \(y'\). Differentiate each term:
Collect the terms containing \(y'\): \(2x+y+(x+2y)y'=0\). At \((1,2)\), this gives \(4+5y'=0\), so \(y'=-4/5\).
Step 3: Find \(y''\). Differentiate the uncollected equation. The derivative of \(xy'\) is \(y'+xy''\), and that of \(2yy'\) is \(2((y')^2+yy'')\). Thus
Substituting the point and slope gives
Therefore \(y''=-42/125\). Check: \(2-8/5+32/25+5(-42/125)=42/25-42/25=0\). The negative second derivative means the local branch bends downward at \((1,2)\).
How to Interpret the Sign Locally
A positive value of \(y''\) at a point means the slope is increasing as \(x\) passes through that point; a negative value means the slope is decreasing there. In the examples above, the positive value \(12/125\) corresponds to upward bending, while the negative values correspond to downward bending.
To assert concavity on an entire interval, the second derivative must have the corresponding sign throughout that interval. For a smooth implicit curve with \(F_y\ne 0\), the local branch has a continuous second derivative. Therefore, if \(y''\) is strictly positive or strictly negative at a point, it retains that sign on some sufficiently small interval around that point. This justifies describing the bending as local. A zero value alone does not determine concavity; the sign may vary on either side.
A Common Sign Error
A frequent mistake is to rearrange the second-derivative equation with the wrong sign. For the circle, the correct equation was \(1+(y')^2+yy''=0\). A mistaken rearrangement might claim
At \((3,4)\), that incorrect calculation gives \(y''=25/64\), the opposite sign from the correct value. The error is that the first two terms must be subtracted when isolating \(yy''\):
Substituting \(y=4\) and \(y'=-3/4\) gives \(y''=-(25/16)/4=-25/64\). Substitution into the original second-derivative equation confirms the correction: \(1+9/16+4(-25/64)=0\). Keeping the equation intact until the final algebra step helps prevent this sign reversal.
Why This Method Matters
Implicit curves arise naturally when a relationship between two variables is known even though \(y\) has not been isolated. Finding \(y''\) reveals how the local slope changes without first solving the equation explicitly for \(y\), which may be difficult or impossible in a useful form.
The method also reinforces a key idea in higher derivatives: every occurrence of \(y\) changes with \(x\). In a second differentiation, product and chain rules produce terms involving both \(y'\) and \(y''\). A dependable calculation therefore identifies the local branch, finds its slope, differentiates carefully, substitutes all known values, and checks that the resulting values satisfy the derivative equation.
Check Your Understanding
For each question, identify the derivative equation needed before substituting numerical values.
- For \(x^2+y^2=13\) at \((2,3)\), find \(y'\) and \(y''\), then state the local bending.
- Why must the point be checked in the original implicit equation before calculating derivatives there?
- What condition allows the formula \(y'=-F_x/F_y\) to be used at a point?
- For \(xy+y^2=6\) at \((1,2)\), verify the second-derivative equation using \(y'=-2/5\) and \(y''=12/125\).
- If \(y''\) is positive at one point, what additional idea justifies describing the curve as concave up on a nearby interval?
