Problem 29 - AP Calculus with 8 Practice Tests 15th

Question

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) \(-\frac{1}{5}\)

Step-by-Step Solution

Answer

The answer is (D) \(-\frac{1}{5}\).

Step 1: 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.

Step 2: 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: \(\frac{dy}{dx} = -\frac{x}{y}\).

Step 3: Find the first derivative at point \((0, 5)\)

Substitute \(x = 0\) and \(y = 5\) into the first derivative \(\frac{dy}{dx} = -\frac{x}{y}\): \(\frac{dy}{dx} = -\frac{0}{5} = 0\).

Step 4: Differentiate the first derivative implicitly

Differentiate \(\frac{dy}{dx} = -\frac{x}{y}\) implicitly to find \(\frac{d^2y}{dx^2}\). Differentiate both sides: \(\frac{d^2y}{dx^2} = -\left(\frac{d}{dx}\frac{x}{y}\right)\).

Step 5: 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 \(\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y(1) + x\left(\frac{dy}{dx}\right)}{y^2}\).

Step 6: 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: \(\frac{d^2y}{dx^2} = \frac{-5(1) + 0(0)}{5^2} = \frac{-5}{25} = -\frac{1}{5}\).

Key Concepts

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:

• \(2x + 2y \frac{dy}{dx} = 0\)

We can solve for \(\frac{dy}{dx}\):

• \(\frac{dy}{dx} = -\frac{x}{y}\)

In simple terms, \(\frac{dy}{dx}\) tells us how much \(y\) changes when \(x\) changes, specifically for points lying on our circle. At the specific point \((0,5)\), the first derivative becomes 0, indicating that there is no change in \(y\) with respect to \(x\) at this location.

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:

• \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]

In our context, \(u = x\) and \(v = y\), making:

• \(u' = 1\) and \(v' = \frac{dy}{dx} = -\frac{x}{y}\)

Substituting these values into our formula gives:

• \[\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y(1) + x\left(-\frac{x}{y}\right)}{y^2}\]

Evaluating this at \((0, 5)\), where we know \(\frac{dy}{dx} = 0\), simplifies the expression to:

• \(\frac{d^2y}{dx^2} = \frac{-5}{25} = -\frac{1}{5}\)

The negative value of the second derivative indicates that the curve is concave downward at this point.

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.

• The radius = 5 (since \(\sqrt{25} = 5\))
• Center = \((0,0)\)

Within this framework, every point on the circle's circumference adheres to this fixed mathematical relationship, which is crucial when applying calculus methods to understand how points on this curve move or change. Moreover, identifying this equation helps clarify why implicit differentiation is needed, as \(x\) and \(y\) are interdependent.