Chapter 11: Problem 39
Find parametric equations for the semicircle $$x^{2}+y^{2}=a^{2}, \quad y>0$$ using as parameter the slope \(t=d y / d x\) of the tangent to the curve at \((x, y) .\)
Short Answer
Expert verified
Parametric equations: \(x(t) = \frac{a t}{\sqrt{t^2 + 1}},\ y(t) = \frac{a}{\sqrt{t^2 + 1}}\).
Step by step solution
01
Understanding the Implicit Function
The equation of the semicircle is given as \(x^2 + y^2 = a^2\) with \(y > 0\). This describes the upper half of a circle centered at the origin with radius \(a\). We are asked to use the slope \(t = \frac{dy}{dx}\) as the parameter for the parametric equations.
02
Differentiate to Find the Slope
Differentiate the equation \(x^2 + y^2 = a^2\) implicitly with respect to \(x\): \[2x + 2y \frac{dy}{dx} = 0\]Solving for the derivative \(\frac{dy}{dx}\), we get:\[\frac{dy}{dx} = -\frac{x}{y}\]Here, \(t = \frac{dy}{dx} = -\frac{x}{y}\).
03
Express x and y in Terms of the Parameter t
Using the relationship \(t = -\frac{x}{y}\), express \(y\): \[y = -\frac{x}{t}\]Substitute \(y\) into the semicircle equation \(x^2 + y^2 = a^2\): \[x^2 + \left(-\frac{x}{t}\right)^2 = a^2\]This simplifies to:\[x^2 + \frac{x^2}{t^2} = a^2\] Factor out \(x^2\):\[x^2\left(1 + \frac{1}{t^2}\right) = a^2\]Thus, find \(x\):\[x^2 = \frac{a^2 t^2}{t^2 + 1}\]\[x = \frac{a t}{\sqrt{t^2 + 1}}\] (assuming \(x\) shares the sign with \(t\)).
04
Substitute x to Find y in terms of t
Substitute the expression for \(x\) back into \(y = -\frac{x}{t}\): \[y = -\frac{\frac{a t}{\sqrt{t^2 + 1}}}{t}\]=\[-\frac{a}{\sqrt{t^2 + 1}}\]Since \(y > 0\) for the upper semicircle, the negative sign adjusts for the definition of \(t = -\frac{x}{y}\). Thus,\[y = \frac{a}{\sqrt{t^2 + 1}}\].
05
Write the Parametric Equations
The parametric equations for the semicircle using the slope \(t\) as parameter are:\[x(t) = \frac{a t}{\sqrt{t^2 + 1}}\]\[y(t) = \frac{a}{\sqrt{t^2 + 1}}\] These equations describe points on the semicircle \(x^2 + y^2 = a^2\) with \(y > 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a technique used in calculus to find the derivative of functions that are not easily solved for one variable in terms of another. Traditional differentiation requires an explicit function, like \( y = f(x) \). However, equations like \( x^2 + y^2 = a^2 \), which describe a circle, are implicit because \( y \) is not directly solved for \( x \).
To use implicit differentiation, we differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). For example, differentiating \( x^2 + y^2 = a^2 \) with respect to \( x \) gives \( 2x + 2y \frac{dy}{dx} = 0 \).
This involves applying the chain rule for differentiation since \( y \) is dependent on \( x \). Here, \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \), and solving for it gives \( \frac{dy}{dx} = -\frac{x}{y} \). This result is crucial when using the slope of the tangent as a parameter in our parametric equations.
To use implicit differentiation, we differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). For example, differentiating \( x^2 + y^2 = a^2 \) with respect to \( x \) gives \( 2x + 2y \frac{dy}{dx} = 0 \).
This involves applying the chain rule for differentiation since \( y \) is dependent on \( x \). Here, \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \), and solving for it gives \( \frac{dy}{dx} = -\frac{x}{y} \). This result is crucial when using the slope of the tangent as a parameter in our parametric equations.
Semicircle Equation
The semicircle equation \( x^2 + y^2 = a^2 \) is a special case of a circle equation which depicts the top half of a circle when we include the condition \( y > 0 \). This equation represents all the points \( (x, y) \) that are at a distance \( a \) from a central point, which in this case is the origin \((0,0)\).
The condition \( y > 0 \) is essential here as it restricts the circle equation to only the upper semicircle. This effectively means that any point \( (x, y) \) we consider must have a positive \( y \)-coordinate, helping us to focus just on the top half.
The task involves parameterizing this semicircle using the slope \( t \) of the tangent line to the curve. This results in expressing \( x \) and \( y \) separately as functions of \( t \), ultimately revealing the parametric form of the semicircle.
The condition \( y > 0 \) is essential here as it restricts the circle equation to only the upper semicircle. This effectively means that any point \( (x, y) \) we consider must have a positive \( y \)-coordinate, helping us to focus just on the top half.
The task involves parameterizing this semicircle using the slope \( t \) of the tangent line to the curve. This results in expressing \( x \) and \( y \) separately as functions of \( t \), ultimately revealing the parametric form of the semicircle.
Tangents to Curves
Tangents to curves are straight lines that just touch the curve at a single point without crossing it. The slope of a tangent line at any point of a curve is given by the derivative \( \frac{dy}{dx} \) at that point.
For the semicircle, the tangent slope \( t = \frac{dy}{dx} = -\frac{x}{y} \) provides a different perspective when parameterizing the curve. By expressing the derivatives and using \( t \) as a parameter, we derive equations dependent on \( t \).
To find parametric equations, first find \( x \) and \( y \) in terms of \( t \):
For the semicircle, the tangent slope \( t = \frac{dy}{dx} = -\frac{x}{y} \) provides a different perspective when parameterizing the curve. By expressing the derivatives and using \( t \) as a parameter, we derive equations dependent on \( t \).
To find parametric equations, first find \( x \) and \( y \) in terms of \( t \):
- For \( x \), we arrive at \( x = \frac{a t}{\sqrt{t^2 + 1}} \) after manipulation involving the original semicircle equation.
- For \( y \), substituting \( x \) back gives \( y = \frac{a}{\sqrt{t^2 + 1}} \).