Problem 6
Question
Find the slope of the line tangent to the following polar curves at the given points. At the points where the curve intersects the origin (when this occurs), find the equation of the tangent line in polar coordinates. $$r=4 \cos \theta ;\left(2, \frac{\pi}{3}\right)$$
Step-by-Step Solution
Verified Answer
Answer: The derivative of r w.r.t θ is $$\frac{dr}{d\theta} = -4\sin\theta.$$
2. What is the slope of the tangent line at the point (2, π/3)?
Answer: The slope of the tangent line at the point (2, π/3) is $$m = \frac{-\sqrt{3} + 2}{1 - 2 \sqrt{3}}$$.
3. What is the equation of the tangent line in polar coordinates when the curve intersects the origin?
Answer: The equation of the tangent line in polar coordinates when the curve intersects the origin is $$\tan(\alpha) = 4$$, with $$\alpha = \arctan(4)$$.
1Step 1: Calculate the derivative of r w.r.t θ
We are given the polar curve $$r=4 \cos \theta.$$ To find the derivative of r w.r.t θ, let's differentiate it:
$$\frac{dr}{d\theta}=-4\sin\theta.$$
2Step 2: Plug in the given point and solve for the slope of the tangent line
We have the point $$\left(2, \frac{\pi}{3}\right)$$, and we want to find the slope of the tangent line at this point. We can use the formula for the slope of the tangent line in polar coordinates, which is:
$$m = \frac{r\frac{dr}{d\theta} + r^2}{r \cos \theta - r^2 \sin\theta}$$
Now substitute the given values:
$$m = \frac{2(-4 \sin(\frac{\pi}{3})) + 2^2 }{2 \cos(\frac{\pi}{3}) - 2^2 \sin(\frac{\pi}{3})}$$
Calculate the slope:
$$m = \frac{2(-4 \frac{\sqrt{3}}{2}) + 4}{2 \frac{1}{2} - 4 \frac{\sqrt{3}}{2}} =\frac{-\sqrt{3} + 2}{1 - 2 \sqrt{3}}$$
3Step 3: Find the equation of the tangent line in polar coordinates when the curve intersects the origin
To find the equation of the tangent line in polar coordinates when the curve intersects the origin, we first need to find the value of θ when r = 0:
$$0=4\cos\theta$$
Solve for θ:
$$\theta= \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \ldots$$
Now, we need to find the equation of the tangent line in polar coordinates when $$\theta = \frac{\pi}{2}$$ :
$$\tan(\alpha) = \frac{r \frac{dr}{d\theta} + r^2 \sin\theta }{r \cos\theta - r^2 \sin\theta}$$
Plug in the values of r and θ when r = 0 and $$\theta = \frac{\pi}{2}$$:
$$\tan(\alpha) = \lim\limits_{r\to0}\frac{r (-4\sin(\frac{\pi}{2})) + r^2 \sin(\frac{\pi}{2})}{r \cos(\frac{\pi}{2}) - r^2 \sin(\frac{\pi}{2}) } = \lim\limits_{r\to0}\frac{-4r + r^2}{-r^2} = 4$$
The equation of the tangent line in polar coordinates when the curve intersects the origin is:
$$\tan(\alpha)= 4, \text{ where } \alpha = \arctan(4)$$
Key Concepts
Slope of Tangent LineDifferentiation of Polar EquationsTangent Line EquationsLimits in Polar Coordinates
Slope of Tangent Line
When analyzing curves in polar coordinates, the slope of a tangent line can reveal a lot about the behavior of the curve at a specific point. To calculate the slope of the tangent line to a polar curve at a particular point, one must first differentiate the polar equation to find
Once you have the derivative, the slope
\( \frac{dr}{d\theta} \), which represents how the radius \( r \) changes with respect to the angle \( \theta \).Once you have the derivative, the slope
\( m \) at a point \( (r, \theta) \) is given by the formula \( m = \frac{r\frac{dr}{d\theta} + r^2}{r \cos \theta - r^2 \sin\theta} \). Using this formula, you plug in the coordinates of the point and the calculated derivative to obtain the slope. In the given exercise, the slope was found by substituting the coordinates \( \left(2, \frac{\pi}{3}\right) \) and the derivative \( -4\sin(\theta) \) into the formula, yielding \( m = \frac{-\sqrt{3} + 2}{1 - 2 \sqrt{3}} \).Differentiation of Polar Equations
The process of finding the derivative of a polar equation is key in determining the behavior of polar curves, such as the slope of a tangent or the rate of change of the radius. Differentiation in polar coordinates follows similar principles to differentiation in cartesian coordinates, but the key difference lies in the variable being with respect to the polar angle
For any polar function
\( \theta \) instead of \( x \).For any polar function
\( r(\theta) \), the first derivative \( \frac{dr}{d\theta} \) is obtained by taking the derivative of the radius as a function of \( \theta \). In our example, the first derivative of \( r = 4 \cos \theta \) is \( \frac{dr}{d\theta} = -4 \sin \theta \). This derivative is essentially the rate at which the radius changes as the angle \( \theta \) increases.Tangent Line Equations
The equation of a tangent line to a polar curve can be approached in different ways depending on the context. For polar curves, tangent lines still represent the instantaneous direction of the curve at a given point, but the equation is derived considering polar coordinates. When a polar curve passes through the origin, the tangent at the origin can be found using limits and understanding the behavior of the curve as it approaches the origin.
In the provided exercise, the slope of the tangent line when the curve
In the provided exercise, the slope of the tangent line when the curve
\( r = 4 \cos \theta \) intersects the origin is deduced using the limit definition as \( \theta \) approaches \( \frac{\pi}{2} \) and \( r \) approaches zero. The resulting slope \( \tan(\alpha) = 4 \) forms the basis of the equation for the tangent line in polar coordinates, where \( \alpha \) is the angle the tangent makes with the polar axis.Limits in Polar Coordinates
In polar coordinates, expressing limits and analyzing them is crucial for understanding the behavior of functions at specific points, particularly when curves approach the origin or infinity. Limits help in finding asymptotes, tangents, or understanding the behavior of curves near singular points.
For the intersecting points at the origin, you set
For the intersecting points at the origin, you set
\( r = 0 \) and solve for \( \theta \). After establishing these points, you can use limits to determine the slope of the tangent line by plugging in values that approximate the origin. The limit operation becomes handy when dealing with indeterminate forms as you approach the origin. In our example, \( \tan(\alpha) \) is found using a limit as \( r \) approaches zero, which is a unique approach in polar coordinates compared to cartesian coordinates where the limit would typically approach a point along the \( x \)-axis.Other exercises in this chapter
Problem 6
What is the equation of the standard parabola with its vertex at the origin that opens downward?
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What is the polar equation of the horizontal line \(y=5 ?\)
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Describe the similarities and differences between the parametric equations \(x=t, y=t^{2}\) and \(x=-t, y=t^{2},\) where \(t \geq 0\) in each case.
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What is the equation of the standard ellipse with vertices at \((\pm a, 0)\) and foci at \((\pm c, 0) ?\)
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