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Chapter 3: Problem 15

Let \(r\) be a positive real number. The equation for a circle of radius \(r\) whose center is the origin is \(x^{2}+y^{2}=r^{2}\) (a) Use implicit differentiation to determine \(\frac{d y}{d x}\). (b) Let \((a, b)\) be a point on the circle with \(a \neq 0\) and \(b \neq 0 .\) Determine the slope of the line tangent to the circle at the point \((a, b)\). (c) Prove that the radius of the circle to the point \((a, b)\) is perpendicular to the line tangent to the circle at the point \((a, b) .\) Hint: Two lines (neither of which is horizontal) are perpendicular if and only if the products of their slopes is equal to -1 .

Short Answer

Expert verified
In summary, the derivative \(\frac{dy}{dx} = -\frac{x}{y}\), the slope of the tangent line at point (a, b) is \(-\frac{a}{b}\), and we have proved that the radius of the circle to the point (a, b) is perpendicular to the tangent line at the point (a, b) using the fact that the product of their slopes is -1.

Step by step solution

01

(a) Find \(\frac{dy}{dx}\) using implicit differentiation

We begin by applying implicit differentiation to the equation of the circle, differentiating both sides with respect to x: \[\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(r^2)\] Now we differentiate each term: \[\frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}r^2\] Since r is a constant, its derivative with respect to x is 0: \[\frac{d}{dx}r^2 = 0\] Now we apply the power rule for differentiation to the x and y terms. We also apply the chain rule to the y term, since we are differentiating it with respect to x: \[2x + 2y\frac{dy}{dx} = 0\] We can solve for the derivative \(\frac{dy}{dx}\) by isolating it: \[\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\] So, \(\frac{dy}{dx} = -\frac{x}{y}\).
02

(b) Calculate the slope of the tangent line at point (a, b)

To find the slope of the tangent line at point (a, b), we need to find the value of \(\frac{dy}{dx}\) at this point. We substitute a for x and b for y in the expression for \(\frac{dy}{dx}\): Slope of tangent line at (a,b) = \(-\frac{a}{b}\)
03

(c) Prove that the radius of the circle to the point(a, b) is perpendicular to the tangent line at the point (a, b)

To prove this, we need to show that the product of the slopes of the radius and the tangent line is equal to -1. The radius goes from the center of the circle at the origin (0, 0) to the point (a, b) on the circle. Therefore, the slope of the radius is: Slope of radius = \(\frac{b - 0}{a - 0} = \frac{b}{a}\) Now let's multiply the slope of the radius by the slope of the tangent line: \(\frac{b}{a} \times -\frac{a}{b} = -1\) Since the product of the slopes of the radius and the tangent line is -1, they are perpendicular. This proves that the radius of the circle to the point (a, b) is perpendicular to the line tangent to the circle at the point (a, b).

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Key Concepts

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

Equation of a Circle
An equation of a circle provides a mathematical description of a circle. Specifically, it allows us to understand where every point on the circle is in relation to its center. Consider a circle with center at the origin (0, 0) and a radius denoted by \(r\). For such a circle, the equation is given by:
  • \(x^2 + y^2 = r^2\)
This equation is based on the Pythagorean theorem. Each point \((x, y)\) on the circle maintains a consistent distance, exactly \(r\), from the center. By graphing this equation, you observe all points equidistant from the origin, forming the circle.
Additionally, when working with this equation, if you need to find how \(y\) changes with \(x\), implicit differentiation is the way. It helps to find the derivative \(\frac{dy}{dx}\) even when \(y\) isn鈥檛 expressed explicitly.
Slope of Tangent Line
The slope of a tangent line at any point on a circle is vital for understanding the circle's geometry. A tangent line touches the circle at exactly one point and represents the line closest to the circle at that point.
To find the slope of this line, suppose you have a point \((a, b)\) on the circle. The previous step using implicit differentiation gives the derivative \(\frac{dy}{dx} = -\frac{x}{y}\) for any point \((x, y)\) on the circle.
  • By substituting \((a, b)\) into this expression, the slope at \((a, b)\) becomes: \(-\frac{a}{b}\).
This value describes the steepness and direction of the tangent at \((a, b)\). Along with understanding its geometric interpretation, knowing the slope allows one to draw the precise tangent or calculate its equation.
Perpendicular Lines
In geometry, two lines are perpendicular if they intersect at a right angle, or 90 degrees. This concept is crucial, especially with circles, to identify specific relationships between the radius of a circle and its tangent line.
At any given point \((a, b)\) on the circle, you have two lines:
  • The radius, which goes from the origin to \((a, b)\), having a slope of \(\frac{b}{a}\).
  • The tangent line at \((a, b)\), having a slope of \(-\frac{a}{b}\).
For two lines to be perpendicular, the product of their slopes should be \(-1\). Calculating the product for the radius and tangent line confirms it: - The product is \(\frac{b}{a} \times -\frac{a}{b} = -1\).
Thus, confirming their perpendicular nature, meeting through a perfect right angle. This perpendicular relationship is fundamental in many geometric constructions and proofs.

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Most popular questions from this chapter

(a) Let \(a, b,\) and \(c\) be real numbers with \(a \neq 0 .\) Explain how to use a part of the quadratic formula (called the discriminant) to determine if the quadratic equation \(a x^{2}+b x+c=0\) has two real number solutions, one real number solution, or no real number solutions. (See Exercise ( 11 ) in Section 1.2 for a statement of the quadratic formula.) (b) Prove that if \(a, b,\) and \(c\) are real numbers for which \(a>0\) and \(c<0\), then one solution of the quadratic equation \(a x^{2}+b x+c=0\) is a positive real number. (c) Prove that if \(a, b,\) and \(c\) are real numbers, if \(a \neq 0, b>0\) and \(\frac{b}{2}<\sqrt{a c},\) then the quadratic equation \(a x^{2}+b x+c=0\) has no real number solution.

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