Chapter 3: Problem 29
If a point moves on the curve \(x^{2}+y^{2}=25,\) then, at \((0,5), \frac{d^{2} y}{d x^{2}}\) is (A) 0 (B) \(\frac{1}{5}\) (C) -5 (D) \(-\frac{1}{5}\)
Short Answer
Expert verified
The answer is (D) \(-\frac{1}{5}\).
Step by step solution
01
Identify the given curve
The curve given is a circle with the equation \(x^{2} + y^{2} = 25\). This is a circle centered at the origin with a radius of 5.
02
Differentiate the equation implicitly
To find \(\frac{d^2y}{dx^2}\), start by differentiating the equation \(x^2 + y^2 = 25\) implicitly with respect to \(x\). This yields: \(2x + 2y \frac{dy}{dx} = 0\). Simplifying gives: \(\frac{dy}{dx} = -\frac{x}{y}\).
03
Find the first derivative at point \((0, 5)\)
Substitute \(x = 0\) and \(y = 5\) into the first derivative \(\frac{dy}{dx} = -\frac{x}{y}\): \(\frac{dy}{dx} = -\frac{0}{5} = 0\).
04
Differentiate the first derivative implicitly
Differentiate \(\frac{dy}{dx} = -\frac{x}{y}\) implicitly to find \(\frac{d^2y}{dx^2}\). Differentiate both sides: \(\frac{d^2y}{dx^2} = -\left(\frac{d}{dx}\frac{x}{y}\right)\).
05
Apply the quotient rule
The quotient rule states \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). Let \(u = x\) and \(v = y\). So, \(u' = 1\) and \(v' = \frac{dy}{dx}\). Substituting in gives \(\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y(1) + x\left(\frac{dy}{dx}\right)}{y^2}\).
06
Substitute known values to find \(\frac{d^2y}{dx^2}\) at \((0,5)\)
We know \(\frac{dy}{dx} = 0\) at \((0, 5)\). Substitute \(x = 0, y = 5\) into the second derivative equation: \(\frac{d^2y}{dx^2} = \frac{-5(1) + 0(0)}{5^2} = \frac{-5}{25} = -\frac{1}{5}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
The first derivative represents the rate at which a function is changing at any given point. In this problem, we start by implicitly differentiating the circle equation \(x^2 + y^2 = 25\) with respect to \(x\). This differentiation gives us the first derivative \(\frac{dy}{dx}\). Implicit differentiation is necessary here because the relationship between \(x\) and \(y\) is not explicit; \(y\) is not isolated on one side of the equation.
When we differentiate \(x^2 + y^2 = 25\), we treat \(y\) as a function of \(x\), applying the chain rule to account for \(y\)'s dependence on \(x\). This gives us:
When we differentiate \(x^2 + y^2 = 25\), we treat \(y\) as a function of \(x\), applying the chain rule to account for \(y\)'s dependence on \(x\). This gives us:
- \(2x + 2y \frac{dy}{dx} = 0\)
- \(\frac{dy}{dx} = -\frac{x}{y}\)
Second Derivative
Once we have the first derivative, \(\frac{dy}{dx} = -\frac{x}{y}\), finding the second derivative helps us understand how the rate of change itself is changing. The second derivative, \(\frac{d^2y}{dx^2}\), is about changes in the first derivative.
To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(-\frac{x}{y}\) again, using the quotient rule. The quotient rule is useful when differentiating the ratio of two functions:
To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(-\frac{x}{y}\) again, using the quotient rule. The quotient rule is useful when differentiating the ratio of two functions:
- \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]
- \(u' = 1\) and \(v' = \frac{dy}{dx} = -\frac{x}{y}\)
- \[\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y(1) + x\left(-\frac{x}{y}\right)}{y^2}\]
- \(\frac{d^2y}{dx^2} = \frac{-5}{25} = -\frac{1}{5}\)
Circle Equation
The equation \(x^2 + y^2 = 25\) represents a circle centered at the origin and having a radius of 5. A circle equation like this one describes a set of points that are all equidistant from the center point \((0,0)\).
This distance, known as the radius, is 5 in this example, since whenever you set \(x = 0\) or \(y = 0\), the remaining variable must reach 5 to satisfy the equation. Understanding the geometric nature of the circle helps in visualizing how \(x\) and \(y\) vary.
This distance, known as the radius, is 5 in this example, since whenever you set \(x = 0\) or \(y = 0\), the remaining variable must reach 5 to satisfy the equation. Understanding the geometric nature of the circle helps in visualizing how \(x\) and \(y\) vary.
- The radius = 5 (since \(\sqrt{25} = 5\))
- Center = \((0,0)\)