Problem 1
Question
The equation \(x^{2}+y^{2}=169\) describes a circle with radius 13 centered at the origin. (a) Solve explicitly for \(y\) in terms of \(x\). Is \(y\) a function of \(x\) ? (b) Differentiate your expression(s) from part (a) to find \(\frac{d y}{d x}\). (c) Now use implicit differentiation on the original equation to find \(\frac{d y}{d x}\). (d) Which method of differentiation (that used in part(a), or that used in part (b)) was easier? Why? (e) What is the slope of the tangent line to the circle at the point \((5,12) ?\) At \((5,-12)\) ?
Step-by-Step Solution
Verified Answer
\(y\) is not a function of \(x\). The results for \(\frac{d y}{d x}\) are \(\frac{-x}{\sqrt{169 - x^{2}}}\) and \(\frac{x}{\sqrt{169 - x^{2}}}\). Explicit differentiation is easier. The slopes at points (5, 12) and (5, -12) are -5/12 and 5/12 respectively.
1Step 1: Solve for \(y\)
Solving the circle equation \(x^{2}+y^{2}=169\) for \(y\), we get \(y=\sqrt{169 - x^{2}}\) and \(y=-\sqrt{169 - x^{2}}\). So, \(y\) is not a function of \(x\) because for each \(x\), there are two possible \(y\) values.
2Step 2: Explicit Differentiation
Differentiating both equations of \(y\) with respect to \(x\), we get \(\frac{d y}{d x} = \frac{-x}{\sqrt{169 - x^{2}}}\) and \(\frac{d y}{d x} = \frac{x}{\sqrt{169 - x^{2}}}\) respectively.
3Step 3: Implicit Differentiation
Taking the derivative of both sides of the circle equation with respect to \(x\), we get \(2x+2y\frac{d y}{d x} = 0\). Solving for \(\frac{d y}{d x}\), we get \(\frac{d y}{d x} = \frac{-x}{y}\). Substituting \(y=\sqrt{169 - x^{2}}\) and \(y=-\sqrt{169 - x^{2}}\) respectively gives us same result as in Step 2.
4Step 4: Comparison of Differentiation Methods
The method used in part (a) (explicit differentiation) is easier. This is because after solving the equation explicitly for \(y\), it's straightforward to obtain \(\frac{d y}{d x}\).
5Step 5: Find Slopes
Substituting \(x = 5\) and \(y = 12\) into the derivative expressions, we get \(\frac{d y}{d x} = -\frac{5}{12}\). This is slope of the tangent at point (5, 12). Similarly for point (5, -12), the slope is \(\frac{5}{12}\).
Key Concepts
Implicit DifferentiationExplicit DifferentiationEquations of CirclesTangent Lines
Implicit Differentiation
Implicit differentiation is a powerful technique used when dealing with equations where it’s challenging to solve explicitly for one variable in terms of another. In the context of the circle equation \(x^2 + y^2 = 169\), implicit differentiation allows us to find the derivative of \(y\) with respect to \(x\) without isolating \(y\) first. This is particularly useful when the relationship between the variables isn't straightforward.
To perform implicit differentiation, we differentiate both sides of the equation with respect to \(x\). Each term involving \(y\), such as \(y^2\), is treated as a function of \(x\). Applying the chain rule, we get \(2x + 2y\frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\), we find \(\frac{dy}{dx} = -\frac{x}{y}\).
This method circumvents the need to explicitly find \(y\), although once \(\frac{dy}{dx}\) is expressed in terms of \(x\) and \(y\), additional substitution might still be necessary to obtain numerical values.
To perform implicit differentiation, we differentiate both sides of the equation with respect to \(x\). Each term involving \(y\), such as \(y^2\), is treated as a function of \(x\). Applying the chain rule, we get \(2x + 2y\frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\), we find \(\frac{dy}{dx} = -\frac{x}{y}\).
This method circumvents the need to explicitly find \(y\), although once \(\frac{dy}{dx}\) is expressed in terms of \(x\) and \(y\), additional substitution might still be necessary to obtain numerical values.
Explicit Differentiation
Explicit differentiation involves solving an equation to explicitly isolate one variable, then differentiating the resulting function. For the given circle equation \(x^2 + y^2 = 169\), we aim to solve for \(y\) in terms of \(x\) directly.
By rearranging, we find \(y^2 = 169 - x^2\), and taking the square root gives \(y = \sqrt{169 - x^2}\) or \(y = -\sqrt{169 - x^2}\). These expressions show two branches, indicating \(y\) is not a function of \(x\) since each \(x\) corresponds to two \(y\) values.
After isolating \(y\), differentiating straightforwardly gives us \(\frac{dy}{dx} = \frac{-x}{\sqrt{169 - x^2}}\) for the positive branch and \(\frac{dy}{dx} = \frac{x}{\sqrt{169 - x^2}}\) for the negative. Despite offering a clear path to differentiation, it is only usable when \(y\) can be neatly solved in terms of \(x\).
By rearranging, we find \(y^2 = 169 - x^2\), and taking the square root gives \(y = \sqrt{169 - x^2}\) or \(y = -\sqrt{169 - x^2}\). These expressions show two branches, indicating \(y\) is not a function of \(x\) since each \(x\) corresponds to two \(y\) values.
After isolating \(y\), differentiating straightforwardly gives us \(\frac{dy}{dx} = \frac{-x}{\sqrt{169 - x^2}}\) for the positive branch and \(\frac{dy}{dx} = \frac{x}{\sqrt{169 - x^2}}\) for the negative. Despite offering a clear path to differentiation, it is only usable when \(y\) can be neatly solved in terms of \(x\).
Equations of Circles
Equations of circles, such as \(x^2 + y^2 = 169\), are fundamental in geometry and calculus. This specific equation represents a circle centered at the origin with a radius of 13.
The general form \((x-h)^2 + (y-k)^2 = r^2\) describes circles, where \((h, k)\) is the center and \(r\) is the radius. For a circle centered at the origin, the equation simplifies since \(h = 0\) and \(k = 0\).
Understanding these equations is crucial when working with problems involving distances and symmetries. They allow us to apply various calculus techniques, including differentiation, to explore properties like tangent lines and their slopes at specific points.
The general form \((x-h)^2 + (y-k)^2 = r^2\) describes circles, where \((h, k)\) is the center and \(r\) is the radius. For a circle centered at the origin, the equation simplifies since \(h = 0\) and \(k = 0\).
Understanding these equations is crucial when working with problems involving distances and symmetries. They allow us to apply various calculus techniques, including differentiation, to explore properties like tangent lines and their slopes at specific points.
Tangent Lines
Tangent lines to a curve at a given point are straight lines that just touch the curve at that point. They have the same slope as the curve at the point of tangency. For the circle described by \(x^2 + y^2 = 169\), the slope of the tangent line can be found using the derivative \(\frac{dy}{dx}\).
At the point \((5, 12)\), plugging into the derivative expression gives slope \(-\frac{5}{12}\). Similarly, for \((5, -12)\), the slope is \(\frac{5}{12}\).
These slope values indicate the direction of the tangent line at those specific points. Knowing the slope, one can further write the equation of the tangent line using point-slope form. Tangent lines are useful for linear approximations and help to understand how a curve behaves locally around a specific point.
At the point \((5, 12)\), plugging into the derivative expression gives slope \(-\frac{5}{12}\). Similarly, for \((5, -12)\), the slope is \(\frac{5}{12}\).
These slope values indicate the direction of the tangent line at those specific points. Knowing the slope, one can further write the equation of the tangent line using point-slope form. Tangent lines are useful for linear approximations and help to understand how a curve behaves locally around a specific point.
Other exercises in this chapter
Problem 1
Differentiate the following. (a) \(y=3^{x}\) (b) \(y=x^{3}\) (c) \(y=x^{x}\), where \(x>0\).
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Find \(f^{\prime}(x)\). (a) \(f(x)=2 x^{x}\), where \(x>0\) (b) \(f(x)=5\left(x^{2}+1\right)^{x}\) (c) \(f(x)=\left(2 x^{4}+5\right)^{3 x+1}\)
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Suppose we toss a rock into a pond causing a circular ripple. If the radius is increasing at a rate of 3 feet per second when the diameter is 4 , how fast is th
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