/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 If \(x^{3}+y^{3}=16,\) find the ... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(x^{3}+y^{3}=16,\) find the value of \(d^{2} y / d x^{2}\) at the point \((2,2).\)

Short Answer

Expert verified
The value of \(d^2 y/dx^2\) at (2,2) is -2.

Step by step solution

01

Implicit Differentiation

We start by differentiating the equation with respect to \(x\). Given the equation \(x^3 + y^3 = 16\), differentiate each term. For \(x^3\), the derivative is \(3x^2\). For \(y^3\), apply the chain rule to get \(3y^2 \frac{dy}{dx}\). Differentiating the constant 16 results in 0. Hence, the differentiated equation is \(3x^2 + 3y^2 \frac{dy}{dx} = 0\).
02

Solve for \(\frac{dy}{dx}\)

From the differentiated equation \(3x^2 + 3y^2 \frac{dy}{dx} = 0\), solve for \(\frac{dy}{dx}\). Subtract \(3x^2\) from both sides to get \(3y^2 \frac{dy}{dx} = -3x^2\). Divide both sides by \(3y^2\) to isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = -\frac{x^2}{y^2} \].
03

Find \(\frac{d^2 y}{dx^2}\) using Second Implicit Differentiation

To find \(\frac{d^2 y}{dx^2}\), differentiate \(\frac{dy}{dx} = -\frac{x^2}{y^2}\) again with respect to \(x\). Using the quotient rule \[ \frac{d}{dx} \left(-\frac{x^2}{y^2}\right) = \left( -\frac{d}{dx}[x^2] \cdot y^2 - x^2 \cdot \frac{d}{dx}[y^2] \right) / (y^2)^2 \]. Calculate derivatives: \(d/dx[x^2] = 2x \) and \(d/dx[y^2] = 2y \frac{dy}{dx} = 2y (-\frac{x^2}{y^2})\). Combine these to get the expression for \(\frac{d^2 y}{dx^2}\).
04

Simplify and Substitute the Known Values

Substitute \(\frac{dy}{dx} = -\frac{x^2}{y^2}\) and known values \((x = 2, y = 2)\) into the equation for \(\frac{d^2 y}{dx^2}\). Replace \(d/dx [y^2] = 2y \frac{dy}{dx}\) with \(2 \cdot 2 \cdot (-\frac{2^2}{2^2}) = -4\). Simplifying, the expression becomes:\[ \begin{align*}\frac{d^2 y}{dx^2} &= \left(- 2x y^2 + x^2 (-4) \right) / (y^2)^2\ &= \left(-4 y^2 - 4x^2 \right) / 16 \\end{align*} \].Substitute \(x = 2\) and \(y = 2\) to get \(-16 - 16 = -32 / 16 = -2\).
05

Final Answer

The second derivative \(\frac{d^2 y}{dx^2}\) at the point \((2,2)\) is \(-2\). Thus, the value is \(-2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
When looking for the second derivative, you're finding how the rate of change of a function's slope (or first derivative) is itself changing. In other words, it measures the curvature of the graph. For example, if you have a graph that bends upwards, the second derivative will be positive. If the graph bends downwards, the second derivative will be negative.
The process begins once you've found the first derivative, in this case using implicit differentiation to get \( \frac{dy}{dx} = -\frac{x^2}{y^2} \). To get the second derivative, \( \frac{d^2y}{dx^2} \), you'll differentiate this result with respect to \( x \) again. This further differentiation reveals deeper insights into how the original curve changes as \( x \) changes.
Quotient Rule
The quotient rule is a very handy tool in calculus for finding the derivative of a function that is the division of two more basic functions. If you have a fraction where one function is divided by another, such as \( u(x) / v(x) \), the quotient rule states:
  • The derivative of the entire fraction is \( \frac{v(x) \, u'(x) - u(x) \, v'(x)}{(v(x))^2} \).
In our solution, we used the quotient rule to differentiate \( \frac{-x^2}{y^2} \). This involves calculating each part:
  • The derivative of the numerator \(-x^2\) is \( -2x \).
  • The derivative of the denominator \(y^2\) requires applying the chain rule, as we'll discuss in the next section.
This method is crucial whenever you have a situation where one variable is influencing another in a fractional relationship. By using the quotient rule, we were able to find \( \frac{d^2 y}{dx^2} \) effectively.
Chain Rule
The chain rule is a powerful method for differentiating compositions of functions. Essentially, it lets you take the derivative of a function that is nested within another function, or a function of a function. It reflects how a change in one variable propagates through a sequence of dependent changes.
In parts of the exercise, we encounter expressions like \( y^3 \) or \( y^2 \). Here, you can't directly differentiate because \( y \) is itself a dependent variable on \( x \). Thus, apply the chain rule. For instance:
  • In differentiating \( y^3 \), you get \( 3y^2 \frac{dy}{dx} \), multiplying the power rule result by the derivative of \( y \) because \( y \) changes as \( x \) changes.
In the final derivative calculations, you'd see terms like \( 2y \frac{dy}{dx} \), which applies the chain rule by respecting the inner changes \( \frac{dy}{dx} \). Understanding this rule deeply helps unravel complex functions and interactions within different parts of mathematics.

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