Derivatives of Polar Curves (College Board AP® Calculus BC): Revision Note

Derivatives of polar curves

How do I find dr/dθ of a polar curve?

• To find $\frac{dr}{d\theta }$ of a polar curve, $r\left(\theta \right)$, differentiate it with respect to $\theta$

  • e.g. if $r=1+\cos \theta$ then $\frac{dr}{d\theta }=-\sin \theta$

• The sign of the derivative can be interpreted as follows:

  • $\frac{dr}{d\theta }>0$ means $r$ is increasing with respect to $\theta$

    • Points on the curve are moving further away from the origin as $\theta$ increases

  • $\frac{dr}{d\theta }<0$ means $r$ is decreasing with respect to $\theta$

    • Points on the curve are moving closer to the origin as $\theta$ increases

• For certain polar curves, the point at which $\frac{dr}{d\theta }=0$is the point that has the greatest distance from the origin

• If two polar curves are given, $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$, then

  • $\frac{d}{d\theta }\left(r_1-r_2\right)$ is the rate at which the distance between the two curves is changing with respect to $\theta$

How do I find dx/dθ of a polar curve?

• To find $\frac{dx}{d\theta }$ of a polar curve, $r\left(\theta \right)$,

  • either differentiate the trigonometric relationship $x=r\cos \theta$ using implicit differentiation (as $r$ is not a constant - it varies with $\theta$) and the product rule

    • $\frac{dx}{d\theta }=\frac{dr}{d\theta }\cos \theta -r\sin \theta$

    • Then substitute in $r\left(\theta \right)$ and $\frac{dr}{d\theta }$

  • or substitute the curve equation, $r\left(\theta \right)$, directly into $x=r\cos \theta$ to get $x=r\left(\theta \right)\cos \theta$

    • Then differentiate this to find $\frac{dx}{d\theta }$

• The sign of the derivative can be interpreted as follows:

  • $\frac{dx}{d\theta }>0$ means the $x$-coordinate is increasing with respect to $\theta$

    • Points on the curve are moving in the positive $x$-direction as $\theta$ increases

  • $\frac{dx}{d\theta }<0$ means the $x$-coordinate is decreasing with respect to $\theta$

    • Points on the curve are moving in the negative $x$-direction as $\theta$ increases

How do I find dy/dθ of a polar curve?

• To find $\frac{dy}{d\theta }$ of a polar curve, $r\left(\theta \right)$

  • either differentiate the trigonometric relationship $y=r\sin \theta$ using the product rule (as $r$ is not a constant - it varies with $\theta$)

    • $\frac{dy}{d\theta }=\frac{dr}{d\theta }\sin \theta +r\cos \theta$

    • Then substitute in $r\left(\theta \right)$ and $\frac{dr}{d\theta }$

  • or substitute the curve equation, $r\left(\theta \right)$, directly into $y=r\sin \theta$ to get $y=r\left(\theta \right)\sin \theta$

    • Then differentiate this to find $\frac{dy}{d\theta }$

• The sign of the derivative can be interpreted as follows:

  • $\frac{dy}{d\theta }>0$ means the $y$-coordinate is increasing with respect to $\theta$

    • Points on the curve are moving upwards as $\theta$ increases

  • $\frac{dy}{d\theta }<0$ means the $y$-coordinate is decreasing with respect to $\theta$

    • Points on the curve are moving downwards as $\theta$ increases

How do I find the slope of a polar curve, dy/dx?

• The formula for the slope of a polar curve, $\frac{dy}{dx}$, is

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

• The derivatives $\frac{dy}{d\theta }$ and $\frac{dx}{d\theta }$ come from above

• The formula itself comes from the chain rule:

  • $\frac{dy}{dx}=\frac{dy}{d\theta }·\frac{d\theta }{dx}=\frac{dy}{d\theta }·\frac{1}{{\displaystyle \frac{dx}{d\theta }}}$

How do I find the second derivative of a polar curve with respect to x?

• The formula for the second derivative of $y$ with respect to $x$, $\frac{d^{2}y}{d{x}^2}$, for polar equations in terms of $\theta$ is

$\frac{d^{2}y}{d{x}^2}=\frac{{\displaystyle \frac{d}{d\theta }}\left({\displaystyle \frac{dy}{dx}}\right)}{{\displaystyle \frac{dx}{d\theta }}}$

• $\frac{d}{d\theta }\left(\frac{dy}{dx}\right)$ means differentiate the expression for the slope of the polar curve, $\frac{dy}{dx}$, with respect to $\theta$

  • Recall that $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{\frac{dr}{d\theta }\sin \theta +r\cos \theta }{\frac{dr}{d\theta }\cos \theta -r\sin \theta }$

  • and that $\frac{dx}{d\theta }=\frac{dr}{d\theta }\cos \theta -r\sin \theta$ from above

• The formula for the second derivative comes from the chain rule:

  • $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)·\frac{d\theta }{dx}=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)·\frac{1}{{\displaystyle \frac{dx}{d\theta }}}$

How do I find derivatives of polar curves with respect to time, t?

• To find derivatives of a polar curve with respect to time, $t\ge 0$, use the chain rule (related rates of change)

  • These are used in questions about a particle moving around a polar curve

• For example, if $\frac{dr}{dt}=2$ at $\theta =\frac{\pi }{2}$ on the curve $r=10\cos \theta$, then to find $\frac{d\theta }{dt}$ at $\theta =\frac{\pi }{2}$:

  • Use that $\frac{d\theta }{dt}=\frac{d\theta }{dr}·\frac{dr}{dt}$

    • Substitute in $\theta =\frac{\pi }{2}$ and $\frac{dr}{dt}=2$

    • Use that$\frac{d\theta }{dr}=\frac{1}{{\displaystyle \frac{dr}{d\theta }}}$ where $\frac{dr}{d\theta }=-10\sin \theta$ (by differentiating $r=10\cos \theta$)

  • This gives $\frac{1}{-10\sin \left({\displaystyle \frac{\pi }{2}}\right)}·2=-\frac{1}{5}$

• If you are given a relationship between $\theta$ and $t$, you can substitute this into the equation of the polar curve

  • e.g. If $r=10\cos \theta$ and $\theta =t^3$, then $r=10\cos \hspace{0.17em}\left(t^3\right)$

  • The position vector is then given by $<x,y>=<r\cos \theta ,r\sin \theta >=<10{\cos }^2\left(t^3\right),10\cos \left(t^3\right)\sin \left(t^3\right)>$

    • See the study guide on 'Motion with Vector-Valued Functions'