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Chapter 3: Problem 8

If \(x^{2} y^{3}=4 / 27\) and \(d y / d t=1 / 2,\) then what is \(d x / d t\) when \(x=2 ?\)

Short Answer

Expert verified
\(\frac{dx}{dt} = -\frac{9}{2}\).

Step by step solution

01

Understand the Problem

We need to find the rate of change of \(x\) with respect to time \(t\) given the relationship \(x^2 y^3 = \frac{4}{27}\) and \(\frac{dy}{dt} = \frac{1}{2}\). We also know that \(x=2\).
02

Implicit Differentiation

Differentiate the equation \(x^2 y^3 = \frac{4}{27}\) with respect to time \(t\). Using the product rule, we get: \[ 2x y^3 \frac{dx}{dt} + 3x^2 y^2 \frac{dy}{dt} = 0 \]
03

Plug in Known Values

Substitute the known values \(x=2\) and \(\frac{dy}{dt} = \frac{1}{2}\) into the differentiated equation: \[ 2 \, (2) \, y^3 \frac{dx}{dt} + 3 \, (2)^2 \, y^2 \, \left(\frac{1}{2}\right) = 0 \] which simplifies to \[ 4y^3 \frac{dx}{dt} + 6y^2 = 0 \]
04

Solve for \(y\)

Using the original equation \(x^2 y^3 = \frac{4}{27}\), substitute \(x = 2\): \[ (2)^2 y^3 = \frac{4}{27} \rightarrow 4y^3 = \frac{4}{27} \] Solve for \(y^3\): \[ y^3 = \frac{1}{27} \] Then \(y = \frac{1}{3}\).
05

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

Substitute \(y^3 = \frac{1}{27}\) and \(y = \frac{1}{3}\) into the setup from Step 3: \[ 4(\frac{1}{27}) \frac{dx}{dt} + 6(\frac{1}{3})^2 = 0 \] Simplify to get \[ \frac{4}{27} \frac{dx}{dt} + \frac{2}{3} = 0 \]. Solve for \(\frac{dx}{dt}\): \[ \frac{4}{27} \frac{dx}{dt} = -\frac{2}{3} \rightarrow \frac{dx}{dt} = -\frac{2}{3} \cdot \frac{27}{4} = -\frac{9}{2} \].

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Key Concepts

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

Rate of Change
The rate of change describes how one quantity changes in relation to another. In this exercise, we are looking at how the variable \(x\) changes over time, that is \(\frac{dx}{dt}\). When dealing with rate of change in equations, especially those involving more than one independent variable, we often need to use implicit differentiation to handle the concurrent changes.
- **Why is it important?** Knowing how \(x\) or \(y\) changes with time can tell us much about the system's dynamics. For example, if \(\frac{dy}{dt} = \frac{1}{2}\), it means that \(y\) increases by \(0.5\) units per time unit.
- **Application:** In practical situations, such as physics or engineering, understanding these rates can help predict future behavior or outcomes. It allows us to model situations and make decisions based on the expected changes.
Product Rule
The product rule is essential when differentiating products of functions. When two variables multiply to form a term, as in \(x^2 y^3\), the product rule assists in differentiating. The product rule states that \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of \(t\).
- **Applying the Product Rule:** For the equation \(x^2 y^3 = \frac{4}{27}\), we need to differentiate each part. Here, \(u = x^2\) and \(v = y^3\). Using the product rule, \(u' = 2x \frac{dx}{dt}\) and \(v' = 3y^2 \frac{dy}{dt}\).
- **Combination in Differentiation:** The differentiation of the equation gives us \(2x y^3 \frac{dx}{dt} + 3x^2 y^2 \frac{dy}{dt} = 0\). This product rule application is crucial in setting up the relationship needed to find \(\frac{dx}{dt}\).
Related Rates
Related rates involve finding the rate at which one quantity changes with respect to another, given some relation between them. Typically, problems involve time as the common factor, and they require differentiating implicitly using the given equation.
- **Understanding Through Example:** In this problem, \(x^2 y^3 = \frac{4}{27}\) relates \(x\) and \(y\). Given \(\frac{dy}{dt} = \frac{1}{2}\), we find how \(x\) changes with time when \(x=2\).
- **Steps to Solve:** Use implicit differentiation to get a related rates equation. Evaluate the unknowns by substituting known values like \(x\) and \(\frac{dy}{dt}\).
- **Using Known Values:** Substituting these values into \(2xy^3 \frac{dx}{dt} + 3x^2 y^2 \frac{dy}{dt} = 0\), simplifies processing. Solving this gives \(\frac{dx}{dt} = -\frac{9}{2}\), meaning \(x\) decreases at a rate of \(4.5\) units per time unit when \(x=2\), and \(y=\frac{1}{3}\).

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Most popular questions from this chapter

Particle acceleration A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\) .

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

Cardiac output In the late 1860 \(\mathrm{s}\) , Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min.}\) If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min} .\) $$\begin{array}{c}{\text { Your cardiac output can be calculated with the formula }} \\ {y=\frac{Q}{D}}\end{array}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L},\) $$y=\frac{233 \mathrm{ml} / \min }{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min},$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

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The edge \(x\) of a cube is measured with an error of at most 0.5\(\%\) . What is the maximum corresponding percentage error in computing the cube's $$\text{a. surface area?} \quad \text{b. volume?}$$
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