Question

Use implicit differentiation twice to find \(y^{\prime \prime}\) at \((3,4)\) if \(x^{2}+y^{2}=25\)

Short answer

The second derivative \(y''\) at (3,4) is \(-\frac{25}{64}\).

Step-by-step solution

Differentiate Implicitly Once

We are given the equation \(x^{2} + y^{2} = 25\). To find the first derivative (\(y'\)), differentiate both sides of the equation with respect to \(x\). The derivative of \(x^{2}\) is \(2x\). The derivative of \(y^{2}\) (using the chain rule) is \(2y\cdot y'\). Therefore, the first differentiation gives us:\[2x + 2yy' = 0\] Solving for \(y'\) gives:\[y' = -\frac{x}{y}\]

Evaluate the First Derivative at (3,4)

Substitute \(x = 3\) and \(y = 4\) into the equation for \(y'\):\[y' = -\frac{3}{4}\]This gives the slope of the tangent at the point (3,4).

Differentiate Implicitly Again

Differentiate the equation \(2x + 2yy' = 0\) again, with respect to \(x\), to find \(y''\).The derivative of \(2x\) is \(2\). For \(2yy'\), use the product rule:\[(2yy')' = 2y' + 2y'y' + 2yy'' = 2y' + 2yy'\cdot y' + 2yy''\] The result of the second differentiation is:\[2 + 2y'y' + 2yy'' = 0\]

Solve for Second Derivative \(y''\)

From the equation \(2 + 2y'y' + 2yy'' = 0\), substitute \(y' = -\frac{x}{y}\):\[2 + 2\left(-\frac{x}{y}\right)^2 + 2y\cdot y'' = 0\]We need to simplify and solve for \(y''\):\[2 + 2\frac{x^2}{y^2} + 2yy'' = 0\]Rearranging gives:\[2yy'' = -2 - 2\frac{x^2}{y^2}\]And solving for \(y''\), we have:\[y'' = -\frac{1}{y}\left(1 + \frac{x^2}{y^2}\right)\]

Evaluate the Second Derivative at (3,4)

Substitute \(x = 3\) and \(y = 4\) into the expression for \(y''\):\[y'' = -\frac{1}{4}\left(1 + \frac{3^2}{4^2}\right)\]Calculate the terms inside the parentheses first:\[1 + \frac{9}{16} = \frac{16}{16} + \frac{9}{16} = \frac{25}{16}\]Then the value of \(y''\) will be:\[y'' = -\frac{1}{4} \times \frac{25}{16} = -\frac{25}{64}\]

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus which helps to differentiate composite functions. It is particularly useful in implicit differentiation, where one cannot separate variables easily. When you have a variable, such as the \(y\) squared in the context of the equation \((x^2 + y^2 = 25)\), the chain rule steps in. We need to differentiate \(y^2\) with respect to \(x\), so using the chain rule:

• The outer function (squaring) gives \(2y\)
• The inner function (differentiating \(y\) with respect to \(x\)) gives \(y'\)

Resulting in the product \(2y\cdot y'\). This allows us to bridge variables and solve effectively for related rates or slopes in functions where direct differentiation is not possible.

Product Rule

The product rule is another crucial derivative rule used when you are dealing with two functions multiplied together. In the exercise, when differentiating \(2yy'\), we used the product rule to find its derivative with respect to \(x\). Here's how we applied the product rule:

• The derivative of the first function \(2y\) with respect to \(x\) is \(2y'\).
• Then, we multiply by the second unchanged function \(y'\), leading to: \(2y' \).
• Next, leave the first function \(2y\) unchanged, and take the derivative of the second function \(y'\), leading to \(2yy''\) when combined using the product rule.
• The sum of these gives us \(2y' + 2yy''\).

Applying the product rule helps maintain accuracy in calculations where functions are intertwined.

Second Derivative

Finding the second derivative involves differentiating the first derivative of a function, giving us insights into the function's concavity and the nature of turning points. In our exercise, after obtaining the first derivative \(y' = -\frac{x}{y}\), we needed to differentiate again to find \(y''\).
Given by the equation \(2 + 2\left(-\frac{x}{y}\right)^2 + 2yy'' = 0\), we solve for \(y''\) by:

• Substituting the value of \(y'\)
• Algebraically rearranging and simplifying
• Solving to eventually obtain \(y'' = -\frac{1}{y}\left(1 + \frac{x^2}{y^2}\right)\)

This shows how we can find the rate at which the slope of the function is changing, an indicator of its shape or behavior in different regions.

Tangent Line

A tangent line represents the best linear approximation to a curve at a given point. It touches the curve at one point and provides a perfect slope match there. When you find a first derivative, like \(y' = -\frac{3}{4}\) at \((3,4)\), you are calculating the slope of this tangent line. Tangent lines have applications in determining instantaneous rates of change. In this context:

• The slope \(m= -\frac{3}{4}\) tells us how steep the line is at the point identified.
• The point-slope form of a line, \((y - y_1) = m(x - x_1)\), can be used for finding the equation of the tangent line.

This concept is essential in many areas, including physics and engineering, to represent real-world changing states.

Coordinate Geometry

Coordinate geometry connects algebra and geometry, using coordinates (like \(x, y\) pairs) to understand geometrical shapes and their properties. For the equation \(x^2 + y^2 = 25\), this represents a circle centered at the origin with a radius of 5. When using implicit differentiation:

• Points such as \(3,4\) are specific locations on the circle.
• The slopes derived, \(y'\) and \(y''\), describe the behavior of the line tangent to the circle at those points.
• Coordinate geometry bridges the gap, taking algebraic concepts into geometric interpretation, like plotting derivatives as tangents.

Understanding coordinate geometry gives insights into how equations translate into shapes, aiding in visualizing solutions and geometrical interpretations.