Calculus in Polar Coordinates: Videos & Practice Problems

Understanding polar coordinates involves recognizing the relationship between radial distance r and angle θ. To differentiate polar curves, convert r into parametric form, where x = f(θ) cos(θ) and y = f(θ) sin(θ). The slope of the tangent line can be found using the formula $dy/dx=$ dy/d\theta /dx/d\theta $.\; For\; area\; under\; polar\; curves,\; use\; the\; integral$ A=1/2\int {\alpha }_{}{\beta }_{}{r}^{2}d$$θ.

Slope of Polar Curves

In the study of polar coordinates, we often represent a point in the plane using a radial distance \( r \) and an angle \( \theta \). When analyzing polar curves, it becomes essential to differentiate these curves to find the slope of the tangent line. To do this, we can express the polar coordinates in parametric form, where \( x \) and \( y \) are defined as:

\( x = r(\theta) \cos(\theta) \)

\( y = r(\theta) \sin(\theta) \)

Here, \( r \) can be expressed as a function of \( \theta \), denoted as \( f(\theta) \). This transformation allows us to apply techniques similar to those used in parametric differentiation.

To find the derivative \( \frac{dy}{dx} \) for a polar curve, we utilize the derivatives of \( y \) and \( x \) with respect to \( \theta \). The formula for the derivative is given by:

\( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \)

It is crucial to ensure that \( \frac{dx}{d\theta} \neq 0 \) to avoid division by zero. The derivatives \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) can be computed using the product rule. For example, if we have a polar curve defined by \( r = 2 \sin(\theta) \), we can find \( y \) and \( x \) as follows:

\( y = 2 \sin^2(\theta) \)

\( x = 2 \sin(\theta) \cos(\theta) = \sin(2\theta) \)

Next, we differentiate \( y \) and \( x \) with respect to \( \theta \):

\( \frac{dy}{d\theta} = 4 \sin(\theta) \cos(\theta) = 2 \sin(2\theta) \)

\( \frac{dx}{d\theta} = 2 \cos(2\theta) \)

Substituting these derivatives into the formula for \( \frac{dy}{dx} \) gives:

\( \frac{dy}{dx} = \frac{2 \sin(2\theta)}{2 \cos(2\theta)} = \tan(2\theta) \)

To find the slope of the tangent line at a specific angle, such as \( \theta = \frac{\pi}{6} \), we substitute this value into our derivative:

\( \frac{dy}{dx} = \tan\left(2 \cdot \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{3}\right) \)

From trigonometric identities, we know that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \). Thus, the slope of the tangent line at \( \theta = \frac{\pi}{6} \) is \( \sqrt{3} \).

This process illustrates how to differentiate polar curves effectively, allowing for the determination of slopes and tangent lines in polar coordinates. Understanding these concepts is vital for further exploration in calculus and analytical geometry.

Slope of Polar Curves Example 1

To find the derivative \(\frac{dy}{dx}\) for the polar curve defined by \(r = 2 \sin(\theta)\), we start by recalling the formula for the derivative of a polar curve. The formula states that:

\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{f'(\theta) \sin(\theta) + f(\theta) \cos(\theta)}{f'(\theta) \cos(\theta) - f(\theta) \sin(\theta)}\]

In this context, \(f(\theta)\) represents \(r\), which is \(2 \sin(\theta)\). To apply the formula, we first need to compute \(f'(\theta)\), the derivative of \(f(\theta)\). The derivative of \(2 \sin(\theta)\) is:

\[f'(\theta) = 2 \cos(\theta)\]

Now, substituting \(f(\theta)\) and \(f'(\theta)\) into the formula, we can calculate the numerator and denominator separately. The numerator becomes:

\[2 \cos(\theta) \sin(\theta) + (2 \sin(\theta) \cos(\theta)) = 2 \cos(\theta) \sin(\theta) + 2 \sin(\theta) \cos(\theta) = 2 \sin(2\theta)\]

For the denominator, we have:

\[2 \cos(\theta) \cos(\theta) - 2 \sin(\theta) \sin(\theta) = 2 \cos^2(\theta) - 2 \sin^2(\theta) = 2(\cos^2(\theta) - \sin^2(\theta)) = 2 \cos(2\theta)\]

Now, substituting the simplified numerator and denominator back into the derivative formula gives us:

\[\frac{dy}{dx} = \frac{2 \sin(2\theta)}{2 \cos(2\theta)} = \frac{\sin(2\theta)}{\cos(2\theta)} = \tan(2\theta)\]

Thus, the derivative \(\frac{dy}{dx}\) for the polar curve \(r = 2 \sin(\theta)\) is:

\[\frac{dy}{dx} = \tan(2\theta)\]

Find the slope of the tangent line of the polar curve $r=2\cos \theta$ at $\theta =\frac{\pi }{3}$.

• A. $\sqrt{3}$
• B. $-\frac{\sqrt{3}}{3}$
• C. $\frac{\sqrt{3}}{3}$
• D. $-\sqrt{3}$

Answer: (C)

Area of Polar Regions

To find the area of polar regions, we can apply a strategy similar to calculating the area under a curve using definite integrals. In polar coordinates, the radial distance \( r \) can be expressed as a function of the angle \( \theta \). The area of a polar region is determined by summing the areas of infinitely many sectors, rather than rectangles as in Cartesian coordinates.

The formula for the area \( A \) of a polar region is given by:

\[A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\]

Here, \( \alpha \) and \( \beta \) are the bounds for the angle \( \theta \), and \( r \) is the function representing the radial distance. This integral effectively sums the areas of the sectors formed by the polar curve.

For example, to find the area of one petal of the rose defined by \( r = 4 \sin(2\theta) \), we first identify the bounds for \( \theta \). Since we are interested in the first quadrant, we set the bounds from \( 0 \) to \( \frac{\pi}{2} \). The area can then be calculated as follows:

\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (4 \sin(2\theta))^2 \, d\theta\]

Calculating \( r^2 \) gives us \( 16 \sin^2(2\theta) \), leading to:

\[A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 16 \sin^2(2\theta) \, d\theta = 8 \int_{0}^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta\]

To evaluate the integral of \( \sin^2(2\theta) \), we can use the trigonometric identity:

\[\sin^2(x) = \frac{1 - \cos(2x)}{2}\]

Applying this identity, we rewrite the integral:

\[A = 8 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(4\theta)}{2} \, d\theta = 4 \int_{0}^{\frac{\pi}{2}} (1 - \cos(4\theta)) \, d\theta\]

Now, we can integrate each term separately:

\[A = 4 \left[ \theta - \frac{1}{4} \sin(4\theta) \right]_{0}^{\frac{\pi}{2}}\]

Evaluating this expression at the bounds, we find:

\[A = 4 \left[ \frac{\pi}{2} - 0 \right] = 2\pi\]

Thus, the area of one petal of the rose is \( 2\pi \). This method illustrates how to find the area of polar regions by leveraging the concept of sector areas and definite integrals, providing a clear pathway to solving similar problems in polar coordinates.

Area of Polar Regions Example 2

To find the area enclosed by the inner loop of the limacon defined by the polar equation \( r = 1 - 2 \cos \theta \), we start by understanding the shape of the curve. The limacon has an outer loop that reaches an \( r \) value of -3 and an inner loop that reaches an \( r \) value of -1. The area we want to calculate is specifically within this inner loop.

The formula for the area \( A \) of a polar region is given by:

\( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \)

Here, \( r \) is the polar function, and \( \alpha \) and \( \beta \) are the bounds of integration that correspond to the angles where the inner loop is traced.

To determine the bounds \( \alpha \) and \( \beta \), we set \( r = 0 \) to find the angles where the curve intersects the origin:

\( 1 - 2 \cos \theta = 0 \)

Solving this gives:

\( 2 \cos \theta = 1 \Rightarrow \cos \theta = \frac{1}{2} \)

The angles where \( \cos \theta = \frac{1}{2} \) are \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \). Thus, we set \( \alpha = \frac{\pi}{3} \) and \( \beta = \frac{5\pi}{3} \).

Now, substituting \( r \) into the area formula, we have:

\( A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (1 - 2 \cos \theta)^2 \, d\theta \)

Next, we expand the integrand:

\( (1 - 2 \cos \theta)^2 = 1 - 4 \cos \theta + 4 \cos^2 \theta \)

Thus, the area becomes:

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

To integrate \( 4 \cos^2 \theta \), we use the power-reducing identity:

\( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \)

Substituting this into the integral gives:

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

Combining terms results in:

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

Now we can integrate each term separately:

1. The integral of \( 3 \) is \( 3\theta \).

2. The integral of \( -4 \cos \theta \) is \( -4 \sin \theta \).

3. The integral of \( 2 \cos(2\theta) \) is \( \sin(2\theta) \).

Thus, the antiderivative is:

\( \frac{1}{2} \left[ 3\theta - 4 \sin \theta + \sin(2\theta) \right] \Bigg|_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \)

Evaluating this at the bounds involves substituting \( \frac{5\pi}{3} \) and \( \frac{\pi}{3} \) into the antiderivative and simplifying. After performing the calculations, we find that the area enclosed by the inner loop of the limacon is:

\( A = 2\pi + \frac{3\sqrt{3}}{2} \)

Alternatively, this can be approximated numerically to about 8.88. This result represents the area of the inner loop of the limacon defined by the given polar equation.

Area of Polar Regions Example 3

To find the area enclosed by one loop of the Lemniscate defined by the equation \( r^2 = 4 \sin(2\theta) \), we start by recognizing that this is a polar curve. The area \( A \) of a polar region can be calculated using the formula:

\( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \)

In this case, we need to determine the bounds \( \alpha \) and \( \beta \) for the integral. Observing the graph of the Lemniscate, we see that one complete loop occurs in the first quadrant, which corresponds to the angle range from \( 0 \) to \( \frac{\pi}{2} \).

Substituting \( r^2 = 4 \sin(2\theta) \) into the area formula, we have:

\( A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 4 \sin(2\theta) \, d\theta \)

This simplifies to:

\( A = 2 \int_{0}^{\frac{\pi}{2}} \sin(2\theta) \, d\theta \)

To solve the integral, we can use the substitution \( u = 2\theta \), which gives \( du = 2 \, d\theta \) or \( d\theta = \frac{du}{2} \). Changing the limits accordingly, when \( \theta = 0 \), \( u = 0 \) and when \( \theta = \frac{\pi}{2} \), \( u = \pi \). Thus, the integral becomes:

\( A = 2 \int_{0}^{\pi} \sin(u) \cdot \frac{1}{2} \, du = \int_{0}^{\pi} \sin(u) \, du \)

The integral of \( \sin(u) \) is \( -\cos(u) \), so we evaluate:

\( A = -\cos(u) \bigg|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \)

Thus, the area enclosed by one loop of the Lemniscate is \( 2 \). This method illustrates how to find the area of polar regions by applying integration techniques and understanding the geometry of the curve.

Find the area enclosed by the cardioid $r=2-2\sin \theta$ between $\theta =\frac{\pi }{6}$ and $\theta =\frac{2\pi }{3}$.

• A. $1.685$
• B. $0.1143$
• C. $2.551$
• D. $0.5693$

Answer: (B)

Here’s what students ask on this topic:

How do you find the slope of a tangent line to a polar curve?

To find the slope of a tangent line to a polar curve, you first convert the polar equation into parametric form. If the polar curve is given as $r=f(\theta )$, then the parametric equations are $x=f(\theta )cos(\theta )$ and $y=f(\theta )sin(\theta )$. The slope is given by $\frac{\frac{d}{y}}{\frac{d}{x}}$, where $\frac{d}{y}=\frac{d}{\theta }(f(\theta )sin(\theta ))$ and $\frac{d}{x}=\frac{d}{\theta }(f(\theta )cos(\theta ))$. Use the product rule to differentiate and then evaluate at the given angle $\theta$.

What is the formula for finding the area of a polar region?

The formula for finding the area of a polar region is $A=\frac{1}{2}\int {\alpha }_{}{\beta }_{}{r}^{2}d\theta$, where $r$ is the radial distance as a function of $\theta$, and ${\alpha }_{}$ and ${\beta }_{}$ are the bounds of integration corresponding to the angles that define the region. This formula sums up the areas of infinitely many small sectors within the polar curve.

How do you convert a polar equation to parametric form?

To convert a polar equation $r=f(\theta )$ to parametric form, use the relationships $x=rcos(\theta )$ and $y=rsin(\theta )$. Substitute $r$ with $f(\theta )$ to get $x=f(\theta )cos(\theta )$ and $y=f(\theta )sin(\theta )$. These parametric equations represent the same curve in Cartesian coordinates.

What are common applications of calculus in polar coordinates?

Calculus in polar coordinates is commonly used in physics, engineering, and mathematics to analyze problems involving circular or radial symmetry. Applications include finding the slope of tangent lines to polar curves, calculating areas of polar regions (e.g., petals of a rose curve), and solving problems in electromagnetism and fluid dynamics where polar coordinates simplify the equations. It is also used in computer graphics for rendering circular or spiral patterns.

What is the significance of the double-angle identity in polar calculus?

The double-angle identity, such as $\sin 2^{\theta }=\frac{1}{2}(1-\cos (2\theta ))$, is significant in polar calculus because it simplifies integrals involving squared trigonometric functions. For example, when finding the area of a polar region, the formula often involves $r^2$, which may include terms like $\sin 2^{\theta }$. Using the identity allows you to rewrite the integral in a solvable form, facilitating the calculation of areas.