Question

The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π/3.

(a) Let R be the shaded region that is inside the graph of r = 4 and also outside the graph of r = 3 + 2cosθ, as shown in the figure above. Write an expression involving an integral for the area of R.

(b) Find the slope of the line tangent to the graph of r = 3 + 2cosθ at θ = π/2.

(c) A particle moves along the portion of the curve r = 3 + 2cosθ for 0 < θ < π/2. The particle moves in such a way that the distance between the particle and the origin increases at a constant rate of 3 units per second. Find the rate at which the angle θ changes with respect to time at the instant when the position of the particle corresponds θ = π/3. Indicate units of measure.

Answer 1

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

or, via symmetry

$\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

____________

(b)

By the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}$

For polar coordinates, x = rcosθ and y = rsinθ. Since
r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

$\begin{aligned} \displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\ &= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\ \left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}} &= 2/3 \end{aligned}$

2/3 is the slope

____________

(c)

"distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

Angle θ and r are related via r = 3 + 2cosθ, so implicitly differentiating with respect to time

$\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$

Answer 2

a) The expression involving an integral for the area of R is $A = \int\limits_{\frac{\pi}{3} }^{\frac{5\pi}{3}} \int\limits_{3 + 2\cdot \cos \theta}^{4} r\,dr\,d\theta$.

b) The slope of the line tangent to the graph of $r = 3 + 2\cdot \cos \theta$ is $\frac{2}{3}$.

c) The rate of change of the angle is $-\sqrt{3}$ radians per second.

Procedure - Integrals and derivatives in polar coordinates

a) Area of the shaded region as a double integral

Given two curves in polar coordinates with the same reference framework, the area between these curves is represented by the following definite double integral:

$A =\int\limits_{\theta_{1}}^{\theta_{2}} \int\limits_{f(\theta)}^{g(\theta)} \,r\,dr\,d\theta$ (1)

Where:

• $f(\theta)$ - "Lower" function
• $g(\theta)$ - "Upper" function
• $\theta_{1}$ - Lower bound, in radians.
• $\theta_{2}$ - Upper bound, in radians.

Determination of the double integral formula

If we know that $f(\theta) = 3 + 2\cdot \cos \theta$, $g(\theta) = 4$, $\theta_{1} = \frac{\pi}{3}$ and $\theta_{2} = \frac{5\pi}{3}$, then the integral equation for the shaded region is:

$A = \int\limits_{\frac{\pi}{3} }^{\frac{5\pi}{3}} \int\limits_{3 + 2\cdot \cos \theta}^{4} r\,dr\,d\theta$ (2)

The expression involving an integral for the area of R is $A = \int\limits_{\frac{\pi}{3} }^{\frac{5\pi}{3}} \int\limits_{3 + 2\cdot \cos \theta}^{4} r\,dr\,d\theta$. $\blacksquare$

b) Determination of the slope of a tangent line

The slope of a line tangent to a curve, expressed in rectangular coordinates, is represented by the following expression:

$\frac{dy}{dx} = \frac{r(\theta) \cdot \cos \theta + \frac{dr}{d\theta}\cdot \sin \theta }{-r(\theta)\cdot \sin \theta + \frac{dr}{d\theta}\cdot \cos \theta }$ (3)

Where $\theta$ is expressed in radians.

If we know that $\theta = \frac{\pi}{2}$, $r = 3 + 2\cdot \cos \theta$ and $\frac{dr}{d\theta} = -2\cdot \sin \theta$, then the slope of the line tangent to the graph is:

$\cos \frac{\pi}{2} = 0$, $\sin \frac{\pi}{2} = 1$

$r\left(\frac{\pi}{2} \right) = 3 + 2\cdot \cos \frac{\pi}{2}$

$r = 3$

$\frac{dr}{d\theta} = -2\cdot \sin \frac{\pi}{2}$

$\frac{dr}{d\theta} = -2$

$\frac{dy}{dx} = \frac{3\cdot (0)-2\cdot (1)}{-3\cdot (1) -2\cdot (0)}$

$\frac{dy}{dx} = \frac{2}{3}$

The slope of the line tangent to the graph of $r = 3 + 2\cdot \cos \theta$ is $\frac{2}{3}$. $\blacksquare$

c) Rate of change of a function in polar coordinates

Mathematically speaking, the rate of change of a variable is the instantaneous change of this variable in time.

If we know that $r = 3 + 2\cdot \cos \theta$, $\theta = \frac{\pi}{3}$ and $\frac{dr}{dt} = 3$, then the rate of change experimented by the angle is:

$\frac{dr}{dt} = -2\cdot \sin \theta \cdot \frac{d\theta}{dt}$

$\frac{d\theta\heta}{dt} = -\left(\frac{1}{2\cdot \sin \theta}\right)\cdot \frac{dr}{dt}$ (3)

Where $\frac{d\theta}{dt}$ is the rate of change of the angle, measured in radians per second.

$\frac{d\theta}{dt} = -\left(\frac{1}{2\cdot \sin \frac{\pi}{3} } \right)\cdot (3)$

$\frac{d\theta}{dt} = -\sqrt{3}\,\frac{rad}{s}$

The rate of change of the angle is $-\sqrt{3}$ radians per second. $\blacksquare$

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