/*! 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 36 Suppose that \(x\) and \(y\) are... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 36

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\), and \(\frac{d x}{d t}\). $$x^{2} y^{2}=2 y^{3}+1$$

Short Answer

Expert verified
\(\frac{dy}{dt} = \frac{-2xy^2 \frac{dx}{dt}}{2x^2 y - 6y^2}\)

Step by step solution

01

Differentiate both sides with respect to t

To find \(\frac{d y}{d t}\), differentiate both sides of the equation with respect to \(t\): \( x^2 y^2 = 2 y^3 + 1 \). Use the chain rule for each term.
02

Apply the product rule

Apply the product rule to \( x^2 y^2 \): \[\frac{d}{dt} (x^2 y^2) = x^2 \frac{d}{dt}(y^2) + y^2 \frac{d}{dt}(x^2) = x^2 (2y \frac{dy}{dt}) + y^2 (2x \frac{dx}{dt}) \]
03

Differentiate the right-hand side

Differentiate the right-hand side \(2y^3 + 1\) with respect to \(t\): \[ \frac{d}{dt}(2y^3 + 1) = 2 \cdot 3y^2 \frac{dy}{dt} \]
04

Combine the derivatives

Equate the differentiated left-hand side and right-hand side: \[ x^2 (2y \frac{dy}{dt}) + y^2 (2x \frac{dx}{dt}) = 6y^2 \frac{dy}{dt} \]
05

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

Rearrange the equation to isolate \( \frac{dy}{dt} \): \[ 2x^2 y \frac{dy}{dt} + 2xy^2 \frac{dx}{dt} = 6y^2 \frac{dy}{dt} \] \[ 2x^2 y \frac{dy}{dt} - 6y^2 \frac{dy}{dt} = -2xy^2 \frac{dx}{dt} \] \[\frac{dy}{dt}(2x^2 y - 6y^2) = -2xy^2 \frac{dx}{dt} \] \[\frac{dy}{dt} = \frac{-2xy^2 \frac{dx}{dt}}{2x^2 y - 6y^2} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differentiable Functions
Differentiable functions are functions that have a derivative at each point in their domain.
This means that the function's rate of change can be calculated at any given point.
In our exercise, both functions \(x(t)\) and \(y(t)\) are differentiable with respect to \(t\).
This is crucial for using the methods of implicit differentiation, as it ensures that their derivatives exist and can be calculated.
When working with differentiable functions, remember:
  • The function must be continuous.

  • It must have a well-defined slope (derivative) at every point.
Chain Rule
The chain rule helps to differentiate composite functions.
If a function is composed of other functions, the derivative of the composed function depends on the derivatives of the inner functions.
Mathematically, if we have a function \(z = f(g(t))\), the chain rule tells us that:
\ \frac{dz}{dt} = \frac{df}{dg} \frac{dg}{dt} \
In our problem, to differentiate terms like \(x^2y^2\) and \(2y^3\), we apply the chain rule to account for the changing \(x\) and \(y\).
For example, differentiating \(y^2\) using the chain rule gives us \(2y \frac{dy}{dt}\) because we treat \(y\) as a function of \(t\).
Product Rule
The product rule is essential when differentiating products of two or more functions.
Given two functions \(u(t)\) and \(v(t)\), the derivative of their product is given by:
\ \frac{d}{dt}[u(t) \times v(t)] = u(t) \frac{dv}{dt} + v(t) \frac{du}{dt} \
In our problem, we have \(x^2 y^2\), a product that needs the product rule.
Applying the product rule yields:
\ x^2 \frac{d}{dt}(y^2) + y^2 \frac{d}{dt}(x^2) \
Which further simplifies to:
\ x^2 (2y \frac{dy}{dt}) + y^2 (2x \frac{dx}{dt}) \
Remember, for each function in the product, differentiate one while keeping the other constant, then sum the results.
Implicit Differentiation
Implicit differentiation is useful when dealing with equations not easily solved for one variable in terms of another.
Instead of explicitly solving for \(y\) in terms of \(x\) (or vice versa), we differentiate both sides with respect to the independent variable.
In our exercise, we differentiate the given equation \(x^2 y^2 = 2 y^3 + 1\) with respect to \(t\) directly.
This requires the use of both the chain rule and the product rule.
Steps involved include:
  • Differentiate both sides with respect to \(t\).

  • Apply product rule and chain rule where needed.

  • Combine and rearrange derivatives to solve for \(\frac{dy}{dt}\).
This gives us: \ \frac{dy}{dt} = \frac{-2xy^2 \frac{dx}{dt}}{2x^2 y - 6y^2} \
By implicitly differentiating, we can find \(\frac{dy}{dt}\) without rearranging \(y\) explicitly in terms of \(x\) and \(t\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use implicit differentiation of the equations to determine the slope of the graph at the given point. $$y^{2}=3 x y-5 ; x=2, y=1$$

Suppose that the price \(p\) (in dollars) and the demand \(x\) (in thousands of units) of a commodity satisfy the demand equation $$ 6 p+x+x p=94 $$ How fast is the demand changing at a time when \(x=4\), \(p=9\), and the price is rising at the rate of $$\$ 2$$ per week?

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\), and \(\frac{d x}{d t}\). $$y^{4}-x^{2}=1$$

Many relations in biology are expressed by power functions, known as allometric equations, of the form \(y=k x^{a}\), where \(k\) and \(a\) are constants. For example, the weight of a male hognose snake is approximately \(446 x^{3}\) grams, where \(x\) is its length in meters. If a snake has length \(.4\) meters and is growing at the rate of \(.2\) meters per year, at what rate is the snake gaining weight? (Source: Museum of Natural History.)

A manufacturer of microcomputers estimates that \(t\) months from now it will sell \(x\) thousand units of its main line of microcomputers per month, where \(x=.05 t^{2}+2 t+5 .\) Because of economies of scale, the profit \(P\) from manufacturing and selling \(x\) thousand units is estimated to be \(P=.001 x^{2}+.1 x-.25\) million dollars. Calculate the rate at which the profit will be increasing 5 months from now.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.