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Chapter 10: Problem 33

Determine whether the point lies on the curve. $$r^{2} \cos \theta=1 ; \quad[1, \pi]$$

Short Answer

Expert verified
No, the point [1, \pi] does not lie on the curve \(r^{2} \cos \theta = 1\).

Step by step solution

01

Substitution

Substitute \(r = 1\) and \(\theta = \pi\) into the polar equation. The equation becomes \((1)^{2} \cos(\pi) = 1\).
02

Evaluation

Evaluate the left-hand side of the equation. \(1^{2} \cos(\pi)\) gives -1.
03

Verification

Verify if the evaluated left-hand side (-1) is equal to the right-hand side of the equation (1). Clearly, -1 is not equal to 1.

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Key Concepts

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

Equation Evaluation
When working with polar equations, evaluating the equation correctly is essential. In our exercise, we have the equation \( r^{2} \cos \theta = 1 \). To determine if a specific point, given in the polar coordinates \([r, \theta] = [1, \pi]\), lies on the curve represented by this equation, we need to evaluate this expression with the substituted values. Let's break it down:
  • First, substitute \( r = 1 \) and \( \theta = \pi \) into the equation.
  • The equation transforms into \( (1)^{2} \cos(\pi) = 1 \).
  • Compute \( (1)^{2} \), resulting in \( 1 \).
  • Next, calculate \( \cos(\pi) \), which equals \( -1 \).
  • Multiply these results: \( 1 \times -1 = -1 \).
This step is crucial to check if both sides of the equation are equal under the given conditions. In this case, the left-hand side equals \(-1\), while the right-hand side is \(1\). Hence, they are not equal.
Point on Curve
To determine if a point lies on a curve described by a polar equation, you substitute the coordinates of the point into the equation and check for equality. In the given problem, we are checking the point \([1, \pi]\) against the curve \( r^{2} \cos \theta = 1 \). Let's understand how this process works:
  • The point \([1, \pi]\) is expressed in polar coordinates, where \(1\) is the radius \(r\), and \(\pi\) is the angle \(\theta\).
  • When substituting these values into the equation, we check if the resulting expression holds true (both sides equals).
  • In our specific example, after substitution and evaluation, the left-hand side totaled to \(-1\), while the right-hand remained \(1\).
  • Since \(-1 eq 1\), the point \([1, \pi]\) does not lie on the curve.
This approach helps in verifying whether any point is on a particular curve, thus ensuring accurate plotting or characterizing of points with respect to polar curves.
Trigonometric Substitution
Trigonometric substitution is a method commonly used to simplify and solve functions involving trigonometric expressions. In polar coordinates, this often involves substituting specific trigonometric values for angles like \(\theta\). In our problem, understanding trigonometric values becomes essential.Here's how we can apply these concepts:
  • We substitute \( \theta = \pi \) into the trigonometric function \( \cos(\theta) \).
  • Remember that \( \cos(\pi) = -1 \). Knowing this helps simplify calculations directly within the equation \(r^{2} \cos(\theta) = 1\).
  • Using known trigonometric values can streamline solving and verifying equations, especially in relation to specific polar curves.
Trigonometric substitution can significantly reduce equation complexity and ensure the accurate evaluation of points within polar coordinates.

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Most popular questions from this chapter

Determine the eccentricity of the ellipse. $$x^{2} / 16+y^{2} / 25=1$$

Find the points \((x, y)\) at which the curve has: (a) a horizontal tangent: (b) a vertical tangent. Then sketch the curve. $$x(t)=3 t-t^{3}, \quad y(t)=t+1$$

A particle moves along the curve described by the parametric equations \(x=f(t), y=g(t) .\) Use a graphing utility to draw the path of the particle and describe the notion of the particle as it moves along the curve. $$x=2 t, \quad y=4 t-t^{2} \quad 0 \leq t \leq 6$$

Use a CAS to find an equation in \(x\) and \(y\) for the line tangent to the polar curve $$r=\frac{4}{2+\sin \theta} \quad \text { at } \theta=\frac{1}{3} \pi$$ Then use a graphing utility to sketch a figure that shows the curve and the tangent line.

Use a graphing utility to draw the curve $$x(\theta)=\cos \theta(a-b \sin \theta), \quad y(\theta)=\sin \theta(a-b \sin \theta)$$ from \(\theta=0\) to \(\theta=2 \pi\) given that (a) \(a=1 . b=2\) (b) \(a=2, b=2\) (c) \(a=2, b=1\) (d) In general, what can you say about the curve if \(ab ?\) (c) Express the curve in the form \(r=\rho(\theta)\)
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