Find polar tangents through Cartesian derivatives

Try

A polar equation tells you a signed radial coordinate r at an angle θ; it does not directly give the slope on Cartesian axes. For the curve r=2+cos θ, predict the tangent direction at θ=0. The radius is 3 and its rate of change is zero there. Does that make the curve’s tangent horizontal? Plot nearby points before answering: the angle is changing even when the radius momentarily stops changing.

Use radians throughout this lesson. Write x=r cos θ and y=r sin θ, so the curve is a parametrized Cartesian path. A tangent direction comes from changes in both coordinates, not from r′ alone. Here a prime on r means differentiation with respect to θ.

Predict

Predict the signs of the coordinate rates at θ=π/2 for the same curve. A small increase in angle moves the point left; the decreasing radius also moves it downward near that angle. Should the ratio dy/dx be positive or negative? Then predict whether dy/dx=0 and dx/dθ=0 describe the same geometric situation. Record which denominator must remain nonzero before using a quotient.

Experiment

Differentiate the two coordinate products. Since the derivative of cos θ is −sin θ and that of sin θ is cos θ,

x′=r′ cos θ−r sin θ, and y′=r′ sin θ+r cos θ.

Where x′ is nonzero, dy/dx=y′/x′. Thus the polar slope is

(r′ sin θ+r cos θ)/(r′ cos θ−r sin θ).

For r=2+cos θ, substitute r′=−sin θ. This gives x′=−2 sin θ(1+cos θ) and y′=2 cos²θ+2 cos θ−1. Inspect the following coordinate-and-direction table, then recompute one row without looking at its last column.

Angle	Cartesian point	x′	y′	Tangent information
0	(3,0)	0	3	Vertical
π/2	(0,2)	−2	−1	Slope 1/2
π	(−1,0)	0	−1	Vertical

At π/2 the coordinates both decrease; their ratio is positive, despite downward motion. A tangent slope describes a line, whereas the signs of the two coordinate rates also describe the direction of traversal. Draw arrows left and down at (0,2), and sketch the line y−2=x/2 through that point. This gives a visual check on the quotient.

Understand

A regular parametrized point has (x′,y′) different from (0,0). At such a point, a horizontal tangent requires y′=0 and x′ nonzero. A vertical tangent requires x′=0 and y′ nonzero. A zero denominator alone does not establish a vertical tangent: the numerator must be checked too.

Find all horizontal tangents of this curve for 0≤θ<2π. Put c=cos θ and solve 2c²+2c−1=0. The roots are (−1+√3)/2 and (−1−√3)/2. The latter is outside [−1,1], so it cannot be a cosine. The valid root lies strictly between 0 and 1 and gives two angles in one full turn. At each, sin θ is nonzero and 1+cos θ is nonzero, so x′ is nonzero. Both are genuinely horizontal tangents. The vertical cases are θ=0 and π; the table checks the required nonzero y′.

Now examine r=1−cos θ at θ=0. Here r=r′=0, so both coordinate derivatives vanish. The quotient is 0/0, and the regular-point criteria give no answer. Additional analysis is possible: near zero, r behaves like θ²/2, so x behaves like θ²/2 and y like θ³/2. More directly, for nearby nonzero angles y/x=tan θ tends to zero. The secant lines through the origin therefore approach the horizontal axis, even though the parametrization has zero velocity at the cusp. This limiting tangent is not obtained by declaring a zero denominator “vertical.”

This example also separates differentiability of a parametrization from regularity. The coordinate functions are smooth at zero; their derivative vector vanishes there. If you need to classify another simultaneous-zero case, inspect higher-order terms or an appropriate limit rather than canceling without a stated domain.

Apply

Write a three-line tangent test: calculate the point, calculate both coordinate derivatives, then state which regularity case applies. Apply it at θ=3π/2 for r=2+cos θ. You obtain (0,−2), x′=2, y′=−1 and slope −1/2. Explain why opposite vertical positions need not give the same slope.

Repair the statements “r′=0 means horizontal tangent” and “any undefined slope means a vertical tangent.” Use one regular point from the table and the cusp example as different counterexamples. Keep exact angles and radicals until the final interpretation; a rounded value near zero is not an exact zero test.

Cartesian concavity is not a second angular derivative

Let r be twice differentiable near the chosen angle. Differentiate the coordinate rates once more:

x″=(r″−r)cos θ−2r′sin θ, and y″=(r″−r)sin θ+2r′cos θ.

Where x′≠0, the chain rule gives d/dx=(1/x′)d/dθ. Applying the quotient rule to y′/x′ therefore proves

d²y/dx²=(y″x′−y′x″)/(x′)³.

The final division by x′ is essential: d/dθ(dy/dx) is not d²y/dx². This formula applies to a local graph y(x), not to a vertical tangent or a singular parameter value without further analysis.

For the existing curve r=2+cos θ at π/2, r=2, r′=−1, r″=0. Hence x′=−2, y′=−1, x″=2, y″=−2. The numerator is 4+2=6 and the denominator is −8, giving d²y/dx²=−3/4. The Cartesian graph is locally concave down even though its first derivative is positive. Checking r″ alone would have missed this information.

Unfamiliar transfer. Use r=3+2sin θ at θ=0. Before calculating, decide whether the local y(x) criterion holds. Find its slope and second derivative and explain why differentiating the slope with respect to θ alone gives the wrong answer. A classmate claims a step fails at θ=π/2. Test that claim; if it is false, find a genuine vertical-tangent angle.

Review the coordinate checks and chain factor

At zero, (r,r′,r″)=(3,2,0), so (x′,y′)=(2,3) and (x″,y″)=(−3,4). Thus dy/dx=3/2 and d²y/dx²=(8+9)/8=17/8. The angular derivative of the slope is 17/4; division by x′=2 converts it to the Cartesian second derivative. At π/2, (x′,y′)=(−5,0), so the criterion does not fail: the curve has a regular horizontal tangent and a valid second derivative. A prompt suggesting failure is not evidence of failure. At 3π/2, (x′,y′)=(1,0), likewise regular. To exhibit a failure, solve x′=2cos²θ−(3+2sin θ)sin θ=2−3sin θ−4sin²θ=0, which has the admissible value sin θ=(−3+√41)/8; at either corresponding angle y′=(3+4sin θ)cos θ≠0. These are vertical tangents, where the displayed y(x) formula is unavailable.

A complete response checks x′ before division, derives both second coordinate rates, retains the cube in the denominator, and rejects the false premise about π/2. No decimal zero test is needed.

Continue

Use the written practice to derive and check tangent criteria on a fresh curve. The parametric derivatives lesson provides the underlying coordinate-rate rule. The next polar investigation uses the same Cartesian conversion to check intersections before selecting an area integral.