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Chapter 4: Problem 42

Here \(x\) and \(y\) are functions of \(t\) and are related as indicated. Obtain the desired derivative from the information given. \(-2 x y^{2} \quad y=22 .\) Given that \(\frac{d y}{d t}=-2\) when \(x=3\) and \(y=\) 2, find \(\frac{d x}{d t}\)

Short Answer

Expert verified
\(\frac{d x}{d t} = 12\)

Step by step solution

01

Differentiation of the equation

Start by differentiating the equation \(-2 xy^{2}\) with respect to \(t\) using both product rule and chain rule. So, we get \( -2 \frac{d x}{d t} y^{2} - 4 x y \frac{d y}{d t} = 0\).
02

Substitute the given values into the equation

Now insert the values of \(x\), \(y\), and \(\frac{d y}{d t}\) into the equation. That means replacing \(x\) by 3, \(y\) by 2, and \(\frac{d y}{d t}\) by -2. We then have \( -2 \frac{d x}{d t} * 2^{2} - 4 * 3 * 2 * -2 = 0\)
03

Solve for \(\frac{d x}{d t}\)

After simplifying the above equation, you get \( -4 \frac{d x}{d t} + 48 = 0\). Now you can solve for \(\frac{d x}{d t}\) by simply isolating \(\frac{d x}{d t}\) on one side of the equation. You will have \( \frac{d x}{d t} = \frac{48}{4} = 12 \). Thus, the desired derivative is 12.

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Key Concepts

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

Product Rule
When it comes to taking derivatives of functions that are the product of two or more functions, the product rule is essential. The product rule states that if you have a function which can be expressed as the product of two other functions, let's call them u(t) and v(t), then the derivative of this product with respect to t is given by u'(t)v(t) + u(t)v'(t).

For our given problem, we applied the product rule to differentiating -2xy^2 with respect to t. Since both x and y are functions of t, we treated x as u(t) and y^2 as v(t). This allowed us to split the differentiation process into manageable parts, making it easier to find the resultant derivative with respect to t.
Chain Rule
The chain rule is a powerful tool for computing the derivative of a composite function. In simple terms, if a variable u is a function of v, and v is in turn a function of t, the derivative of u with respect to t is the product of the derivative of u with respect to v and the derivative of v with respect to t. Symbolically, this can be denoted as (du/dt) = (du/dv) * (dv/dt).

In our exercise, the derivative of y^2 with respect to t is an application of the chain rule, as y is a function of t. This is why we calculate dy/dt and then multiply it by the derivative of y^2 with respect to y, which is 2y. By combining the product and chain rules, we were able to compute the derivatives needed to solve the problem.
Derivative Calculation
Derivative calculation is the process of finding the rate at which a function is changing at any given point, commonly denoted as (df/dx) or f'(x). It's a fundamental concept in calculus that represents an instantaneous rate of change, similar to finding the velocity of an object when its position changes over time.

In our demonstrated problem, we were asked to find the derivative of -2xy^2 with respect to time t. By using the product and chain rules to calculate the derivatives of the individual components of the equation, and then applying basic algebra, we efficiently found the desired derivative, contributing to a deeper understanding of how the variables x and y change with time.
Related Rates
Related rates problems involve finding the rate at which one quantity changes with respect to time when given the rate at which another related quantity changes. They often require implicit differentiation and are solved using the chain rule.

In the given example, we knew the rate of change of y with respect to t, and we were asked to find the rate of change of x with respect to t. By differentiating the given equation implicitly and substituting the known rates, we found a relation between the rates of change of x and y. Solving this allowed us to conclude the rate at which x changes with respect to time when x equals 3, and y equals 2.

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Most popular questions from this chapter

The function \(f(x)=x^{4}-2 x^{2}-\frac{17}{16}\) has two zeros, one at a point \(a\) between \(0\) and \(2,\) and the other at \(-a . C\) is an even function.) (a) Show that the Newton-Raphson method fails in the search for \(a\) if we start at \(x=\frac{1}{2} .\) What are the outputs \(x_{1}, x_{2}, x_{3}, \ldots\) in this case? (b) Estimate \(a\) by starting at \(x_{1}=2 .\) Determine \(x_{4}\) rounded off to five decimal places and evaluate \(f\left(x_{4}\right)\)

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