One Curve Can Have Many Slopes
A curve does not generally have one slope that describes it everywhere. Its tangent direction can change as you move from point to point. With an implicit equation, you can find the slope at each specified point by differentiating the relation and substituting that point’s coordinates.
Comparing the resulting values reveals two different features. The sign tells whether \(y\) locally increases or decreases as \(x\) increases. The magnitude tells how steep the tangent is: a larger absolute value means a steeper tangent, while a value closer to zero means a flatter one.
A Reliable Comparison Method
Begin by checking that every listed point lies on the stated curve. Then differentiate the relation, solve for \(dy/dx\) when possible, and substitute each point carefully. Keep the slopes as exact fractions while comparing them. A decimal approximation can hide useful exact relationships or make close values harder to distinguish.
Substitute each coordinate pair into the original equation to confirm it belongs to the curve.
Use implicit differentiation, treating \(y\) as a function of \(x\), and isolate \(dy/dx\) where its coefficient is nonzero.
Substitute both coordinates into the derivative formula, preserving negative signs.
Compare signs to describe direction and absolute values to compare steepness. Check each result in the differentiated equation.
Why the Formula Gives the Local Slope
Suppose a differentiable curve is described near a point by \(F(x,y)=0\), and suppose \(y\) can locally be regarded as a differentiable function of \(x\). Along that local part of the curve, the expression \(F(x,y(x))\) stays equal to zero. Its derivative with respect to \(x\) must therefore also be zero.
If \(F_y(x,y)\ne 0\), rearranging gives \( \frac{dy}{dx}=-\frac{F_x}{F_y} \). This is a local slope: it describes the tangent direction at the point under consideration, not a single direction shared by the entire curve. The formula also explains why evaluating at different coordinates can produce different results.
Worked Example: Comparing Slopes on a Circle
Worked Example: Comparing Slopes on a Circle
For the circle \(x^2+y^2=25\), compare the slopes at \((3,4)\), \((-3,4)\), and \((4,3)\).
Step 1: Check the points. At \((3,4)\), \(3^2+4^2=9+16=25\). At \((-3,4)\), \((-3)^2+4^2=9+16=25\). At \((4,3)\), \(4^2+3^2=16+9=25\). All three points lie on the circle.
Step 2: Differentiate.
Rearrange to obtain \(2y\frac{dy}{dx}=-2x\), so \( \frac{dy}{dx}=-\frac{x}{y} \), provided \(y\ne0\). Each listed point has a nonzero \(y\)-coordinate.
Step 3: Evaluate. At \((3,4)\), the slope is \(-\frac{3}{4}\). At \((-3,4)\), it is \(-\frac{-3}{4}=\frac{3}{4}\). At \((4,3)\), it is \(-\frac{4}{3}\).
Step 4: Compare and check. The first two slopes have equal magnitude, \(\frac34\), but opposite signs. At the first point, the curve locally decreases as \(x\) increases; at the second, it locally increases. The third slope is negative and has magnitude \(\frac43\), which is greater than \(\frac34\), so its tangent is steeper than the first tangent.
Substitution into the differentiated relation confirms each value. At \((3,4)\), \(2(3)+2(4)(-\frac34)=6-6=0\). At \((-3,4)\), \(2(-3)+2(4)(\frac34)=-6+6=0\). At \((4,3)\), \(2(4)+2(3)(-\frac43)=8-8=0\).
Signed Slope and Steepness Are Different Comparisons
A signed comparison asks which tangent slopes are positive or negative. A steepness comparison uses absolute values. For example, \(-\frac43\) is less than \(-\frac34\) as a number, but its absolute value is larger: \(\frac43>\frac34\). So the tangent with slope \(-\frac43\) is steeper, even though its signed slope is smaller.
For a finite slope, a positive value means that locally \(y\) rises as \(x\) increases; a negative value means that \(y\) falls. A zero slope means the tangent is horizontal. These statements concern the local behavior at each point. They do not mean that \(y\) must keep rising or falling throughout the whole curve.
Worked Example: Two Points with Negative Slopes
Worked Example: Comparing Slopes on a Symmetric Curve
For the curve \(x^2+xy+y^2=7\), compare the slopes at \((1,2)\) and \((2,1)\).
Step 1: Check membership. At \((1,2)\), \(1^2+(1)(2)+2^2=1+2+4=7\). At \((2,1)\), \(2^2+(2)(1)+1^2=4+2+1=7\). Both points are on the curve.
Step 2: Differentiate. Apply the product rule to \(xy\): its derivative is \(y+x\frac{dy}{dx}\).
Collecting the derivative terms gives \( (x+2y)\frac{dy}{dx}=-(2x+y) \). Thus \( \frac{dy}{dx}=-\frac{2x+y}{x+2y} \) when \(x+2y\ne0\). At the two points here, the denominators are \(5\) and \(4\), so the formula applies.
Step 3: Evaluate and compare. At \((1,2)\), the numerator \(2x+y=2(1)+2=4\), and the denominator \(x+2y=1+2(2)=5\). The slope is \(-\frac45\). At \((2,1)\), the numerator is \(2(2)+1=5\), and the denominator is \(2+2(1)=4\). The slope is \(-\frac54\). Both tangents have negative slope, but \(\left|-\frac54\right|=\frac54>\frac45=\left|-\frac45\right|\), so the tangent at \((2,1)\) is steeper.
Step 4: Verify. At \((1,2)\), the differentiated equation becomes \(2+2+(1+4)(-\frac45)=4-4=0\). At \((2,1)\), it becomes \(4+1+(2+2)(-\frac54)=5-5=0\). Both computed slopes satisfy the equation.
Worked Example: Opposite Directions on One Curve
Worked Example: Comparing the Two Branches of \(y^2=x^3\)
Compare the slopes at \((1,1)\) and \((1,-1)\) on \(y^2=x^3\).
Step 1: Check the points. For either point, \(y^2=1\) and \(x^3=1\). Thus both satisfy the relation.
Step 2: Differentiate.
At points with \(y\ne0\), solve to get \( \frac{dy}{dx}=\frac{3x^2}{2y} \). Both listed points have nonzero \(y\)-coordinates.
Step 3: Substitute. At \((1,1)\), the slope is \(\frac{3(1)^2}{2(1)}=\frac32\). At \((1,-1)\), the slope is \(\frac{3(1)^2}{2(-1)}=-\frac32\). The slopes have equal magnitude and opposite signs: at one point the curve locally rises as \(x\) increases, while at the other it locally falls.
Step 4: Verify. At \((1,1)\), the differentiated equation gives \(2(1)(\frac32)=3\), matching \(3(1)^2=3\). At \((1,-1)\), its left side is \(2(-1)(-\frac32)=3\), again matching \(3(1)^2=3\).
A Common Sign Error
When substituting a negative coordinate, preserve its sign in the formula. On the circle \(x^2+y^2=25\), a mistaken calculation at \((-3,4)\) might say \( -\frac{x}{y}=-\frac{3}{4} \), as if the \(x\)-coordinate were positive. That gives the wrong slope.
A second comparison error is to rank steepness using the signed values alone. For negative slopes, a more negative number has a greater absolute value and therefore represents a steeper tangent. State whether you are comparing signed slope or steepness, and use absolute values for the latter.
Why Comparing Points Matters
Several evaluations show how one implicit relation can describe changing local behavior. A curve may have slopes with opposite signs at different points, or slopes with the same sign but different steepness. The original equation identifies the curve; the differentiated equation gives a point-specific slope condition. Keeping these roles distinct helps make comparisons both accurate and meaningful.
For each point, retain a short verification: confirm it belongs to the original curve, then substitute the proposed slope into the differentiated relation. This catches coordinate and sign errors without replacing the comparison itself. If a slope formula has a denominator, also check that the denominator is nonzero at the points where you use it.
Check Your Understanding
Use implicit differentiation and keep your slopes exact.
- On \(x^2+y^2=25\), compare the slopes at \((3,4)\) and \((4,3)\). Which tangent is steeper?
- For \(x^2+xy+y^2=7\), verify that \((1,2)\) and \((2,1)\) lie on the curve, then compare the signs and magnitudes of their slopes.
- On \(y^2=x^3\), calculate and compare the slopes at \((1,1)\) and \((1,-1)\).
- Explain why comparing the signed values \(-\frac34\) and \(-\frac43\) is not the same as comparing tangent steepness.
- For any one of the examples, substitute each computed slope into the differentiated relation to verify it.
