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The cissoid of Diocles (from about 200 \(\mathrm{B.C.}\)). Find equations for the tangent and normal to the cissoid of Diocles \(y^{2}(2-x)=x^{3}\) at \((1,1)\).

Short Answer

Expert verified
Tangent: \( y = 2x - 1 \); Normal: \( y = -\frac{1}{2}x + \frac{3}{2} \).

Step by step solution

01

Identify the Implicit Differentiation

The equation of the cissoid of Diocles is given by \(y^{2}(2-x) = x^{3}\). To find the tangent and normal lines, we first need to determine the derivative \(\frac{dy}{dx}\) at the given point. This requires implicit differentiation of the given equation.
02

Differentiate Implicitly

Differentiate both sides of the equation \(y^{2}(2-x) = x^{3}\). Using the product rule and implicit differentiation, we have:\[ \frac{d}{dx}[y^2(2-x)] = \frac{d}{dx}[x^3] \]\[ 2y \cdot \frac{dy}{dx} (2-x) - y^2 = 3x^2 \]
03

Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\):\[ 2y (2-x) \cdot \frac{dy}{dx} = y^2 + 3x^2 \]Solving for \(\frac{dy}{dx}\), we get:\[ \frac{dy}{dx} = \frac{y^2 + 3x^2}{2y(2-x)} \]
04

Evaluate the Derivative at the Given Point

Substitute \(x = 1\) and \(y = 1\) in the derivative equation:\[ \frac{dy}{dx} \bigg|_{(1,1)} = \frac{1^2 + 3 \cdot 1^2}{2 \cdot 1 (2-1)} = \frac{4}{2} = 2 \]Thus, the slope of the tangent line at \((1, 1)\) is 2.
05

Write the Equation of the Tangent Line

Using the point-slope form of a line \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point \((1, 1)\):\[ y - 1 = 2(x - 1) \]\[ y = 2x - 1 \]This is the equation of the tangent line.
06

Find the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent's slope. Since the tangent slope is 2, the normal slope is \(-\frac{1}{2}\).
07

Write the Equation of the Normal Line

Using the point-slope form again, with the normal slope \(-\frac{1}{2}\):\[ y - 1 = -\frac{1}{2}(x - 1) \]\[ y = -\frac{1}{2}x + \frac{3}{2} \]This is the equation of the normal line.

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Key Concepts

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

Cissoid of Diocles
The Cissoid of Diocles is an ancient curve named after the Greek mathematician Diocles, who studied it around 200 B.C. This curve has intriguing properties and historical significance in the context of classical geometry. The equation for the cissoid of Diocles is given by:\[y^2(2-x) = x^3\]This equation describes a curve that can be used to solve problems related to doubling the cube and geometric constructions. The shape of the cissoid features a loop, and its unique form results from the combination of a parabola and a line's movement around a common base point. Understanding the cissoid offers insights into early mathematical thought and significant applications in the study of tangents and normals in calculus.
Tangent Line Equation
The concept of a tangent line is fundamental in calculus and helps to understand how a curve behaves at a particular point. For the cissoid of Diocles, the goal is to determine the equation of the tangent line at the point \((1, 1)\). The equation of the tangent line is derived using the slope point form:\[y - y_1 = m(x - x_1)\]where \(m\) is the slope at the point, and \((x_1, y_1)\) is the point of tangency.To find the slope \((m)\), implicit differentiation is used. For the Cissoid of Diocles, the derivative \(\frac{dy}{dx}\) was evaluated to be 2 when \(x = 1\) and \(y = 1\). Substituting this slope and the point into the point-slope formula gives the tangent line equation:
  • \(y = 2x - 1\)
This equation tells us that at the specific point \((1,1)\), the curve of the cissoid is best approximated by this linear function.
Normal Line Equation
The normal line to a curve at a given point is perpendicular to the tangent line at that same point. Calculating the normal line involves understanding the relationship between slopes of perpendicular lines. Specifically, the normal line's slope is the negative reciprocal of the tangent line's slope. For the cissoid of Diocles, once the slope of the tangent line is found as 2, the slope of the normal line is calculated as:
  • \(-\frac{1}{2}\)
Using this slope and the known point \((1,1)\), the point-slope form yields the equation of the normal line:
  • \(y = -\frac{1}{2}x + \frac{3}{2}\)
This equation describes the line that cuts perpendicularly across the tangent line at the point, pointing out how the cissoid shifts direction at that exact spot. Understanding the normal line helps visualize the curve's geometric properties beyond just its path of movement.
Implicit Differentiation Process
Implicit differentiation is a powerful technique used to differentiate equations that are not easily solved for one variable in terms of another. In the case of the cissoid of Diocles, which is given by \(y^2(2-x) = x^3\), implicit differentiation is necessary since the equation is not solved explicitly for \(y\).To apply implicit differentiation:
  • Differential equations of both sides of the equation: \[ \frac{d}{dx}[y^2(2-x)] = \frac{d}{dx}[x^3] \]
  • Use the product rule and chain rule where needed. For the product \(y^2(2-x)\), the product rule applies.
  • Rearrange terms to solve for \(\frac{dy}{dx}\) where required.
Ultimately, this process results in finding \(\frac{dy}{dx}\), the slope of the tangent at the given point, which in this case evaluates to 2 at \((1, 1)\). Implicit differentiation allows us to analyze the behavior of curves not easily expressed in traditional function form, expanding the toolkit for solving complex geometric problems.

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Most popular questions from this chapter

A draining hemispherical reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius 13 \(\mathrm{m}\) , shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) meters deep. a. At what rate is the water level changing when the water is 8 \(\mathrm{m}\) deep? b. What is the radius \(r\) of the water's surface when the water is y \(\mathrm{m}\) deep? c. At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

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Radians versus degrees: degree mode derivatives What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$ f(h)=\frac{\sin h}{h} $$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$ \lim _{h \rightarrow 0} \frac{\cos h-1}{h} $$ c. Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos \(x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin \(x\) and \(\cos x ?\)

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