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Polar Calculus and Applications · Tutorial 974 of 998

Deriving the Polar Slope Formula

Derive the polar slope formula from the chain rule, then use it carefully by checking the differentiability and denominator conditions.

Advanced 10 min read

What You'll Learn

  • Differentiate the Cartesian coordinates of a polar curve with respect to its angle parameter.
  • Obtain the polar formula for \(dy/dx\) as a quotient of parameter derivatives.
  • State the conditions needed to use the formula at a parameter value.
  • Distinguish a zero denominator from a zero-over-zero expression.
  • Check polar slope calculations by evaluating the coordinate derivatives directly.

Why a polar slope needs a derivation

A polar equation describes a curve using a radius \(r\) that may depend on the angle \(\theta\). But the slope \(dy/dx\) is defined using Cartesian coordinates. To find it, we must connect the polar description to the Cartesian coordinates of the point.

The key is to regard \(\theta\) as a parameter. As \(\theta\) changes, both Cartesian coordinates change: \(x=r(\theta)\cos\theta\) and \(y=r(\theta)\sin\theta\). The ratio of their rates of change gives the slope when the rate of change of \(x\) is nonzero.

Definition: Suppose \(r\) is a function of \(\theta\). The corresponding polar curve has Cartesian parametrization \(x(\theta)=r(\theta)\cos\theta\), \(y(\theta)=r(\theta)\sin\theta\). At a parameter value where the derivatives exist and \(x'(\theta)\ne 0\), its slope is \(dy/dx=y'(\theta)/x'(\theta)\).

Here, primes on \(x\), \(y\), and \(r\) mean derivatives with respect to \(\theta\). The condition \(x'(\theta)\ne 0\) matters: the quotient \(y'/x'\) is not defined when its denominator is zero.

Deriving the formula

Start with the Cartesian coordinates of a point described in polar form. Differentiate each coordinate with respect to \(\theta\). Both \(r\) and the trigonometric factor depend on \(\theta\), so the product rule is required in each calculation.

$$ x=r\cos\theta \qquad\text{and}\qquad y=r\sin\theta. $$

For \(x\), the product rule gives the derivative of \(r\) times \(\cos\theta\), plus \(r\) times the derivative of \(\cos\theta\). For \(y\), it gives the derivative of \(r\) times \(\sin\theta\), plus \(r\) times the derivative of \(\sin\theta\).

$$ \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta $$
$$ \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta. $$

When \(dx/d\theta\ne 0\), the parametric slope rule says to divide \(dy/d\theta\) by \(dx/d\theta\). Substituting the two derivatives gives the polar slope formula.

Theorem: If \(r\) is differentiable at \(\theta\) and \(\frac{dr}{d\theta}\cos\theta-r\sin\theta\ne 0\) there, then the slope of the polar curve at that parameter value is
$$ \frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta} {\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}. $$

This derivation explains why both terms appear in the numerator and denominator. The terms involving \(dr/d\theta\) account for the changing radius; the other terms account for the changing angle. Neither change can generally be ignored.

When the formula is defined

The formula requires that \(r\) have a derivative at the parameter value and that the denominator be nonzero. In coordinate terms, the denominator is exactly \(dx/d\theta\). Thus the formula is undefined whenever

$$ \frac{dr}{d\theta}\cos\theta-r\sin\theta=0. $$

This condition says the quotient \(dy/d\theta\) divided by \(dx/d\theta\) cannot be evaluated there. If the numerator is nonzero, the quotient has a nonzero numerator and zero denominator, so it does not give a finite value for \(dy/dx\). If both numerator and denominator are zero, the expression is \(0/0\), which is also undefined; that alone does not determine whether a slope can be found by some other analysis.

These are statements about this formula at a particular parameter value. A curve may pass through the same Cartesian point at more than one parameter value, and the derivatives must be evaluated at the parameter value being considered. Also, a point with \(r=0\) does not automatically make the formula undefined: the denominator condition, not the value of \(r\) alone, determines whether this quotient can be used.

Worked Example: The slope of a circle

Worked Example: The slope of a circle

Let \(r=3\), so the curve is a circle of radius \(3\) centered at the pole. Find its slope at \(\theta=\pi/4\).

Since \(r=3\) is constant, its derivative is \(dr/d\theta=0\). At \(\theta=\pi/4\), \(\sin\theta=\cos\theta=\sqrt{2}/2\). Substitute these values into the coordinate derivatives:

$$ \frac{dx}{d\theta} = 0\cdot\frac{\sqrt{2}}{2} - 3\cdot\frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}, $$
$$ \frac{dy}{d\theta} = 0\cdot\frac{\sqrt{2}}{2} + 3\cdot\frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}. $$

The denominator is nonzero, so the slope is

$$ \frac{dy}{dx} = \frac{3\sqrt{2}/2}{-3\sqrt{2}/2} = -1. $$

Check: directly, \(x(\theta)=3\cos\theta\), so \(dx/d\theta=-3\sin\theta\); and \(y(\theta)=3\sin\theta\), so \(dy/d\theta=3\cos\theta\). At \(\pi/4\), these are \(-3\sqrt{2}/2\) and \(3\sqrt{2}/2\), respectively. Their quotient is indeed \(-1\).

Worked Example: A varying radius

Worked Example: A varying radius

Let \(r=2+\cos\theta\). Find the slope at \(\theta=\pi/3\).

First evaluate the radius and its derivative. Since \(dr/d\theta=-\sin\theta\), at \(\pi/3\) we have \(r=2+1/2=5/2\) and \(dr/d\theta=-\sqrt{3}/2\). Also \(\sin(\pi/3)=\sqrt{3}/2\) and \(\cos(\pi/3)=1/2\). Therefore,

$$ \frac{dy}{d\theta} = \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{5}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{4}+\frac{5}{4} = \frac{1}{2}, $$
$$ \frac{dx}{d\theta} = \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{5}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{4}-\frac{5\sqrt{3}}{4} = -\frac{3\sqrt{3}}{2}. $$

The denominator is not zero. Thus

$$ \frac{dy}{dx} = \frac{1/2}{-3\sqrt{3}/2} = -\frac{1}{3\sqrt{3}} = -\frac{\sqrt{3}}{9}. $$

Check: differentiating \(x=(2+\cos\theta)\cos\theta\) by the product rule gives \(dx/d\theta=-\sin\theta\cos\theta-(2+\cos\theta)\sin\theta\). At \(\pi/3\), this is \(-\sqrt{3}/4-5\sqrt{3}/4=-3\sqrt{3}/2\). Differentiating \(y=(2+\cos\theta)\sin\theta\) gives \(dy/d\theta=-\sin^2\theta+(2+\cos\theta)\cos\theta\). At \(\pi/3\), this is \(-3/4+5/4=1/2\). Their quotient agrees with the result.

Worked Example: A spiral at a specific angle

Worked Example: A spiral at a specific angle

Let \(r=\theta\). Find the slope at \(\theta=\pi/2\).

Here \(dr/d\theta=1\), \(r=\pi/2\), \(\sin(\pi/2)=1\), and \(\cos(\pi/2)=0\). The coordinate derivatives are

$$ \frac{dy}{d\theta} = 1\cdot 1+\frac{\pi}{2}\cdot 0 = 1, \qquad \frac{dx}{d\theta} = 1\cdot 0-\frac{\pi}{2}\cdot 1 = -\frac{\pi}{2}. $$

Since \(dx/d\theta=-\pi/2\ne0\), divide:

$$ \frac{dy}{dx} = \frac{1}{-\pi/2} = -\frac{2}{\pi}. $$

Check: the Cartesian coordinates are \(x(\theta)=\theta\cos\theta\) and \(y(\theta)=\theta\sin\theta\). Their derivatives are \(dx/d\theta=\cos\theta-\theta\sin\theta\) and \(dy/d\theta=\sin\theta+\theta\cos\theta\). At \(\pi/2\), they equal \(-\pi/2\) and \(1\), so their quotient is \(-2/\pi\), as found.

A common mistake: dropping the radius derivative

A frequent error is to differentiate \(x=r\cos\theta\) as though \(r\) were constant, even when \(r\) depends on \(\theta\). That loses the product-rule term \((dr/d\theta)\cos\theta\). The same error in \(y\) loses \((dr/d\theta)\sin\theta\).

For the curve \(r=2+\cos\theta\) at \(\theta=\pi/3\), the incorrect calculation that treats \(r\) as constant would use \(dx/d\theta=-r\sin\theta=-5\sqrt{3}/4\) and \(dy/d\theta=r\cos\theta=5/4\). It would give the incorrect quotient

$$ \frac{5/4}{-5\sqrt{3}/4} = -\frac{1}{\sqrt{3}}. $$

The omitted contributions are \((dr/d\theta)\cos\theta=-\sqrt{3}/4\) in \(dx/d\theta\) and \((dr/d\theta)\sin\theta=-3/4\) in \(dy/d\theta\). Including them gives \(dx/d\theta=-3\sqrt{3}/2\) and \(dy/d\theta=1/2\), so the correct slope is \(-1/(3\sqrt{3})\). The discrepancy is why the radius must be differentiated whenever it varies with the parameter.

Why this formula is useful

The formula converts a polar description into a Cartesian slope without first solving for \(y\) as an explicit function of \(x\). This is especially helpful when the radius has a simple expression in \(\theta\), but the Cartesian equations are not easy to combine into a single equation \(y=f(x)\).

Its practical use depends on careful bookkeeping: differentiate both coordinate products, evaluate the derivatives at the same angle, and check the denominator before dividing. A zero denominator means the displayed slope quotient is undefined there; it is not permission to divide by zero or to cancel a term. If both derivatives vanish, the quotient is indeterminate, and further analysis is needed rather than a value inferred from this formula alone.

1
Differentiate the radius.
Find \(dr/d\theta\) from the given polar equation.
2
Compute both coordinate derivatives.
Use \(dx/d\theta=(dr/d\theta)\cos\theta-r\sin\theta\) and \(dy/d\theta=(dr/d\theta)\sin\theta+r\cos\theta\).
3
Check before dividing.
Use the quotient only if \(dx/d\theta\ne0\), then evaluate \(dy/dx=(dy/d\theta)/(dx/d\theta)\).

Check Your Understanding

Use the coordinate derivatives, not a memorized shortcut, to justify each answer.

  1. For \(r=4\), find \(dx/d\theta\), \(dy/d\theta\), and \(dy/dx\) at \(\theta=\pi/6\).
  2. For \(r=1+\sin\theta\), write the numerator and denominator of the polar slope formula at a general angle \(\theta\).
  3. What condition on \(r\) and \(dr/d\theta\) makes the polar slope formula undefined?
  4. If both \(dx/d\theta\) and \(dy/d\theta\) are zero at a parameter value, what can the formula alone establish about the slope?
  5. Explain which product-rule term is missing if one differentiates \(x=r\cos\theta\) as \(dx/d\theta=-r\sin\theta\) for a varying radius.

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