Problem 26
Question
An equation is given. Do the following in each of these problems: (a) Find two functions defined by the equation, and state their domains. (b) Draw a sketch of the graph of each of the functions obtained in part (a). (c) Draw a sketch of the graph of the equation. (d) Find the derivative of each of the functions obtained in part (a) and state the domains of the derivatives. (e) Find \(D_{x} y\) by implicit differentiation from the given equation, and verify that the result so obtained agrees with the results in part (d). (f) Find an equation of each tangent line at the given value of \(x_{1}\). $$ x^{2}+y^{2}=25 ; x_{1}=4 $$
Step-by-Step Solution
Verified Answer
Two functions: \( y_1 = \sqrt{25 - x^2} \), \( y_2 = -\sqrt{25 - x^2} \) with domains \( -5 \leq x \leq 5 \). Tangent lines when \( x_1 = 4 \) are \( y = -\frac{4}{3}x + \frac{25}{3} \) and \( y = \frac{4}{3}x - \frac{25}{3} \).
1Step 1: Rewrite the equation
Given the equation \[ x^2 + y^2 = 25 \] represents a circle with radius 5 centered at the origin. Express y in terms of x.
2Step 2: Find two functions
Solve for y: \( y = \pm \sqrt{25 - x^2} \). These represent the two functions \( y_1 = \sqrt{25 - x^2} \) and \( y_2 = -\sqrt{25 - x^2} \). For both functions, the domain is \( -5 \leq x \leq 5 \).
3Step 3: Sketch graph of functions
Draw the graphs of \( y_1 = \sqrt{25 - x^2} \) and \( y_2 = -\sqrt{25 - x^2} \). These correspond to the upper and lower semicircles, respectively, of the circle with radius 5.
4Step 4: Sketch graph of the equation
Sketch the full circle \( x^2 + y^2 = 25 \). This circle has a radius of 5 and is centered at the origin.
5Step 5: Find derivatives
Find the derivatives of \( y_1 \) and \( y_2 \). For \( y_1 \): \( y_1' = \frac{d}{dx}(\sqrt{25 - x^2}) \), which by chain rule is \( -\frac{x}{\sqrt{25 - x^2}} \). For \( y_2 \): \( y_2' = \frac{d}{dx}(-\sqrt{25 - x^2}) \), which is \( \frac{x}{\sqrt{25 - x^2}} \). Both derivatives are defined for \( -5 < x < 5 \).
6Step 6: Verify using implicit differentiation
Differentiate \( x^2 + y^2 = 25 \) implicitly with respect to x: \( 2x + 2y \frac{dy}{dx} = 0 \), which simplifies to \( \frac{dy}{dx} = -\frac{x}{y} \). This agrees with the derivatives found in Step 5.
7Step 7: Find tangent lines
Given \( x_1 = 4 \), find \( y \). Substituting \( x = 4 \) in the original equation: \( 4^2 + y^2 = 25 \), so \( y = 3 \) or \( y = -3 \). For \( y = 3 \), the tangent line equation is \( y - 3 = (-\frac{4}{3})(x - 4) \), which simplifies to \( y = -\frac{4}{3}x + 25/3 \). For \( y = -3 \), the tangent line equation is \( y + 3 = (\frac{4}{3})(x - 4) \), which simplifies to \( y = \frac{4}{3}x - 25/3 \).
Key Concepts
Circle EquationFunction DerivativesDomain of FunctionsTangent Line Equations
Circle Equation
A circle equation in the form \(x^2 + y^2 = r^2\) describes a circle with radius \(r\) centered at the origin. This particular problem involves the equation \(x^2 + y^2 = 25\), indicating a circle with radius 5. To find the functions defined by this equation, we solve for \(y\), getting \(y = \pm\sqrt{25 - x^2}\). These two solutions represent the upper and lower halves of the circle, respectively. The values of \(x\) must satisfy \(25 - x^2 \geq 0\), leading to the domain \(-5 \leq x \leq 5\).
Function Derivatives
Derivatives represent the rate of change of a function. For the functions derived from \(x^2 + y^2 = 25\), we need to differentiate both \(y_1 = \sqrt{25 - x^2}\) and \(y_2 = -\sqrt{25 - x^2}\) with respect to \(x\).
For \(y_1\): \[ y_1' = \frac{d}{dx}(\sqrt{25 - x^2}) \] Applying the chain rule, we get \[ y_1' = -\frac{x}{\sqrt{25 - x^2}} \]
Similarly, for \(y_2\): \[ y_2' = \frac{d}{dx}(-\sqrt{25 - x^2}) \] This yields \[ y_2' = \frac{x}{\sqrt{25 - x^2}} \]
Both derivatives exist for \(-5 < x < 5\) as \(\sqrt{25 - x^2}\) is zero at the endpoints, making the derivatives undefined there.
For \(y_1\): \[ y_1' = \frac{d}{dx}(\sqrt{25 - x^2}) \] Applying the chain rule, we get \[ y_1' = -\frac{x}{\sqrt{25 - x^2}} \]
Similarly, for \(y_2\): \[ y_2' = \frac{d}{dx}(-\sqrt{25 - x^2}) \] This yields \[ y_2' = \frac{x}{\sqrt{25 - x^2}} \]
Both derivatives exist for \(-5 < x < 5\) as \(\sqrt{25 - x^2}\) is zero at the endpoints, making the derivatives undefined there.
Domain of Functions
The domain of a function is the set of all possible input values (x-values) that make the function valid. For the functions \(y = \pm \sqrt{25 - x^2}\), the domain is determined by the condition that the argument of the square root must be non-negative. Thus, we have \(25 - x^2 \geq 0\), simplifying to \(-5 \leq x \leq 5\).
It's important to note that since \(\sqrt{25 - x^2}\) must be real, x is constrained within this interval. Outside this range, the value under the square root becomes negative, making the function undefined.
This domain \( -5 \leq x \leq 5 \) applies to both \(y_1\) and \(y_2\). Similarly, their respective derivatives are defined for \(-5 < x < 5\).
It's important to note that since \(\sqrt{25 - x^2}\) must be real, x is constrained within this interval. Outside this range, the value under the square root becomes negative, making the function undefined.
This domain \( -5 \leq x \leq 5 \) applies to both \(y_1\) and \(y_2\). Similarly, their respective derivatives are defined for \(-5 < x < 5\).
Tangent Line Equations
A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. To find the tangent lines to the circle \(x^2 + y^2 = 25\) at \(x_1 = 4\), we first determine the corresponding y-values: plugging \(x = 4\) into the equation gives us \((4)^2 + y^2 = 25\), simplifying to \(y^2 = 9\), so \(y = 3\) or \(-3\).
Using these points, \( (4, 3) \) and \((4, -3)\), and the slopes from the derivatives, we find the tangent line equations.
For the upper semicircle \( y = 3 \): The slope is \[ m = -\frac{4}{3} \]. The tangent line equation is \[ y - 3 = -\frac{4}{3}(x - 4) \]. Simplifying gives us \[ y = -\frac{4}{3}x + \frac{25}{3} \].
For the lower semicircle \( y = -3 \): The slope is \[ m = \frac{4}{3} \]. The tangent line equation is \[ y + 3 = \frac{4}{3}(x - 4) \]. Simplifying, we get \[ y = \frac{4}{3}x - \frac{25}{3} \].
These lines just touch the circle at the given points, confirming our derivations.
Using these points, \( (4, 3) \) and \((4, -3)\), and the slopes from the derivatives, we find the tangent line equations.
For the upper semicircle \( y = 3 \): The slope is \[ m = -\frac{4}{3} \]. The tangent line equation is \[ y - 3 = -\frac{4}{3}(x - 4) \]. Simplifying gives us \[ y = -\frac{4}{3}x + \frac{25}{3} \].
For the lower semicircle \( y = -3 \): The slope is \[ m = \frac{4}{3} \]. The tangent line equation is \[ y + 3 = \frac{4}{3}(x - 4) \]. Simplifying, we get \[ y = \frac{4}{3}x - \frac{25}{3} \].
These lines just touch the circle at the given points, confirming our derivations.
Other exercises in this chapter
Problem 26
Let \(f\) be a function whose domain is the set of all real numbers and \(f(a+b)=f(a) \cdot f(b)\) for all \(a\) and \(b\). Furthermore, suppose that \(f(0)=1\)
View solution Problem 26
Given \(f(u)=u^{2}+5 u+5\) and \(g(x)=(x+1) /(x-1)\). Find the derivative of \(f \circ g\) in two ways: (a) by first finding \((f \circ g)(x)\) and then finding
View solution Problem 27
A particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from
View solution Problem 27
If \(f\) is differentiable at \(a\), prove that $$ f^{\prime}(a)=\lim _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a-\Delta x)}{2 \Delta x} $$ (HINT: \(f(a+\
View solution