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Chapter 11: Problem 59

Find polar equations for the circles in Exercises \(57-64 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$x^{2}+(y-5)^{2}=25$$

Short Answer

Expert verified
Polar equation: \(r = 5\sin\theta + 5\). Cartesian: \(x^2 + (y-5)^2 = 25\).

Step by step solution

01

Identify the Cartesian Equation

The given Cartesian equation is \(x^2 + (y - 5)^2 = 25\). This represents a circle with center at \((0, 5)\) and radius \(5\).
02

Convert Cartesian Coordinates to Polar Coordinates

In polar coordinates, we use \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute these into the equation: \((r\cos\theta)^2 + (r\sin\theta - 5)^2 = 25\).
03

Expand and Simplify the Expression

Expand the expression: \[r^2\cos^2\theta + (r\sin\theta - 5)^2 = r^2\sin^2\theta - 10r\sin\theta + 25.\] Combine like terms: \[r^2(\cos^2\theta + \sin^2\theta) = 25 + 10r\sin\theta.\] As \(\cos^2\theta + \sin^2\theta = 1\), the equation simplifies to \[r^2 = 25 + 10r\sin\theta.\]
04

Solve for the Polar Equation

Rearrange \(r^2 = 25 + 10r\sin\theta\): \[r^2 - 10r\sin\theta - 25 = 0.\] This can be factored or solved using the quadratic formula in terms of \(r\).
05

Quadratic Formula Application

To solve \[r^2 - 10r\sin\theta - 25 = 0,\] apply the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -10\sin\theta\), \(c = -25\). Compute \[r = \frac{-(-10\sin\theta) \pm \sqrt{(-10\sin\theta)^2 - 4 \cdot 1 \cdot (-25)}}{2 \cdot 1}.\]
06

Simplification to Find r

Simplify: \[r = \frac{10\sin\theta \pm \sqrt{100\sin^2\theta + 100}}{2}.\] Simplifies further to \[r = 5\sin\theta \pm 5\sqrt{\sin^2\theta + 1}.\] The polar equation for the circle becomes \[r = 5\sin\theta + 5.\]
07

Sketch and Label the Circle

Draw a circle on the coordinate plane with center \((0,5)\), radius \(5\). Label it with both its Cartesian equation \(x^2 + (y - 5)^2 = 25\) and its derived polar equation \(r = 5\sin\theta + 5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Equation
The Cartesian equation helps us define shapes like circles on a 2D plane using the familiar x and y axes. In our example, the Cartesian equation is given as \(x^2 + (y - 5)^2 = 25\). This equation represents a circle. Here are some key points:
  • The numbers \(x\) and \(y\) are the coordinates that denote any point on the circle.
  • The part \((y - 5)^2\) suggests that the center of the circle is at \((0, 5)\), because the circle's standard form is \((x - h)^2 + (y - k)^2 = r^2\).
  • The expression \(25\) is \(r^2\), meaning the radius \(r\) of the circle is 5.
Recognizing the components of a Cartesian equation allows us to easily interpret where the circle is located and its size.
Circle Equation
A circle equation is a mathematical expression representing all points that are a fixed distance from a center point. In the Cartesian coordinate system, it is written as:
  • \((x - h)^2 + (y - k)^2 = r^2\), where \(h\) and \(k\) are the coordinates of the center.
  • \(r\) is the radius of the circle.
For instance, our example \(x^2 + (y - 5)^2 = 25\) is a circle with:
  • Center at \((0, 5)\) 鈥� the \(y\) value shifted by 5 units.
  • Radius 5, since \(25 = 5^2\).
The circle equation allows you to visualize and understand circular shapes on a coordinate grid.
Quadratic Formula
The quadratic formula is a tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). For circles, such equations often arise when you're converting between coordinate systems.
  • Our problem involves solving \(r^2 - 10r\sin\theta - 25 = 0\).
  • The quadratic formula is: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Here, \(a = 1, b = -10\sin\theta, c = -25\).
  • Plugging these into the formula resolves \(r\) for polar coordinates.
Quadratic formula applications simplify solving circular equations and transforming them between coordinate systems.
Coordinate Conversion
Coordinate conversion is about transforming coordinates from one system to another, such as from Cartesian to polar.
  • Polar coordinates express a point's position as a distance \(r\) from the origin and an angle \(\theta\).
  • For conversion, use \(x = r\cos\theta\) and \(y = r\sin\theta\).
  • Substituting these in gives a new equation form: \((r\cos\theta)^2 + (r\sin\theta - 5)^2 = 25\).
Coordinate conversion is crucial for understanding and expressing shapes like circles in different mathematical contexts.By mastering this process, complex geometrical problems become much simpler.

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Most popular questions from this chapter

a. Epicycloid $$x=9 \cos t-\cos 9 t, \quad y=9 \sin t-\sin 9 t ; \quad 0 \leq t \leq 2 \pi$$ b. Hypocycloid $$x=8 \cos t+2 \cos 4 t, \quad y=8 \sin t-2 \sin 4 t ; \quad 0 \leq t \leq 2 \pi$$ c. Hypotrochoid $$x=\cos t+5 \cos 3 t, \quad y=6 \cos t-5 \sin 3 t ; \quad 0 \leq t \leq 2 \pi$$

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=2, \quad x=4$$

Use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(11.15 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6 $$

Find all polar coordinates of the origin.

Find polar equations for the circles in Exercises \(57-64 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$x^{2}+y^{2}+y=0$$
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